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ldoatpci | maths | probability | classical-defininition-of-probability | Let A be the event that the absolute difference between two randomly choosen real numbers in the sample space $[0,60]$ is less than or equal to a . If $\mathrm{P}(\mathrm{A})=\frac{11}{36}$, then $\mathrm{a}$ is equal to _______. | [] | null | 10 | $$
\begin{aligned}
& |\mathrm{x}-\mathrm{y}|<\mathrm{a} \Rightarrow-\mathrm{a}<\mathrm{x}-\mathrm{y}<\mathrm{a} \\\\
& \Rightarrow \mathrm{x}-\mathrm{y}<\mathrm{a} \text { and } \mathrm{x}-\mathrm{y}>-\mathrm{a}
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leeeen4s/b4b3dede-ba28-4c8d-af23-57eb35c25759/f55c2dc0-b1fb-11ed-afd2-4f73ef263f1c/file-1leeeen4t.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leeeen4s/b4b3dede-ba28-4c8d-af23-57eb35c25759/f55c2dc0-b1fb-11ed-afd2-4f73ef263f1c/file-1leeeen4t.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 31st January Evening Shift Mathematics - Probability Question 52 English Explanation">
<br><br>$$
\begin{aligned}
& \mathrm{P}(\mathrm{A})=\frac{\operatorname{ar}(\mathrm{OACDEG})}{(\mathrm{OBDF})} \\\\
& =\frac{\operatorname{ar}(\mathrm{OBDF})-\operatorname{ar}(\mathrm{ABC})-\operatorname{ar}(\mathrm{EFG})}{\operatorname{ar}(\mathrm{OBDF})} \\\\
& \Rightarrow \frac{11}{36}=\frac{(60)^2-\frac{1}{2}(60-\mathrm{a})^2-\frac{1}{2}(60-\mathrm{a})^2}{3600} \\\\
& \Rightarrow 1100=3600-(60-\mathrm{a})^2 \\\\
& \Rightarrow (60-\mathrm{a})^2=2500 \Rightarrow 60-\mathrm{a}=50 \\\\
& \Rightarrow \mathrm{a}=10
\end{aligned}
$$ | integer | jee-main-2023-online-31st-january-evening-shift |
ldr0ncc0 | maths | probability | classical-defininition-of-probability | A bag contains six balls of different colours. Two balls are drawn in succession with replacement. The probability that both the balls are of the same colour is p. Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colour is $q$. If $p: q=m: n$, where $m$ and $n$ are coprime, then $m+n$ is equal to : | [] | null | 14 | <p>$$p = {6 \over {36}} = {1 \over 6}$$</p>
<p>$$q = {{{}^6{C_1} \times {}^5{C_1} \times {{4!} \over {3!}}} \over {{6^4}}} = {{120} \over {1296}} = {5 \over {54}}$$</p>
<p>$${p \over q} = {{{1 \over 6}} \over {{5 \over {54}}}} = {{54} \over {6 \times 5}} = {9 \over 5} = {m \over n}$$</p>
<p>$$m + n = 14$$</p> | integer | jee-main-2023-online-30th-january-evening-shift |
1ldr7mspf | maths | probability | classical-defininition-of-probability | <p>If an unbiased die, marked with $$-2,-1,0,1,2,3$$ on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is :</p> | [{"identifier": "A", "content": "$$\\frac{27}{288}$$"}, {"identifier": "B", "content": "$$\\frac{521}{2592}$$"}, {"identifier": "C", "content": "$$\\frac{440}{2592}$$"}, {"identifier": "D", "content": "$$\\frac{881}{2592}$$"}] | ["B"] | null | <p>$${}^5{C_0} \times {3^5} = 243$$</p>
<p>$${}^5{C_2} \times {2^2} \times {3^3} = 1080$$</p>
<p>$${}^5{C_4} \times {2^4}\,.\,3 = 240$$</p>
<p>$$\therefore$$ required probability</p>
<p>$$ = {{243 + 1080 + 240} \over {6 \times 6 \times 6 \times 6 \times 6}} = {{521} \over {2592}}$$</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1ldu4fpco | maths | probability | classical-defininition-of-probability | <p>Let N be the sum of the numbers appeared when two fair dice are rolled and let the probability that $$N-2,\sqrt{3N},N+2$$ are in geometric progression be $$\frac{k}{48}$$. Then the value of k is :</p> | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "4"}] | ["D"] | null | $n-2, \sqrt{3 n}, n+2 \rightarrow$ G.P.
<br/><br/>
$3 n=n^{2}-4$
<br/><br/>
$\Rightarrow n^{2}-3 n-4=0$
<br/><br/>
$\Rightarrow n=4,-1$ (rejected)
<br/><br/>
$P(S=4)=\frac{3}{36}=\frac{1}{12}=\frac{4}{48}$
<br/><br/>
$\therefore k=4$ | mcq | jee-main-2023-online-25th-january-evening-shift |
1ldv1kqtg | maths | probability | classical-defininition-of-probability | <p>Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $$S = \left\{ {x \in \mathbb{Z}:x(66 - x) \ge {5 \over 9}M} \right\}$$ and the event $$\mathrm{A = \{ x \in S:x\,is\,a\,multiple\,of\,3\}}$$. Then P(A) is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{1}{3}$$"}, {"identifier": "B", "content": "$$\\frac{1}{5}$$"}, {"identifier": "C", "content": "$$\\frac{7}{22}$$"}, {"identifier": "D", "content": "$$\\frac{15}{44}$$"}] | ["A"] | null | $x+y=66$
<br/><br/>
$$
\begin{aligned}
& \frac{x+y}{2} \geq \sqrt{x y} \\\\
\Rightarrow & 33 \geq \sqrt{x y} \\\\
\Rightarrow & x y \leq 1089 \\\\
\therefore & M=1089 \\\\
S: & x(66-x) \geq \frac{5}{9} \cdot 1089 \\\\
& 66 x-x^{2} \geq 605 \\\\
\Rightarrow & x^{2}-66 x+605 \leq 0
\end{aligned}
$$
<br/><br/>$\Rightarrow(x-55)(x-11) \leq 0 ; 11 \leq x \leq 55$
<br/><br/>Therefore $S=\{11,12,13 \ldots 55\} $
<br/><br/>$\Rightarrow n(S)=45$
<br/><br/>Elements of $S$ which are multiple of 3 are
<br/><br/>$$
\begin{aligned}
& 12+(n-1) 3=54 \Rightarrow 3(n-1)=42 \Rightarrow n=15 \\\\
& n(A)=15
\end{aligned}
$$
<br/><br/>$\Rightarrow P(A)=\frac{15}{45}=\frac{1}{3}$ | mcq | jee-main-2023-online-25th-january-morning-shift |
1ldyaqkkl | maths | probability | classical-defininition-of-probability | <p>Let N denote the number that turns up when a fair die is rolled. If the probability that the system of equations</p>
<p>$$x + y + z = 1$$</p>
<p>$$2x + \mathrm{N}y + 2z = 2$$</p>
<p>$$3x + 3y + \mathrm{N}z = 3$$</p>
<p>has unique solution is $${k \over 6}$$, then the sum of value of k and all possible values of N is :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "21"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "19"}] | ["C"] | null | For unique solution $\Delta \neq 0$
<br/><br/>
i.e. $\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & N & 2 \\ 3 & 3 & N\end{array}\right| \neq 0$
<br/><br/>
$\Rightarrow\left(N^{2}-6\right)-(2 N-6)+(6-3 N) \neq 0$
<br/><br/>
$\Rightarrow N^{2}-5 N+6 \neq 0$
<br/><br/>
$\therefore N \neq 2$ and $N \neq 3$
<br/><br/>
$\therefore $ Probability of not getting 2 or 3 in a throw of dice $=\frac{2}{3}$
<br/><br/>
As given $\frac{2}{3}=\frac{k}{6} \Rightarrow k=4$
<br/><br/>
$\therefore$ Required value $=1+4+5+6+4=20$
| mcq | jee-main-2023-online-24th-january-morning-shift |
1ldyb4hyp | maths | probability | classical-defininition-of-probability | <p>Let $$\Omega$$ be the sample space and $$\mathrm{A \subseteq \Omega}$$ be an event.</p>
<p>Given below are two statements :</p>
<p>(S1) : If P(A) = 0, then A = $$\phi$$</p>
<p>(S2) : If P(A) = 1, then A = $$\Omega$$</p>
<p>Then :</p> | [{"identifier": "A", "content": "both (S1) and (S2) are true"}, {"identifier": "B", "content": "both (S1) and (S2) are false"}, {"identifier": "C", "content": "only (S2) is true"}, {"identifier": "D", "content": "only (S1) is true"}] | ["B"] | null | $\Omega=$ sample space<br/><br/>
$\mathrm{A}=$ be an event<br/><br/>
$ \Omega$ = A wire of length 1 which starts at point 0 and ends at point 1 on the coordinate axis = $[0,1]$<br/><br/>
$\mathrm{A}=\left\{\frac{1}{2}\right\}$ = Selecting a point on the wire which is at $\left\{\frac{1} {2}\right\}$ or 0.5
<br/><br/>As wire is an 1-D object so from geometrical probability
<br/><br/>$P(A)=\frac{\text { Favourable Length }}{\text { Total Length }}$
<br/><br/>Here total length of wire = 1 unit
<br/><br/>and point has zero length so point A at $\left\{\frac{1} {2}\right\}$ or 0.5 has length = 0.
<br/><br/>$$ \therefore $$ Favorable length = 0
<br/><br/>
$$ \therefore $$ $\mathrm{P}(\mathrm{A})=0 {\text { but }} \mathrm{A} \neq \phi$
<br/><br/>Now $\overline{\mathrm{A}}$ = $[0,1]$ - $\left\{\frac{1} {2}\right\}$
<br/><br/>So, length of $\overline{\mathrm{A}}$ = Length of entire wire - Length of point A = 1
<br/><br/>
$$ \therefore $$ $\mathrm{P}(\overline{\mathrm{A}})=1 {\text { but }} \overline{\mathrm{A}} \neq \Omega$.<br/><br/>
Then both statements are false.
<br/><br/><b>Attention :</b> According to NTA option A is correct. Which is wrong. That is proven here using geometrical probability.
<br/><br/><b>Note :</b>
<br/><br/>Geometrical probability :
<br/><br/>1. For 1-D object, $P(A)=\frac{\text { Favourable Length }}{\text { Total Length }}$
<br/><br/>2. For 2-D object, $P(A)=\frac{\text { Favourable Area }}{\text { Total Area }}$
<br/><br/>3. For 2-D object, $P(A)=\frac{\text { Favourable volume }}{\text { Total volume }}$ | mcq | jee-main-2023-online-24th-january-morning-shift |
1lgutw8tu | maths | probability | classical-defininition-of-probability | <p>Let $$S=\left\{M=\left[a_{i j}\right], a_{i j} \in\{0,1,2\}, 1 \leq i, j \leq 2\right\}$$ be a sample space and $$A=\{M \in S: M$$ is invertible $$\}$$ be an event. Then $$P(A)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{47}{81}$$"}, {"identifier": "B", "content": "$$\\frac{49}{81}$$"}, {"identifier": "C", "content": "$$\\frac{50}{81}$$"}, {"identifier": "D", "content": "$$\\frac{16}{27}$$"}] | ["C"] | null | We have, $S=\left\{M=\left[a_{i j}\right], a_{i j} \in\{0,1,2\}, 1 \leq i, j \leq 2\right\}$
<br/><br/>Let $M=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, where $a, b, c, d \in\{0,1,2\}$
<br/><br/>$$
n(s)=3^4=81
$$
<br/><br/>If $A$ is invertible, then $|A| \neq 0$
<br/><br/>Now, if $|A|=0$, then $|M|=0$
<br/><br/>$\therefore a d-b c=0$ or $a d=b c$
<br/><br/><b>Case I :</b> When $a d=b c=0$, then
<br/><br/>There are five ways when $a d=0$ i.e.,
<br/><br/>$(a, d)=(0,0),(0,1),(0,2),(1,0),(2,0)$
<br/><br/>Similarly, there are again five ways, when $b c=0$
<br/><br/>$\therefore$ There are total $5 \times 5=25$ ways, when $a d=b c=0$
<br/><br/><b>Case II :</b> When $a d=b c=1$
<br/><br/>There is only one way, when $a d=b c=1$
<br/><br/>$$
\text { i.e. } \quad a=b=c=d=1
$$
<br/><br/><b>Case III :</b> When $a d=b c=2$
<br/><br/>There are two ways, when $a d=2$, i.e.
<br/><br/>$$
(a, d)=(1,2) \text { or }(2,1)
$$
<br/><br/>Similarly, there are two ways
<br/><br/>when $b c=2$ i.e., $(b, c)=(1,2)$ or $(2,1)$
<br/><br/><b>Case IV :</b> When $a d-b c=4$
<br/><br/>There is only way, when $a d=b c=4$
<br/><br/>$$
\text { i.e., } a=b=c=d=2
$$
<br/><br/>$\therefore$ Total number of ways, when
<br/><br/>$$
\begin{aligned}
& (\bar{A})=\frac{31}{81}|A|=0 \text { is } 25+1+4+1=31 \\\\
& \text { Hence, } P(A)=1-P(\bar{A})=1-\frac{31}{81}=\frac{50}{81}
\end{aligned}
$$ | mcq | jee-main-2023-online-11th-april-morning-shift |
1lgxt6jpu | maths | probability | classical-defininition-of-probability | <p>Let N denote the sum of the numbers obtained when two dice are rolled. If the probability that $${2^N} < N!$$ is $${m \over n}$$, where m and n are coprime, then $$4m-3n$$ is equal to :</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "10"}] | ["C"] | null | $N$ denote the sum of the numbers obtained when two dice are rolled.
<br/><br/>Such that $2^N < N$!
<br/><br/>$$
\text { i.e., } 4 \leq N \leq 12 \text { i.e., } N \in\{4,5,6, \ldots 12\}
$$
<br/><br/>Now, $P(N=2)+P(N=3)=\frac{1}{36}+\frac{2}{36}=\frac{3}{36}=\frac{1}{12}$
<br/><br/>So, required probability $=1-\frac{1}{12}=\frac{11}{12}=\frac{m}{n}$
<br/><br/>$$
4 m-3 n=4 \times 11-3 \times 12=44-36=8
$$ | mcq | jee-main-2023-online-10th-april-morning-shift |
1lh2ygb1d | maths | probability | classical-defininition-of-probability | <p>Three dice are rolled. If the probability of getting different numbers on the three dice is $$\frac{p}{q}$$, where $$p$$ and $$q$$ are co-prime, then $$q-p$$ is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["B"] | null | Total number of outcomes $=6 \times 6 \times 6=216$
<br/><br/>Number of outcomes getting different numbers on the three dice are ${ }^6 P_3=\frac{6 !}{3 !}=120$
<br/><br/>$\therefore$ Required probability $=\frac{120}{216}=\frac{5}{9}$
<br/><br/>$\therefore p=5$ and $q=9$
<br/><br/>$\therefore q-p=9-5=4$ | mcq | jee-main-2023-online-6th-april-evening-shift |
jaoe38c1lsfkotb7 | maths | probability | classical-defininition-of-probability | <p>An integer is chosen at random from the integers $$1,2,3, \ldots, 50$$. The probability that the chosen integer is a multiple of atleast one of 4, 6 and 7 is</p> | [{"identifier": "A", "content": "$$\\frac{8}{25}$$\n"}, {"identifier": "B", "content": "$$\\frac{9}{50}$$\n"}, {"identifier": "C", "content": "$$\\frac{14}{25}$$\n"}, {"identifier": "D", "content": "$$\\frac{21}{50}$$"}] | ["D"] | null | <p>Given set $$=\{1,2,3, \ldots \ldots . .50\}$$</p>
<p>$$\mathrm{P}(\mathrm{A})=$$ Probability that number is multiple of 4</p>
<p>$$\mathrm{P(B)}=$$ Probability that number is multiple of 6</p>
<p>$$\mathrm{P}(\mathrm{C})=$$ Probability that number is multiple of 7</p>
<p>Now,</p>
<p>$$\mathrm{P}(\mathrm{A})=\frac{12}{50}, \mathrm{P}(\mathrm{B})=\frac{8}{50}, \mathrm{P}(\mathrm{C})=\frac{7}{50}$$</p>
<p>again</p>
<p>$$\begin{aligned}
& \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{4}{50}, \mathrm{P}(\mathrm{B} \cap \mathrm{C})=\frac{1}{50}, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{1}{50} \\
& \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0
\end{aligned}$$</p>
<p>Thus</p>
<p>$$\begin{aligned}
P(A & \cup B \cup C)=\frac{12}{50}+\frac{8}{50}+\frac{7}{50}-\frac{4}{50}-\frac{1}{50}-\frac{1}{50}+0 \\
& =\frac{21}{50}
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
1lsgadxd6 | maths | probability | classical-defininition-of-probability | <p>Two integers $$x$$ and $$y$$ are chosen with replacement from the set $$\{0,1,2,3, \ldots, 10\}$$. Then the probability that $$|x-y|>5$$, is :</p> | [{"identifier": "A", "content": "$$\\frac{31}{121}$$\n"}, {"identifier": "B", "content": "$$\\frac{60}{121}$$\n"}, {"identifier": "C", "content": "$$\\frac{62}{121}$$\n"}, {"identifier": "D", "content": "$$\\frac{30}{121}$$"}] | ["D"] | null | <p>If $$x=0, y=6,7,8,9,10$$</p>
<p>If $$x=1, y=7,8,9,10$$</p>
<p>If $$x=2, y=8,9,10$$</p>
<p>If $$x=3, y=9,10$$</p>
<p>If $$x=4, y=10$$</p>
<p>If $$x=5, y=$$ no possible value</p>
<p>Total possible ways $$=(5+4+3+2+1) \times 2$$</p>
<p>$$=30$$</p>
<p>Required probability $$=\frac{30}{11 \times 11}=\frac{30}{121}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
luxwcbol | maths | probability | classical-defininition-of-probability | <p>If an unbiased dice is rolled thrice, then the probability of getting a greater number in the $$i^{\text {th }}$$ roll than the number obtained in the $$(i-1)^{\text {th }}$$ roll, $$i=2,3$$, is equal to</p> | [{"identifier": "A", "content": "5/54"}, {"identifier": "B", "content": "2/54"}, {"identifier": "C", "content": "1/54"}, {"identifier": "D", "content": "3/54"}] | ["A"] | null | <p>Let's denote the outcomes of the three rolls as $$X_1$$, $$X_2$$, and $$X_3$$, where $$X_i$$ represents the number obtained in the $$i^{\text{th}}$$ roll. We are looking for the probability that:</p>
<p>$$X_2 > X_1 \text{ and } X_3 > X_2$$</p>
<p>The total number of outcomes when rolling a dice three times is: </p>
<p>$$6^3 = 216$$</p>
<p>Let's count the favorable outcomes. For each satisfying outcome, we must ensure that both inequalities are adhered to. Consider the possible sequences where every subsequent roll's number is higher than the previous roll's number. These sequences are:</p>
<ul>
<li>(1, 2, 3)</li>
<li>(1, 2, 4)</li>
<li>(1, 2, 5)</li>
<li>(1, 2, 6)</li>
<li>(1, 3, 4)</li>
<li>(1, 3, 5)</li>
<li>(1, 3, 6)</li>
<li>(1, 4, 5)</li>
<li>(1, 4, 6)</li>
<li>(1, 5, 6)</li>
<li>(2, 3, 4)</li>
<li>(2, 3, 5)</li>
<li>(2, 3, 6)</li>
<li>(2, 4, 5)</li>
<li>(2, 4, 6)</li>
<li>(2, 5, 6)</li>
<li>(3, 4, 5)</li>
<li>(3, 4, 6)</li>
<li>(3, 5, 6)</li>
<li>(4, 5, 6)</li>
</ul>
<p>Clearly, there are 20 such favorable outcomes. Hence, the probability is given by:</p>
<p>$$ \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{20}{216} = \frac{5}{54} $$</p>
<p>Therefore, the correct option is:</p>
<p>Option A: $$\frac{5}{54}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
luy9clq6 | maths | probability | classical-defininition-of-probability | <p>Let $$\mathrm{a}, \mathrm{b}$$ and $$\mathrm{c}$$ denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked $$1,2,3,4$$. If the probability that $$a x^2+b x+c=0$$ has all real roots is $$\frac{m}{n}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$$, then $$\mathrm{m}+\mathrm{n}$$ is equal to _________.</p> | [] | null | 19 | <p>A quadratic equation $$ax^2 + bx + c = 0$$ has real roots if and only if its discriminant is non-negative. The discriminant $$\Delta$$ of the quadratic equation is given by:</p>
<p>$$\Delta = b^2 - 4ac$$</p>
<p>For the quadratic equation to have all real roots, the discriminant must be non-negative:</p>
<p>$$\Delta \geq 0$$</p>
<p>That means:</p>
<p>$$b^2 - 4ac \geq 0$$</p>
<p>Given that $$a, b, c$$ are the outcomes of rolling a fair tetrahedral die, they can each be one of the numbers 1, 2, 3, or 4. Our task is to determine the probability that this condition holds.</p>
<p>We need to analyze the cases where $$b^2 \geq 4ac$$.</p>
<p>Let’s consider all possible values for $$a$$, $$b$$, and $$c$$, and count how many of them satisfy the condition. Since there are 4 choices for each of the variables, there are a total of $$4 \times 4 \times 4 = 64$$ possible combinations.</p>
<p>Now, we count the valid combinations where $$b^2 \geq 4ac$$:</p>
<ul>
<li>For $$a = 1$$: $$b^2 \geq 4c$$</li>
<ul>
<li>$$b = 1: 1 \geq 4c \rightarrow \text{(Not possible since } c \ \text{must be } \geq 1 \text{ and not zero)}$$</li>
<li>$$b = 2: 4 \geq 4c \rightarrow c \leq 1 \rightarrow c = 1$$ (1 case)</li>
<li>$$b = 3: 9 \geq 4c \rightarrow c \leq 2 \rightarrow c = 1 \text{ or } 2$$ (2 cases)</li>
<li>$$b = 4: 16 \geq 4c \rightarrow c \leq 4 \rightarrow c = 1, 2, 3, 4$$ (4 cases)</li>
</ul>
<p>Total for $$a = 1 = 1 + 2 + 4 = 7$$</p>
<li>For $$a = 2$$: $$b^2 \geq 8c$$</li>
<ul>
<li>$$b = 1: 1 \geq 8c \rightarrow \text{(Not possible)}$$</li>
<li>$$b = 2: 4 \geq 8c \rightarrow \text{(Not possible)}$$</li>
<li>$$b = 3: 9 \geq 8c \rightarrow c \leq 1$$ (1 case)</li>
<li>$$b = 4: 16 \geq 8c \rightarrow c \leq 2$$ (2 cases)</li>
</ul>
<p>Total for $$a = 2 = 1 + 2 = 3$$</p>
<li>For $$a = 3$$: $$b^2 \geq 12c$$</li>
<ul>
<li>$$b = 1: 1 \geq 12c \rightarrow \text{(Not possible)}$$</li>
<li>$$b = 2: 4 \geq 12c \rightarrow \text{(Not possible)}$$</li>
<li>$$b = 3: 9 \geq 12c \rightarrow \text{(Not possible)}$$</li>
<li>$$b = 4: 16 \geq 12c \rightarrow c \leq 1$$ (1 case)</li>
</ul>
<p>Total for $$a = 3 = 1$$</p>
<li>For $$a = 4$$: $$b^2 \geq 16c$$</li>
<ul>
<li>$$b = 1: 1 \geq 16c \rightarrow \text{(Not possible)}$$</li>
<li>$$b = 2: 4 \geq 16c \rightarrow \text{(Not possible)}$$</li>
<li>$$b = 3: 9 \geq 16c \rightarrow \text{(Not possible)}$$</li>
<li>$$b = 4: 16 \geq 16c \rightarrow c \leq 1$$ (1 case)</li>
</ul>
<p>Total for $$a = 4 = 1$$</p>
</ul>
<p>Adding up all the valid cases:</p>
<p>$$7 + 3 + 1 + 1 = 12$$</p>
<p>The total number of valid combinations is 12 out of 64. Thus, the probability is:</p>
<p>$$\frac{12}{64} = \frac{3}{16}$$</p>
<p>The value of $$\mathrm{m} = 3$$ and $$\mathrm{n} = 16$$. The sum $$\mathrm{m} + \mathrm{n} = 3 + 16 = 19$$.</p>
<p>Hence, the answer is 19.</p> | integer | jee-main-2024-online-9th-april-morning-shift |
lv5grw62 | maths | probability | classical-defininition-of-probability | <p>Let the sum of two positive integers be 24 . If the probability, that their product is not less than $$\frac{3}{4}$$ times their greatest possible product, is $$\frac{m}{n}$$, where $$\operatorname{gcd}(m, n)=1$$, then $$n$$-$$m$$ equals</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "11"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "8"}] | ["A"] | null | <p>Take two numbers as $$a$$ and $$b$$</p>
<p>$$a+b=24$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw8pero8/601e15c3-bb4a-487e-b3ff-b7b116b4f031/2e363180-1336-11ef-9f8d-838c388c326d/file-1lw8pero9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw8pero8/601e15c3-bb4a-487e-b3ff-b7b116b4f031/2e363180-1336-11ef-9f8d-838c388c326d/file-1lw8pero9.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 8th April Morning Shift Mathematics - Probability Question 8 English Explanation"></p>
<p>For product to be maximum</p>
<p>$$\begin{aligned}
& \frac{a+b}{2} \geq \sqrt{a b} \\
& 144>a b
\end{aligned}$$</p>
<p>Maximum product is 144</p>
<p>Now, $$a b \geq \frac{3}{4} \cdot 144=108$$</p>
<p>Sample space $$=\{(23,1),(22,2), \ldots\}$$</p>
<p>Integer points on line in shaded region</p>
<p>$$\begin{aligned}
& \{(6,18),(7,17),(8,16), \ldots(18,6)\} \\
& P(E)=\frac{n(E)}{n(S)}=\frac{13}{23}=\frac{m}{n} \Rightarrow n-m=10
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
lv7v4fz3 | maths | probability | classical-defininition-of-probability | <p>The coefficients $$a, b, c$$ in the quadratic equation $$a x^2+b x+c=0$$ are chosen from the set $$\{1,2,3,4,5,6,7,8\}$$. The probability of this equation having repeated roots is :</p> | [{"identifier": "A", "content": "$$\\frac{1}{128}$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{64}$$\n"}, {"identifier": "C", "content": "$$\\frac{3}{256}$$\n"}, {"identifier": "D", "content": "$$\\frac{3}{128}$$"}] | ["B"] | null | <p>Given quadratic equation is</p>
<p>$$a x^2+b x+c=0 \text { where } a, b, c \in\{1,2,3, \ldots, 8\}$$</p>
<p>For repeated roots,</p>
<p>$$\begin{aligned}
& b^2-4 a c=0 \\
& \Rightarrow b^2=4 a c
\end{aligned}$$</p>
<p>$$\Rightarrow a c$$ must be a perfect square</p>
<p>$$(a, c) \in\{(1,1),(1,4),(2,2),(2,8),(3,3),(4,1),(4,4),(5,5),(6,6),(7,7),(8,2),(8,8)\}$$</p>
<p>Corresponding $$b$$ must lie in set $$\{1,2,3, \ldots 8\}$$</p>
<p>$$\begin{aligned}
& (a, b, c) \in\{(1,2,1),(1,2,4),(2,4,2),(2,8,8) \\
& (3,6,3),(4,4,1),(4,8,4),(8,8,2)\} \\
& \therefore \text { probability }=\frac{8}{8^3} \\
& =\frac{1}{64} \\
&
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift |
lvb294g8 | maths | probability | classical-defininition-of-probability | <p>If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is :</p> | [{"identifier": "A", "content": "$$\\frac{18}{25}$$\n"}, {"identifier": "B", "content": "$$\\frac{12}{25}$$\n"}, {"identifier": "C", "content": "$$\\frac{6}{25}$$\n"}, {"identifier": "D", "content": "$$\\frac{4}{25}$$"}] | ["B"] | null | <p>We have 3 letters and 5 addresses, where 3 letters are posted to exactly 2 addresses. First, we will select 2 addresses in $${ }^5 C_2$$ ways.</p>
<p>Now, 3 letters can be posted to exactly 2 addresses in 6 ways.</p>
<p>$$\begin{aligned}
\therefore \text { Probability }=\frac{{ }^5 C_2 \times 6}{5^3} & \\
& =\frac{60}{125}=\frac{12}{25}
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
cR0wVgMAi5qt4x9Q | maths | probability | conditional-probability-and-multiplication-theorem | Two aeroplanes $${\rm I}$$ and $${\rm I}$$$${\rm I}$$ bomb a target in succession. The probabilities of $${\rm I}$$ and $${\rm I}$$$${\rm I}$$ scoring a hit correctly are $$0.3$$ and $$0.2,$$ respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is : | [{"identifier": "A", "content": "$$0.2$$ "}, {"identifier": "B", "content": "$$0.7$$ "}, {"identifier": "C", "content": "$$0.06$$ "}, {"identifier": "D", "content": "0.32"}] | ["D"] | null | <p>The desired probability</p>
<p>= (0.7) (0.2) + (0.7) (0.8) (0.7) (0.2) + (0.7) (0.8) (0.7) (0.8) (0.7) (0.2) + .......</p>
<p>= 0.14 [1 + (0.56) + (0.56)<sup>2</sup> + .......]</p>
<p>= 0.14 $$\left( {{1 \over {1 - 0.56}}} \right) = {{0.14} \over {0.44}} = {7 \over {22}}$$ = 0.32</p> | mcq | aieee-2007 |
9tucb0kVsi88la1U | maths | probability | conditional-probability-and-multiplication-theorem | It is given that the events $$A$$ and $$B$$ are such that
<br/>$$P\left( A \right) = {1 \over 4},P\left( {A|B} \right) = {1 \over 2}$$ and $$P\left( {B|A} \right) = {2 \over 3}.$$ Then $$P(B)$$ is : | [{"identifier": "A", "content": "$${1 \\over 6}$$"}, {"identifier": "B", "content": "$${1 \\over 3}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}] | ["B"] | null | Given that,
<br><br>$$P\left( {{A \over B}} \right) = {1 \over 2}$$
<br><br>$$ \Rightarrow $$ $${{P\left( {A \cap B} \right)} \over {P\left( B \right)}}$$ = $${1 \over 2}$$.............. equation (1)
<br><br>$$P\left( {{B \over A}} \right) = {2 \over 3}$$
<br><br>$$ \Rightarrow $$ $${{P\left( {A \cap B} \right)} \over {P\left( A \right)}}$$ = $${2 \over 3}$$.............. equation (2)
<br><br>Dividing equation (1) by equation (2) we get,
<br><br>$${{P\left( A \right)} \over {P\left( B \right)}}$$ = $${3 \over 4}$$
<br><br>$$ \Rightarrow $$ $${P\left( B \right)}$$ = $${4 \over 3}$$ $$ \times $$ $${P\left( A \right)}$$
<br> = $${4 \over 3}$$ $$ \times $$ $${1 \over 4}$$
<br> = $${1 \over 3}$$
<br><br>$$\therefore$$ Option (B) is correct. | mcq | aieee-2008 |
UnpNoXVqt1E3GNHO | maths | probability | conditional-probability-and-multiplication-theorem | One ticket is selected at random from $$50$$ tickets numbered $$00, 01, 02, ...., 49.$$ Then the probability that the sum of the digits on the selected ticket is $$8$$, given that the product of these digits is zer, equals : | [{"identifier": "A", "content": "$${1 \\over 7}$$ "}, {"identifier": "B", "content": "$${5 \\over 14}$$"}, {"identifier": "C", "content": "$${1 \\over 50}$$"}, {"identifier": "D", "content": "$${1 \\over 14}$$"}] | ["D"] | null | Sample space = {00, 01, 02, 03, ..........49} = 50 tickets
<br><br>n(S) = 50
<br><br>n(Sum = 8) = { 08, 17, 26, 35, 44 } = 5
<br><br>n(Product = 0) = { 00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40 } = 14
<br><br>$$\therefore$$ Probability when product is 0 = P(Product = 0) = $${14 \over {50}}$$
<br><br>n(Sum = 8 $$ \cap $$ Product = 0) = { 08 } = 1
<br><br>$$\therefore$$ Probability when sum is 8 and product is 0 = P(Sum = 8 $$ \cap $$ Product = 0) = $${1 \over {50}}$$
<br><br>Required probability,<br> $$P\left( {{{Sum = 8} \over {Product = 0}}} \right)$$ <br><br>= $${{P\left( {Sum = 8 \cap Product = 0} \right)} \over {P\left( {Product = 0} \right)}}$$<br>
<br>= $${{{1 \over {50}}} \over {{{14} \over {50}}}}$$
<br><br>=$${1 \over {14}}$$
<br><br>$$\therefore$$ Option (D) is correct. | mcq | aieee-2009 |
29tfdggsNkdMWL3J | maths | probability | conditional-probability-and-multiplication-theorem | If $$C$$ and $$D$$ are two events such that $$C \subset D$$ and $$P\left( D \right) \ne 0,$$ then the correct statement among the following is : | [{"identifier": "A", "content": "$$P\\left( {{C \\over D}} \\right)$$$$ \\ge P\\left( C \\right)$$ "}, {"identifier": "B", "content": "$$P\\left( {{C \\over D}} \\right)$$$$ < P\\left( C \\right)$$ "}, {"identifier": "C", "content": "$$P\\left( {{C \\over D}} \\right)$$$$ = {{P\\left( D \\right)} \\over {P\\left( C \\right)}}$$ "}, {"identifier": "D", "content": "$$P\\left( {{C \\over D}} \\right)$$$$ = P\\left( C \\right)$$ "}] | ["A"] | null | Given that $$C \subset D$$ means $$C$$ is present entirely inside $$D$$. Which is shown below.
<img class="question-image" src="https://imagex.cdn.examgoal.net/Fyl9Vf0KWiyB0u0XJ/tctGFeitNT0FkdGQUNU4VngNUHo8n/XoAkkZjSps5tTrqaC7VVGN/image.svg" loading="lazy" alt="AIEEE 2011 Mathematics - Probability Question 174 English Explanation 1">
<br><br>$$P\left( {{C \over D}} \right)$$ = $${{P\left( {C \cap D} \right)} \over {P\left( D \right)}}$$ = $${{P\left( C \right)} \over {P\left( D \right)}}$$
<br><br>As $$C \cap D$$ means common part of events C and D which is equal to C.
<img class="question-image" src="https://imagex.cdn.examgoal.net/3cwPhLtSMMbYWfmCP/EDmxTKI6c4GYJb2RPfMLqbjGkhNpF/mgqePz0BAb9zmlk9xrkpWi/image.svg" loading="lazy" alt="AIEEE 2011 Mathematics - Probability Question 174 English Explanation 2">
<br><br> $$0 \le P\left( D \right) \le 1$$
<br><br>$$\therefore$$ $${{P\left( C \right)} \over {P\left( D \right)}} \ge P\left( C \right)$$
<br><br><b>Note:</b> Here we are dividing with $${P\left( D \right)}$$ which is $$ \le 1$$ and $$ \ge 0$$, as we know on dividing with a number n in the range $$0 \le n \le 1$$ we get always more than or equal to the original number. | mcq | aieee-2011 |
DAR08l7NWYHC4QVf | maths | probability | conditional-probability-and-multiplication-theorem | Three numbers are chosen at random without replacement from $$\left\{ {1,2,3,..8} \right\}.$$ The probability that their minimum is $$3,$$ given that their maximum is $$6,$$ is : | [{"identifier": "A", "content": "$${3 \\over 8}$$ "}, {"identifier": "B", "content": "$${1 \\over 5}$$"}, {"identifier": "C", "content": "$${1 \\over 4}$$"}, {"identifier": "D", "content": "$${2 \\over 5}$$"}] | ["B"] | null | Given set S = $$\left\{ {1,2,3,..8} \right\}$$
<br><br>Choosing 3 numbers from 8 numbers can be done $${{}^8{C_3}}$$ ways.
<br><br>Choosing 3 numbers from 8 numbers while minimum no is 3 can be done $$1 \times {}^5{C_2}$$ ways.
<br><br>$$\therefore$$ Probablity P(min = 3) = $${{1 \times {}^5{C_2}} \over {{}^8{C_3}}}\,$$
<br><br>Choosing 3 numbers from 8 numbers while maximum no is 6 can be done $$1 \times {}^5{C_2}$$ ways.
<br><br>$$\therefore$$ Probablity P(max = 6) = $${{1 \times {}^5{C_2}} \over {{}^8{C_3}}}\,$$
<br><br>Choosing 3 numbers from 8 numbers while minimum number 3 and maximum no is 6 can be done $$1 \times {}^2{C_1} \times 1$$ ways.
<br><br>$$\therefore$$ $$P\left( {\min = 3 \cap \max = 6} \right)$$ = $${{1 \times {}^2{C_1} \times 1} \over {{}^8{C_3}}}$$
<br><br>The probability that their minimum is $$3,$$ given that their maximum is $$6,$$ is :
<br>$$P\left( {{{\min = 3} \over {\max = 6}}} \right)$$
<br> <br>= $${{P\left( {\min = 3 \cap \max = 6} \right)} \over {P\left( {\max = 6} \right)}}$$
<br><br>= $${{{{{}^2{C_1}} \over {{}^8{C_3}}}} \over {{{{}^5{C_2}} \over {{}^8{C_3}}}}}$$
<br><br> = $${{{}^2{C_1}} \over {{}^5{C_2}}}$$
<br><br>= $${{1 \over 5}}$$ | mcq | aieee-2012 |
gTCUP1krIS6fyFBA | maths | probability | conditional-probability-and-multiplication-theorem | Let $$A$$ and $$B$$ be two events such that $$P\left( {\overline {A \cup B} } \right) = {1 \over 6},\,P\left( { {A \cap B} } \right) = {1 \over 4}$$ and $$P\left( {\overline A } \right) = {1 \over 4},$$ where $$\overline A $$ stands for the complement of the event $$A$$. Then the events $$A$$ and $$B$$ are : | [{"identifier": "A", "content": "independent but not equally likely. "}, {"identifier": "B", "content": "independent and equally likely. "}, {"identifier": "C", "content": "mutually exclusive and independent."}, {"identifier": "D", "content": "equally likely but not independent."}] | ["A"] | null | <p>$$P(\overline {A \cup B} ) = {1 \over 6}$$</p>
<p>or, $$1 - P(A \cup B) = {1 \over 6}$$</p>
<p>$$\therefore$$ $$P(A \cup B) = 1 - {1 \over 6} = {5 \over 6};$$ $$P(A \cap B) = {1 \over 4}$$; and $$P(\overline A ) = {1 \over 4};$$</p>
<p>$$\therefore$$ $$P(A) = 1 - P(\overline A ) = 1 - {1 \over 4} = {3 \over 4}$$</p>
<p>We know, $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$</p>
<p>or, $${5 \over 6} = {3 \over 4} + P(B) - {1 \over 4}$$</p>
<p>or, $$P(B) = {5 \over 6} - {1 \over 2} = {1 \over 3}$$</p>
<p>Now, $$P(A)\,.\,P(B) = {3 \over 4}.\,{1 \over 3} = {1 \over 4} = P(A \cap B)$$</p>
<p>i.e., events A and B are mutually independent.</p>
<p>Since the probability of A and B are different, so they are not equally likely events.</p>
<p>Therefore, (A) is the correct option.</p> | mcq | jee-main-2014-offline |
IEceFTmzpblWyxmi | maths | probability | conditional-probability-and-multiplication-theorem | Let two fair six-faced dice $$A$$ and $$B$$ be thrown simultaneously. If $${E_1}$$ is the event that die $$A$$ shows up four, $${E_2}$$ is the event that die $$B$$ shows up two and $${E_3}$$ is the event that the sum of numbers on both dice is odd, then which of the following statements is $$NOT$$ true? | [{"identifier": "A", "content": "$${E_1}$$ and $${E_2}$$ are independent."}, {"identifier": "B", "content": "$${E_2}$$ and $${E_3}$$ are independent."}, {"identifier": "C", "content": "$${E_1}$$ and $${E_3}$$ are independent."}, {"identifier": "D", "content": "$${E_1},$$ $${E_2}$$ and $${E_3}$$ are independent."}] | ["D"] | null | Total possible outcome with two six faced dice = 6<sup>2</sup> = 36
<br><br>When dice A shows up 4, the possible cases are
<br>E<sub>1</sub> = { (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) } = 6 cases
<br>$$\therefore$$ $$P\left( {{E_1}} \right) = {6 \over {36}} = {1 \over 6}$$
<br><br> When dice B shows up 2, the possible cases are
<br>E<sub>2</sub> = { (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) } = 6 cases
<br><br>$$P\left( {{E_2}} \right) = {6 \over {36}} = {1 \over 6}$$
<br><br>$${E_1} \cap {E_2}$$ = Common in both in E<sub>1</sub> and E<sub>2</sub> = { (4, 2) }
<br><br>$$P\left( {{E_1} \cap {E_2}} \right) = {1 \over {36}}$$
<br><br>And $$P\left( {{E_1}} \right)$$.$$P\left( {{E_2}} \right)$$ = $${1 \over 6}$$.$${1 \over 6}$$ = $${1 \over 36}$$
<br><br>$$\therefore$$$$P\left( {{E_1} \cap {E_2}} \right)$$ = $$P\left( {{E_1}} \right)$$.$$P\left( {{E_2}} \right)$$
<br><br>$$\therefore$$ E<sub>1</sub> and E<sub>2</sub> are independent.
<br><br>$${E_3}$$ = [ (1, 2), (1, 4), (1, 6),
<br> (2, 1), (2, 3), (2, 5),
<br> (3, 2), (3, 4), (3, 6),
<br> (4, 1), (4, 3), (4, 5),
<br> (5, 2), (5, 4), (5, 6),
<br> (6, 1), (6, 3), (6, 5) ] = 18 cases
<br><br>$${E_1} \cap {E_3}$$ = { (4, 1) (4, 3) (4, 5) } = 3 cases
<br><br>$$\therefore$$ $$P\left( {{E_1} \cap {E_3}} \right) = {3 \over {36}} = {1 \over {12}}$$ = $${1 \over 6} \times {1 \over 2}$$ = $$P\left( {{E_1}} \right)$$ $$ \times $$ $$P\left( {{E_3}} \right)$$
<br><br>$$\therefore$$ E<sub>1</sub> and E<sub>3</sub> are independent.
<br><br>$${E_2} \cap {E_3}$$ = { (1, 2) (3, 2) (5, 2) } = 3 cases
<br><br>$$\therefore$$ $$P\left( {{E_2} \cap {E_3}} \right) = {3 \over {36}} = {1 \over {12}}$$ = $${1 \over 6} \times {1 \over 2}$$ = $$P\left( {{E_2}} \right)$$ $$ \times $$ $$P\left( {{E_3}} \right)$$
<br><br>$$\therefore$$ E<sub>2</sub> and E<sub>3</sub> are independent.
<br><br>$${E_1} \cap {E_2} \cap {E_3}$$ = 0
<br><br>$$\therefore$$ $$P\left( {{E_1} \cap {E_2} \cap {E_3}} \right)$$ = 0
<br><br>$$P\left( {{E_1}} \right) \times $$ $$P\left( {{E_2}} \right) \times $$$$P\left( {{E_3}} \right)$$ = $${1 \over 6}$$ $$ \times $$ $${1 \over 6}$$ $$ \times $$ $${1 \over 2}$$ = $${1 \over {72}}$$
<br><br>$$\therefore$$ $${E_1},$$ $${E_2}$$ and $${E_3}$$ are not independent.
<br><br>$$\therefore$$ Option (D) is correct. | mcq | jee-main-2016-offline |
PTFEW83CW5cQQ7CvoX4Zl | maths | probability | conditional-probability-and-multiplication-theorem | If A and B are any two events such that P(A) = $${2 \over 5}$$ and P (A $$ \cap $$ B) = $${3 \over {20}}$$, hen the conditional probability, P(A $$\left| {} \right.$$(A' $$ \cup $$ B')), where A' denotes the complement of A, is equal to : | [{"identifier": "A", "content": "$${1 \\over 4}$$ "}, {"identifier": "B", "content": "$${5 \\over 17}$$"}, {"identifier": "C", "content": "$${8 \\over 17}$$"}, {"identifier": "D", "content": "$${11 \\over 20}$$"}] | ["B"] | null | $$P\left( {{A \over {A' \cup B'}}} \right)$$
<br><br>$$ = {{P\left[ {A \cap \left( {A' \cup B'} \right)} \right]} \over {P\left( {A' \cap B'} \right)}}$$
<br><br>$$ = {{P\left[ {\left( {A \cap A'} \right) \cup \left( {A \cap B'} \right)} \right]} \over {P\left( {A \cap B} \right)'}}$$
<br><br>$$\left[ \, \right.$$As $$\left. {\left( {A \cap B} \right)' = A' \cap B'} \right]$$
<br><br>$$ = {{P\left( {A \cap B'} \right)} \over {P\left( {A \cap B} \right)'}}$$
<br><br>As $$A \cap A' = \phi $$
<br><br>$$ \therefore $$ $$\phi \cup \left( {A \cap B'} \right) = A \cap B'$$
<br><br>$$ = {{P\left( A \right) - P\left( {A \cap B} \right)} \over {1 - P\left( {A \cap B} \right)}}$$
<br><br>$$\left[ \, \right.$$As $$A \cap B' = A - \left( {A \cap B} \right)$$
<br><br>and $$\left. {\left( {A \cap B} \right)' = 1 - \left( {A \cap B} \right)} \right]$$
<br><br>Given $$P\left( A \right) = {2 \over 5}$$
<br><br>and $$P\left( {A \cap B} \right) = {3 \over {20}}$$
<br><br>$$ = {{{2 \over 5} - {3 \over {20}}} \over {1 - {3 \over {20}}}}$$
<br><br>$$ = {{{{8 - 3} \over {20}}} \over {{{17} \over {20}}}}$$
<br><br>$$ = {5 \over {17}}$$ | mcq | jee-main-2016-online-9th-april-morning-slot |
GzgdjCK46V8fjwqTu9deK | maths | probability | conditional-probability-and-multiplication-theorem | Let E and F be two independent events. The probability that both E and F happen is $${1 \over {12}}$$ and the probability that neither E nor F happens is $${1 \over {2}}$$, then a value of $${{P\left( E \right)} \over {P\left( F \right)}}$$ is : | [{"identifier": "A", "content": "$${4 \\over 3}$$"}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "$${1 \\over 3}$$"}, {"identifier": "D", "content": "$${5 \\over 12}$$"}] | ["A"] | null | <p>Let P(E) = x and P(F) = y</p>
<p>Now, $$P(E \cap F) = {1 \over {12}}$$</p>
<p>$$ \Rightarrow P(E)P(F) = {1 \over {12}}$$</p>
<p>$$ \Rightarrow xy = {1 \over {12}}$$</p>
<p>Also, $$P(E' \cap F') = {1 \over 2}$$</p>
<p>$$ \Rightarrow (1 - P(E))(1 - P(F)) = {1 \over 2}$$</p>
<p>$$ \Rightarrow (1 - x)(1 - y) = {1 \over 2}$$</p>
<p>$$ \Rightarrow 1 - x - y + xy = {1 \over 2}$$</p>
<p>$$ \Rightarrow x + y = 1 + xy - {1 \over 2}$$</p>
<p>$$ \Rightarrow x + y = 1 + {1 \over {12}} - {1 \over 2}$$</p>
<p>$$ \Rightarrow x + y = {1 \over 2} + {1 \over {12}} = {7 \over {12}}$$ ........ (1)</p>
<p>Now,</p>
<p>$${(x - y)^2} = {(x + y)^2} - 4xy$$</p>
<p>$$ \Rightarrow {(x - y)^2} = {{49} \over {144}} - {1 \over 3} \Rightarrow {{49 - 98} \over {144}} \Rightarrow {1 \over {144}}$$</p>
<p>$$ \Rightarrow (x - y) = {1 \over {12}} \Rightarrow x - y = {1 \over 2}$$ ........ (2)</p>
<p>From Eqs. (1) and (2), we get</p>
<p>$$(x + y)(x - y) = {7 \over {12}} + {1 \over {12}}$$</p>
<p>$$ \Rightarrow x = {4 \over {12}};y = {3 \over {12}}$$</p>
<p>$$ \Rightarrow {x \over y} = {4 \over 3} = {{P(E)} \over {P(F)}}$$</p> | mcq | jee-main-2017-online-9th-april-morning-slot |
QpF3mzdVaPsLYadpaz830 | maths | probability | conditional-probability-and-multiplication-theorem | A player X has a biased coin whose probability of showing heads is p and a player Y has a fair coin. They start playing a game with their own coins and play alternately. The player who throws a head first is a winner. If X starts the game, and the probability of winning the game by both the players is equal, then the value of 'p' is : | [{"identifier": "A", "content": "$${1 \\over 5}$$"}, {"identifier": "B", "content": "$${1 \\over 3}$$"}, {"identifier": "C", "content": "$${2 \\over 5}$$"}, {"identifier": "D", "content": "$${1 \\over 4}$$"}] | ["B"] | null | P(X getting head) = p
<br><br>$$ \therefore $$ P(X getting tail) = 1 - p
<br><br>P(Y getting head) = P(Y getting tail) = $${1 \over 2}$$
<br><br>P(X wins) = p + (1 - p)$${1 \over 2}$$p + (1 - p)$${1 \over 2}$$(1 - p)$${1 \over 2}$$p + ...
<br><br>= $${p \over {1 - \left( {{{1 - p} \over 2}} \right)}}$$
<br><br>= $${{2p} \over {1 + p}}$$
<br><br>P(Y win) = (1 - p)$${1 \over 2}$$ + (1 - p)$${1 \over 2}$$(1 - p)$${1 \over 2}$$ + ...
<br><br>= $$\left( {{{1 - p} \over 2}} \right).{p \over {1 - \left( {{{1 - p} \over 2}} \right)}} = {{1 - p} \over {1 + p}}$$
<br><br>According to question,
<br><br>P(X wins) = P(Y wins)
<br><br>$$ \therefore $$ $${{2p} \over {1 + p}}$$ = $${{1 - p} \over {1 + p}}$$
<br><br>$$ \Rightarrow $$ 3p = 1
<br><br>$$ \Rightarrow $$ p = $${1 \over 3}$$ | mcq | jee-main-2018-online-15th-april-evening-slot |
pKPACm03fi1d3b1OacMOk | maths | probability | conditional-probability-and-multiplication-theorem | Let A, B and C be three events, which are pair-wise independent and $$\overrightarrow E $$ denotes the completement of an event E. If $$P\left( {A \cap B \cap C} \right) = 0$$ and $$P\left( C \right) > 0,$$ then $$P\left[ {\left( {\overline A \cap \overline B } \right)\left| C \right.} \right]$$ is equal to : | [{"identifier": "A", "content": "$$P\\left( {\\overline A } \\right) - P\\left( B \\right)$$"}, {"identifier": "B", "content": "$$P\\left( A \\right) + P\\left( {\\overline B } \\right)$$"}, {"identifier": "C", "content": "$$P\\left( {\\overline A } \\right) - P\\left( {\\overline B } \\right)$$"}, {"identifier": "D", "content": "$$P\\left( {\\overline A } \\right) + P\\left( {\\overline B } \\right)$$"}] | ["A"] | null | Here, $$P\left( {\overline A \cap \overline B \left| C \right.} \right) = {{P\left( {\overline A \cap \overline B \cap C} \right)} \over {P\left( C \right)}}$$
<br><br>= $${{P\left[ {\left( {\overline {A \cup B} } \right) \cap C} \right]} \over {P\left( C \right)}}$$
<br><br>= $${{P\left[ {C - \left( {A \cup B} \right)} \right]} \over {P\left( C \right)}}$$
<br><br>= $${{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right) + P\left( {A \cap B \cap C} \right)} \over {P\left( C \right)}}$$
<br><br>= $${{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right)} \over {P\left( C \right)}}$$ ($$ \because $$$$\left. {P\left( {A \cap B \cap C} \right) = 0} \right)$$
<br><br>= $${{P\left( C \right) - P\left( A \right).P(C) - P\left( B \right).P(C)} \over {P\left( C \right)}}$$
<br><br>[$$ \because $$ A, B and C are independent events]
<br><br>= 1 - P(A) - P(B)
<br><br>= $$P\left( {\overline A } \right)$$ - P(B) or $$P\left( {\overline B } \right)$$ - P(A) | mcq | jee-main-2018-online-16th-april-morning-slot |
YsZBablbi4cftZyZiVjNo | maths | probability | conditional-probability-and-multiplication-theorem | An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is : | [{"identifier": "A", "content": "$${{21} \\over {49}}$$"}, {"identifier": "B", "content": "$${{27} \\over {49}}$$"}, {"identifier": "C", "content": "$${{26} \\over {49}}$$"}, {"identifier": "D", "content": "$${{32} \\over {49}}$$"}] | ["D"] | null | 5 Red and 2 green balls
<br><br>P(one red ball) = $${5 \over 7}$$
<br><br>P(one green ball) = $${2 \over 7}$$
<br><br><b>Case I : </b>
<br><br>If drawn ball is green than a red ball is added
<br><br>$$\left( {\matrix{
{6{\mathop{\rm Re}\nolimits} d} \cr
{1\,Green} \cr
} } \right)$$ P (red ball) = $${6 \over 7}$$
<br><br><b>Case II : </b>
<br><br>If drawn ball is red than a green ball is added
<br><br>$$\left( {\matrix{
{4{\mathop{\rm Re}\nolimits} d} \cr
{3\,Green} \cr
} } \right)$$ P (red ball) = $${4 \over 7}$$
<br><br>P (2<sup>nd</sup> red ball) = $${5 \over 7}$$ $$ \times {4 \over 7} + {2 \over 7} \times {6 \over 7}$$ = $${{32} \over {49}}$$ | mcq | jee-main-2019-online-9th-january-evening-slot |
bvonEScgOPynV9McufeFA | maths | probability | conditional-probability-and-multiplication-theorem | An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered 1, 2, 3, ……, 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is : | [{"identifier": "A", "content": "$${{19} \\over {36}}$$"}, {"identifier": "B", "content": "$${{15} \\over {72}}$$"}, {"identifier": "C", "content": "$${{13} \\over {36}}$$"}, {"identifier": "D", "content": "$${{19} \\over {72}}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265335/exam_images/oszgg4ddvwq2jkpn9psx.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Mathematics - Probability Question 153 English Explanation">
<br>$$P\left( A \right) = {1 \over 2} \times {{11} \over {36}} + {1 \over 2} \times {2 \over 9} = {{19} \over {72}}$$ | mcq | jee-main-2019-online-10th-january-morning-slot |
8U4m3K6A2fQNMMmrULIuQ | maths | probability | conditional-probability-and-multiplication-theorem | Two integers are selected at random from the set {1, 2, ...., 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is : | [{"identifier": "A", "content": "$${2 \\over 5}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${7 \\over 10}$$"}, {"identifier": "D", "content": "$${3 \\over 5}$$"}] | ["A"] | null | Since sum of two numbers is even so either both are odd or both are even. Hence number of elements in reduced samples space = <sup>5</sup>C<sub>2</sub> + <sup>6</sup>C<sub>2</sub>
<br><br>So, required probability = $${{{}^5{C_2}} \over {{}^5{C_2} + {}^6{C_2}}}$$ = $${2 \over 5}$$ | mcq | jee-main-2019-online-11th-january-morning-slot |
zrFBvUWrqtdjgdeYy5zha | maths | probability | conditional-probability-and-multiplication-theorem | Let A and B be two non-null events such that
A $$ \subset $$ B . Then, which of the following statements
is always correct? | [{"identifier": "A", "content": "P(A|B) = 1"}, {"identifier": "B", "content": "P(A|B) = P(B) \u2013 P(A)"}, {"identifier": "C", "content": "P(A|B) $$ \\le $$ P(A)\n"}, {"identifier": "D", "content": "P(A|B) $$ \\ge $$ P(A)\n"}] | ["D"] | null | $$P\left( {{A \over B}} \right) = {{P\left( {A \cap B} \right)} \over {P\left( B \right)}}$$
<br><br>As A $$ \subset $$ B,
<br><br>then P(A$$ \cap $$B) = P(A)
<br><br>$$ \therefore $$ $$P\left( {{A \over B}} \right) = {{P\left( A \right)} \over {P\left( B \right)}}$$
<br><br>As P(B) $$ \le $$ 1
<br><br>$$ \therefore $$ $${{P\left( A \right)} \over {P\left( B \right)}}$$ $$ \ge $$ P(A) | mcq | jee-main-2019-online-8th-april-morning-slot |
FtUibW67JlXKrCLOmU18hoxe66ijvwpbb2e | maths | probability | conditional-probability-and-multiplication-theorem | Four persons can hit a target correctly with
probabilities
$${1 \over 2}$$, $${1 \over 3}$$, $${1 \over 4}$$ and
$${1 \over 8}$$ respectively. if all hit
at the target independently, then the probability that
the target would be hit, is : | [{"identifier": "A", "content": "$${{25} \\over {32}}$$"}, {"identifier": "B", "content": "$${{25} \\over {192}}$$"}, {"identifier": "C", "content": "$${{1} \\over {192}}$$"}, {"identifier": "D", "content": "$${{7} \\over {32}}$$"}] | ["A"] | null | Let four persons are A, B, C and D.
<br><br>Probablity of hitting a target by them,
<br><br>P(A) = $${1 \over 2}$$
<br><br>P(B) = $${1 \over 3}$$
<br><br>P(C) = $${1 \over 4}$$
<br><br>P(D) = $${1 \over 8}$$
<br><br>Probablity of hitting target atleast once = 1 - Probablity of not hitting by anybody
<br><br>P(Hit) = 1 - $$P\left( {\overline A \cap \overline B \cap \overline C \cap \overline D } \right)$$
<br><br>= 1 - $$P\left( {\overline A } \right).P\left( {\overline B } \right).P\left( {\overline C } \right).P\left( {\overline D } \right)$$
<br><br>= 1 - $${1 \over 2}.{2 \over 3}.{3 \over 4}.{7 \over 8}$$
<br><br>= $${{25} \over {32}}$$ | mcq | jee-main-2019-online-9th-april-morning-slot |
0lS1ulEsuyoAXjx4Pr3rsa0w2w9jwxkwc5u | maths | probability | conditional-probability-and-multiplication-theorem | Assume that each born child is equally likely to be a boy or a girl. If two families have two children each,
then the conditional probability that all children are girls given that at least two are girls is : | [{"identifier": "A", "content": "$${1 \\over {10}}$$"}, {"identifier": "B", "content": "$${1 \\over {17}}$$"}, {"identifier": "C", "content": "$${1 \\over {11}}$$"}, {"identifier": "D", "content": "$${1 \\over {12}}$$"}] | ["C"] | null | A = At least two girls<br><br>
B = All girls<br><br>
$$P\left( {{B \over A}} \right) = {{P\left( {B \cap A} \right)} \over {P\left( A \right)}}$$<br><br>
$$ \Rightarrow {{P(B)} \over {P(A)}} = {{{{\left( {{1 \over 4}} \right)}^2}} \over {1 - {}^4{C_0}{{\left( {{1 \over 2}} \right)}^4} - {}^4{C_1}{{\left( {{1 \over 2}} \right)}^4}}}$$<br><br>
$$ \Rightarrow {1 \over {16 - 1 - 4}} = {1 \over {11}}$$ | mcq | jee-main-2019-online-10th-april-morning-slot |
WIVZ5wahzNzL3rDRF2jgy2xukfajvp4w | maths | probability | conditional-probability-and-multiplication-theorem | In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws total a of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. The probability of A winning the game is :
| [{"identifier": "A", "content": "$${5 \\over {6}}$$"}, {"identifier": "B", "content": "$${5 \\over {31}}$$"}, {"identifier": "C", "content": "$${31 \\over {61}}$$"}, {"identifier": "D", "content": "$${30 \\over {61}}$$"}] | ["D"] | null | Sum total 6 = {(1,5)(2,4)(3,3)(4,2)(5,1)}
<br><br>$$P(6) = {5 \over 36}$$
<br><br>Sum total 7 = {(1,6)(2,5)(3,4)(4,3)(5,2)(6,1)}
<br><br>$$\,P(7) = {6 \over {36}}$$ = $${1 \over 6}$$
<br><br>Game ends and A wins if A throws 6 in 1<sup>st</sup>
throw or A don’t throw 6 in 1<sup>st</sup> throw, B don’t
throw 7 in 1<sup>st</sup> throw and then A throw 6 in his
2
<sup>nd</sup> chance and so on.
<br><br>P(A) = A + $$\overline A \overline B A$$ + $$\overline A \overline B \overline A \overline B A$$
<br><br>= $${5 \over {36}} + \left( {{{31} \over {36}}} \right)\left( {{{30} \over {36}}} \right)\left( {{5 \over {36}}} \right) + $$ ..... $$\infty $$
<br><br>= $${{{5 \over {36}}} \over {1 - \left( {{{31} \over {36}}} \right)\left( {{{30} \over {36}}} \right)}}$$
<br><br>= $${{5 \times 36} \over {36 \times 36 - 31 \times 30}}$$
<br><br>= $${30 \over {61}}$$ | mcq | jee-main-2020-online-4th-september-evening-slot |
1YfQyWb3UUUTNEknDmjgy2xukezeh1np | maths | probability | conditional-probability-and-multiplication-theorem | Let E<sup>C</sup> denote the complement of an event E.
Let E<sub>1</sub>
, E<sub>2</sub>
and E<sub>3</sub>
be any pairwise independent
events with P(E<sub>1</sub>) > 0
<br/><br>and P(E<sub>1</sub> $$ \cap $$ E<sub>2</sub> $$ \cap $$ E<sub>3</sub>) = 0.
<br/><br>Then P($$E_2^C \cap E_3^C/{E_1}$$) is equal to :</br></br> | [{"identifier": "A", "content": "$$P\\left( {E_3^C} \\right)$$ - P(E<sub>2</sub>)"}, {"identifier": "B", "content": "$$P\\left( {E_2^C} \\right)$$ + P(E<sub>3</sub>)"}, {"identifier": "C", "content": "$$P\\left( {E_3^C} \\right)$$ - $$P\\left( {E_2^C} \\right)$$"}, {"identifier": "D", "content": "P(E<sub>3</sub>) - $$P\\left( {E_2^C} \\right)$$"}] | ["A"] | null | Given E<sub>1</sub> , E<sub>2</sub> , E<sub>3</sub> are pairwise indepedent events
<br><br>so P(E<sub>1</sub> $$ \cap $$ E<sub>2</sub> ) = P(E<sub>1</sub> ).P(E<sub>2</sub> )
<br><br>and P(E<sub>2</sub> $$ \cap $$ E<sub>3</sub> ) = P(E<sub>2</sub> ).P(E<sub>3</sub> )
<br><br>and P(E<sub>3</sub> $$ \cap $$ E<sub>1</sub> ) = P(E<sub>3</sub> ).P(E<sub>1</sub> )
<br><br>and P(E<sub>1</sub> $$ \cap $$ E<sub>2</sub> $$ \cap $$ E<sub>3</sub>) = 0
<br><br>Now $$P\left( {{{E_2^C \cap E_3^C} \over {{E_1}}}} \right)$$<br><br>
$$ = {{P\left[ {{E_1} \cap \left( {E_2^C \cap E_3^C} \right)} \right]} \over {P\left( {{E_1}} \right)}}$$<br><br>
$$ = {{P\left( {{E_1}} \right) - \left[ {P\left( {{E_1} \cap {E_2}} \right) + P\left( {{E_1} \cap {E_3}} \right) - P\left( {{E_1} \cap {E_2} \cap {E_3}} \right)} \right]} \over {P\left( {{E_1}} \right)}}$$<br><br>
$$ = {{P\left( {{E_1}} \right) - P\left( {{E_1}} \right).P\left( {{E_2}} \right) - P\left( {{E_1}} \right).P\left( {{E_3}} \right) - 0} \over {P\left( {{E_1}} \right)}}$$<br><br>
$$ = 1 - P\left( {{E_2}} \right) - P\left( {{E_3}} \right)$$<br><br>
$$ = 1 - P\left( {{E_3}} \right) - P\left( {{E_2}} \right)$$<br><br>
$$ = P\left( {E_3^C} \right) - P\left( {{E_2}} \right)$$
| mcq | jee-main-2020-online-2nd-september-evening-slot |
wcFmUM2A9VVSs7CPqj7k9k2k5gz74fk | maths | probability | conditional-probability-and-multiplication-theorem | Let A and B be two independent events such
that<br/> P(A) = $${1 \over 3}$$ and P(B) = $${1 \over 6}$$.<br/> Then, which of
the following is TRUE? | [{"identifier": "A", "content": "$$P\\left( {{A \\over {A \\cup B}}} \\right) = {1 \\over 4}$$"}, {"identifier": "B", "content": "$$P\\left( {{A \\over B}} \\right) = {2 \\over 3}$$"}, {"identifier": "C", "content": "$$P\\left( {{{A'} \\over {B'}}} \\right) = {1 \\over 3}$$"}, {"identifier": "D", "content": "$$P\\left( {{A \\over {B'}}} \\right) = {1 \\over 3}$$"}] | ["D"] | null | Given P(A) = $${1 \over 3}$$ and P(B) = $${1 \over 6}$$
<br><br>A and B are independent
<br><br>So P(A$$ \cap $$B) = $${1 \over 3} \times {1 \over 6}$$ = $${1 \over {18}}$$
<br><br>P(A$$ \cup $$B) = P(A) + P(B) – P(A $$ \cap $$ B)
<br><br>= $${1 \over 3}$$ + $${1 \over 6}$$ - $${1 \over {18}}$$
<br><br>= $${4 \over 9}$$
<br><br>$$P\left( {{A \over {B'}}} \right)$$
<br><br>= $${{P\left( {A \cap B'} \right)} \over {P\left( {B'} \right)}}$$
<br><br>= $${{P\left( A \right) - P\left( {A \cap B} \right)} \over {1 - P\left( B \right)}}$$
<br><br>= $${{{1 \over 3} - {1 \over {18}}} \over {1 - {1 \over 6}}}$$ = $${{1 \over 3}}$$ | mcq | jee-main-2020-online-8th-january-morning-slot |
bsMRfGHLGyb32Fe3n2jgy2xukezlxh5x | maths | probability | conditional-probability-and-multiplication-theorem | A dice is thrown two times and the sum of the
scores appearing on the die is observed to be
a multiple of 4. Then the conditional probability
that the score 4 has appeared atleast once is : | [{"identifier": "A", "content": "$${1 \\over 8}$$"}, {"identifier": "B", "content": "$${1 \\over 9}$$"}, {"identifier": "C", "content": "$${1 \\over 4}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["B"] | null | Let A is the event for getting score a multiple
of 4.
<br><br>So, A = { (1, 3), (3, 1), (2, 2), (2, 6), (6, 2),
(3, 5), (5, 3), (4, 4), (6, 6) } = 09
<br><br>n(A) = 9
<br><br>B : Score of 4 has appeared at least once.
<br><br>B = {(4, 4)}
<br><br>So, Required probability = $${1 \over 9}$$ | mcq | jee-main-2020-online-3rd-september-morning-slot |
aXHQRXYYpHINc0geBR1kls3ppy8 | maths | probability | conditional-probability-and-multiplication-theorem | When a missile is fired from a ship, the probability that it is intercepted is $${1 \over 3}$$ and the probability that the missile hits the target, given that it is not intercepted, is $${3 \over 4}$$. If three missiles are fired independently from the ship, then the probability that all three hit the target, is : | [{"identifier": "A", "content": "$${3 \\over 4}$$"}, {"identifier": "B", "content": "$${3 \\over 8}$$"}, {"identifier": "C", "content": "$${1 \\over 27}$$"}, {"identifier": "D", "content": "$${1 \\over 8}$$"}] | ["D"] | null | Probability of not getting intercepted = $${2 \over 3}$$<br><br>When it is not intercepted, probability of missile hitting target = $${3 \over 4}$$<br><br>$$\therefore$$ So when such 3 missiles launched
then P (all 3 hitting the target)
<br><br>= $${\left( {{2 \over 3} \times {3 \over 4}} \right)} $$ $$ \times $$ $${\left( {{2 \over 3} \times {3 \over 4}} \right)} $$ $$ \times $$ $${\left( {{2 \over 3} \times {3 \over 4}} \right)} $$
<br><br>$$= {1 \over 8}$$ | mcq | jee-main-2021-online-25th-february-morning-slot |
lNV17llKqZw6mSaHaj1kmhz6wk9 | maths | probability | conditional-probability-and-multiplication-theorem | A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is : | [{"identifier": "A", "content": "$${{39} \\over {50}}$$"}, {"identifier": "B", "content": "$${{3} \\over {4}}$$"}, {"identifier": "C", "content": "$${{22} \\over {425}}$$"}, {"identifier": "D", "content": "$${{52} \\over {867}}$$"}] | ["A"] | null | Consider the events,
<br><br>E<sub>1</sub> = missing card is spade
<br><br>E<sub>2</sub> = missing card is not a spade
<br><br>A = Two spade cards are drawn
<br><br>$$P\left( {{E_1}} \right) = {1 \over 4}$$
<br>$$P\left( {{E_2}} \right) = {3 \over 4}$$
<br><br>$$P\left( {{A \over {{E_1}}}} \right) = {{{}^{12}{C_2}} \over {{}^{51}{C_2}}}$$
<br><br>$$P\left( {{A \over {{E_2}}}} \right) = {{{}^{13}{C_2}} \over {{}^{51}{C_2}}}$$
<br><br>$$P\left( {{{{E_2}} \over A}} \right) = {{P\left( {{A \over {{E_2}}}} \right).P\left( {{E_2}} \right)} \over {P\left( {{A \over {{E_1}}}} \right).P\left( {{E_1}} \right) + P\left( {{A \over {{E_2}}}} \right).P\left( {{E_2}} \right)}}$$
<br><br>$$ = {{{{{}^{13}{C_2}} \over {{}^{51}{C_2}}}.{3 \over 4}} \over {{{{}^{12}{C_2}} \over {{}^{51}{C_2}}}.{1 \over 4} + {{{}^{13}{C_2}} \over {{}^{51}{C_2}}}.{3 \over 4}}}$$
<br><br>= $${{39} \over {50}}$$ | mcq | jee-main-2021-online-16th-march-morning-shift |
1ktbbtchx | maths | probability | conditional-probability-and-multiplication-theorem | Let A and B be independent events such that P(A) = p, P(B) = 2p. The largest value of p, for which P (exactly one of A, B occurs) = $${5 \over 9}$$, is : | [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "$${2 \\over 9}$$"}, {"identifier": "C", "content": "$${4 \\over 9}$$"}, {"identifier": "D", "content": "$${5 \\over 12}$$"}] | ["D"] | null | P (Exactly one of A or B)<br><br>$$ = P\left( {A \cap \overline B } \right) + \left( {\overline A \cap B} \right) = {5 \over 9}$$<br><br>$$ = P(A)P(\overline B ) + P(\overline A )P(B) = {5 \over 9}$$<br><br>$$ \Rightarrow P(A)(1 - P(B)) + (1 - P(A))P(B) = {5 \over 9}$$<br><br>$$ \Rightarrow p(1 - 2p) + (1 - p)2p = {5 \over 9}$$<br><br>$$ \Rightarrow 36{p^2} - 27p + 5 = 0$$<br><br>$$ \Rightarrow p = {1 \over 3}$$ or $${5 \over {12}}$$<br><br>$${p_{\max }} = {5 \over {12}}$$ | mcq | jee-main-2021-online-26th-august-morning-shift |
1ktd01kjf | maths | probability | conditional-probability-and-multiplication-theorem | A fair die is tossed until six is obtained on it. Let x be the number of required tosses, then the conditional probability P(x $$\ge$$ 5 | x > 2) is : | [{"identifier": "A", "content": "$${{125} \\over {216}}$$"}, {"identifier": "B", "content": "$${{11} \\over {36}}$$"}, {"identifier": "C", "content": "$${{5} \\over {6}}$$"}, {"identifier": "D", "content": "$${{25} \\over {36}}$$"}] | ["D"] | null | P(x $$\ge$$ 5 | x > 2) = $${{P(x \ge 5)} \over {P(x > 2)}}$$<br><br>= $${{{{\left( {{5 \over 6}} \right)}^4}.{1 \over 6} + {{\left( {{5 \over 6}} \right)}^5}.{1 \over 6} + ....... + \infty } \over {{{\left( {{5 \over 6}} \right)}^2}.{1 \over 6} + {{\left( {{5 \over 6}} \right)}^3}.{1 \over 6} + ...... + \infty }}$$<br><br>=$${{{{{{\left( {{5 \over 6}} \right)}^4}.{1 \over 5}} \over {1 - {5 \over 6}}}} \over {{{{{\left( {{5 \over 6}} \right)}^2}.{1 \over 6}} \over {1 - {5 \over 6}}}}} = {\left( {{5 \over 6}} \right)^2} = {{25} \over {36}}$$ | mcq | jee-main-2021-online-26th-august-evening-shift |
1ktend64o | maths | probability | conditional-probability-and-multiplication-theorem | When a certain biased die is rolled, a particular face occurs with probability $${1 \over 6} - x$$ and its opposite face occurs with probability $${1 \over 6} + x$$. All other faces occur with probability $${1 \over 6}$$. Note that opposite faces sum to 7 in any die. If 0 < x < $${1 \over 6}$$, and the probability of obtaining total sum = 7, when such a die is rolled twice, is $${13 \over 96}$$, then the value of x is : | [{"identifier": "A", "content": "$${1 \\over 16}$$"}, {"identifier": "B", "content": "$${1 \\over 8}$$"}, {"identifier": "C", "content": "$${1 \\over 9}$$"}, {"identifier": "D", "content": "$${1 \\over 12}$$"}] | ["B"] | null | Probability of obtaining total sum 7 = probability of getting opposite faces.<br><br>Probability of getting opposite faces<br><br>$$ = 2\left[ {\left( {{1 \over 6} - x} \right)\left( {{1 \over 6} + x} \right) + {1 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6}} \right]$$<br><br>$$ \Rightarrow 2\left[ {\left( {{1 \over 6} - x} \right)\left( {{1 \over 6} + x} \right) + {1 \over 6} \times {1 \over 6} + {1 \over 6} \times {1 \over 6}} \right] = {{13} \over {96}}$$ (given)<br><br>$$ \Rightarrow $$ $$x = {1 \over 8}$$ | mcq | jee-main-2021-online-27th-august-morning-shift |
1ktit2n77 | maths | probability | conditional-probability-and-multiplication-theorem | An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is 0.9 and that of the second unit is 0.8. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second unit is functioning is p, then 98 p is equal to _____________. | [] | null | 28 | I<sub>1</sub> = first unit is functioning<br><br>I<sub>2</sub> = second unit is functioning<br><br>P(I<sub>1</sub>) = 0.9, P(I<sub>2</sub>) = 0.8<br><br>P($$\overline {{I_1}} $$) = 0.1, P($$\overline {{I_2}} $$) = 0.2<br><br>$$P = {{0.8 \times 0.1} \over {0.1 \times 0.2 + 0.9 \times 0.2 + 0.1 \times 0.8}} = {8 \over {28}}$$<br><br>$$98P = {8 \over {28}} \times 98 = 28$$ | integer | jee-main-2021-online-31st-august-morning-shift |
1l5aie7xz | maths | probability | conditional-probability-and-multiplication-theorem | <p>Let E<sub>1</sub> and E<sub>2</sub> be two events such that the conditional probabilities $$P({E_1}|{E_2}) = {1 \over 2}$$, $$P({E_2}|{E_1}) = {3 \over 4}$$ and $$P({E_1} \cap {E_2}) = {1 \over 8}$$. Then :</p> | [{"identifier": "A", "content": "$$P({E_1} \\cap {E_2}) = P({E_1})\\,.\\,P({E_2})$$"}, {"identifier": "B", "content": "$$P(E{'_1} \\cap E{'_2}) = P(E{'_1})\\,.\\,P(E{_2})$$"}, {"identifier": "C", "content": "$$P({E_1} \\cap E{'_2}) = P({E_1})\\,.\\,P({E_2})$$"}, {"identifier": "D", "content": "$$P(E{'_1} \\cap {E_2}) = P({E_1})\\,.\\,P({E_2})$$"}] | ["C"] | null | <p>$$P\left( {{{{E_1}} \over {{E_2}}}} \right) = {1 \over 2} \Rightarrow {{P({E_1} \cap {E_2})} \over {P({E_2})}} = {1 \over 2}$$</p>
<p>$$P\left( {{{{E_2}} \over {{E_1}}}} \right) = {3 \over 4} \Rightarrow {{P({E_2} \cap {E_1})} \over {P({E_1})}} = {3 \over 4}$$</p>
<p>$$P({E_1} \cap {E_2}) = {1 \over 8}$$</p>
<p>$$P({E_2}) = {1 \over 4},\,P({E_1}) = {1 \over 6}$$</p>
<p>(A) $$P({E_1} \cap {E_2}) = {1 \over 8}$$ and $$P({E_1})\,.\,P({E_2}) = {1 \over {24}}$$</p>
<p>$$ \Rightarrow P({E_1} \cap {E_2}) \ne P({E_1})\,.\,P({E_2})$$</p>
<p>(B) $$P(E{'_1} \cap E{'_2}) = 1 - P({E_1} \cup {E_2})$$</p>
<p>$$ = 1 - \left[ {{1 \over 4} + {1 \over 6} - {1 \over 8}} \right] = {{17} \over {24}}$$</p>
<p>$$P(E{'_1}) = {3 \over 4} \Rightarrow P(E{'_1})P({E_2}) = {3 \over {24}}$$</p>
<p>$$ \Rightarrow P(E{'_1} \cap E{'_2}) \ne P(E{'_1})\,.\,P({E_2})$$</p>
<p>(C) $$P({E_1} \cap E{'_2}) = P({E_1}) - P({E_1} \cap {E_2})$$</p>
<p>$$ = {1 \over 6} - {1 \over 8} = {1 \over {24}}$$</p>
<p>$$P({E_1})\,.\,P({E_2}) = {1 \over {24}}$$</p>
<p>$$ \Rightarrow P({E_1} \cap E{'_2}) = P({E_1})\,.\,P({E_2})$$</p>
<p>(D) $$P(E{'_1} \cap {E_2}) = P({E_2}) - P({E_1} \cap {E_2})$$</p>
<p>$$ = {1 \over 4} - {1 \over 8} = {1 \over 8}$$</p>
<p>$$P({E_1})P({E_2}) = {1 \over {24}}$$</p>
<p>$$ \Rightarrow P(E{'_1} \cap {E_2}) \ne P({E_1})\,.\,P({E_2})$$</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
1l6f12wqy | maths | probability | conditional-probability-and-multiplication-theorem | <p>If $$A$$ and $$B$$ are two events such that $$P(A)=\frac{1}{3}, P(B)=\frac{1}{5}$$ and $$P(A \cup B)=\frac{1}{2}$$, then $$P\left(A \mid B^{\prime}\right)+P\left(B \mid A^{\prime}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{3}{4}$$"}, {"identifier": "B", "content": "$$\\frac{5}{8}$$"}, {"identifier": "C", "content": "$$\\frac{5}{4}$$"}, {"identifier": "D", "content": "$$\\frac{7}{8}$$"}] | ["B"] | null | <p>$$P(A) = {1 \over 3},\,P(B) = {1 \over 5}$$ and $$P\left( {A \cup B} \right) = {1 \over 2}$$</p>
<p>$$\therefore$$ $$P\left( {A \cap B} \right) = {1 \over 3} + {1 \over 5} - {1 \over 2} = {1 \over {30}}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7buiqub/55e764a3-4414-4eff-841a-7ce0120a8c5f/1bed5420-25ff-11ed-9c74-c5a04899a045/file-1l7buiquc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7buiqub/55e764a3-4414-4eff-841a-7ce0120a8c5f/1bed5420-25ff-11ed-9c74-c5a04899a045/file-1l7buiquc.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Evening Shift Mathematics - Probability Question 66 English Explanation"></p>
<p>Now, $$P\left( {A|B'} \right) + P\left( {B|A'} \right) = {{P\left( {A \cap B'} \right)} \over {P\left( {B'} \right)}} + {{P\left( {B \cap A'} \right)} \over {P\left( {A'} \right)}}$$</p>
<p>$$ = {{{9 \over {30}}} \over {{4 \over 5}}} + {{{5 \over {30}}} \over {{2 \over 3}}} = {5 \over 8}$$</p> | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6m5ors7 | maths | probability | conditional-probability-and-multiplication-theorem | <p>Out of $$60 \%$$ female and $$40 \%$$ male candidates appearing in an exam, $$60 \%$$ candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is :</p> | [{"identifier": "A", "content": "$$\\frac{2}{3}$$"}, {"identifier": "B", "content": "$$\\frac{11}{16}$$"}, {"identifier": "C", "content": "$$\\frac{23}{32}$$"}, {"identifier": "D", "content": "$$\\frac{13}{16}$$"}] | ["A"] | null | <p>P (Female) $$ = {{60} \over {100}} = {3 \over 5}$$</p>
<p>P (Male) $$ = {2 \over 5}$$</p>
<p>P (Female/Qualified) $$ = {{40} \over {60}} = {2 \over 3}$$</p>
<p>P (Male/qualified) $$ = {{20} \over {60}} = {1 \over 3}$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
1l6no0j1z | maths | probability | conditional-probability-and-multiplication-theorem | <p>Let $$\mathrm{A}$$ and $$\mathrm{B}$$ be two events such that $$P(B \mid A)=\frac{2}{5}, P(A \mid B)=\frac{1}{7}$$ and $$P(A \cap B)=\frac{1}{9} \cdot$$ Consider</p>
<p>(S1) $$P\left(A^{\prime} \cup B\right)=\frac{5}{6}$$,</p>
<p>(S2) $$P\left(A^{\prime} \cap B^{\prime}\right)=\frac{1}{18}$$</p>
<p>Then :</p> | [{"identifier": "A", "content": "Both (S1) and (S2) are true"}, {"identifier": "B", "content": "Both (S1) and (S2) are false"}, {"identifier": "C", "content": "Only (S1) is true"}, {"identifier": "D", "content": "Only (S2) is true"}] | ["A"] | null | <p>$$P(A/B) = {1 \over 7} \Rightarrow {{P(A \cap B)} \over {P(B)}} = {1 \over 7}$$</p>
<p>$$ \Rightarrow P(B) = {7 \over 9}$$</p>
<p>$$P(B/A) = {2 \over 5} \Rightarrow {{P(A \cap B)} \over {P(A)}} = {2 \over 5}$$</p>
<p>$$P(A) = {5 \over 2}\,.\,{1 \over 9} = {5 \over {18}}$$</p>
<p>$$S2:P(A' \cap B') = {1 \over {18}}$$</p>
<p>$$S1:$$ and $$P(A' \cup B) = {1 \over 9} + {6 \over 9} + {1 \over {18}} = {5 \over 6}.$$</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
1ldu5z45j | maths | probability | conditional-probability-and-multiplication-theorem | <p>25% of the population are smokers. A smoker has 27 times more chances to develop lung cancer than a non smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is $$\frac{k}{10}%$$. Then the value of k is __________.</p> | [] | null | 9 | Probability of a person being smoker $=\frac{1}{4}$
<br/><br/>
Probability of a person being non-smoker $=\frac{3}{4}$
<br/><br/>
$P\left(\frac{\text { Person is smoker }}{\text { Person diagonsed with cancer }}\right)=\frac{\frac{1}{4} \cdot 27 P}{\frac{1}{4} \cdot 27 P+\frac{3 P}{4}}$
<br/><br/>
$=\frac{9}{10}=\frac{k}{10}$
<br/><br/>
$\Rightarrow k=9$ | integer | jee-main-2023-online-25th-january-evening-shift |
1lgzyazug | maths | probability | conditional-probability-and-multiplication-theorem | <p>In a bolt factory, machines $$A, B$$ and $$C$$ manufacture respectively $$20 \%, 30 \%$$ and $$50 \%$$ of the total bolts. Of their output 3, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. If the bolt drawn is found the defective, then the probability that it is manufactured by the machine $$C$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{2}{7}$$"}, {"identifier": "B", "content": "$$\\frac{9}{28}$$"}, {"identifier": "C", "content": "$$\\frac{5}{14}$$"}, {"identifier": "D", "content": "$$\\frac{3}{7}$$"}] | ["C"] | null | Given : $P(A)=\frac{20}{100}=\frac{2}{10}$
<br/><br/>$$
P(B)=\frac{30}{100}=\frac{3}{10} $$
<br/><br/>$$ P(C)=\frac{50}{100}=\frac{5}{10}
$$
<br/><br/>Let $\mathrm{E} \rightarrow$ Event that the bolt is defective.
<br/><br/>$$
\text { So, } P(E / A)=\frac{3}{100}, $$
<br/><br/>$$P\left(\frac{E}{B}\right)=\frac{4}{100}, P\left(\frac{E}{C}\right)=\frac{2}{100}
$$
<br/><br/>So, $\mathrm{P}(\mathrm{C} / \mathrm{E})$
<br/><br/>$$
\begin{aligned}
& =\frac{P\left(\frac{E}{C}\right) \times P(C)}{P\left(\frac{E}{A}\right) \times P(A)+P\left(\frac{E}{B}\right) \times P(B)+P\left(\frac{E}{C}\right) \times P(C)} \\\\
& =\frac{\frac{5}{10} \times \frac{2}{100}}{\frac{3}{100} \times \frac{2}{10}+\frac{4}{100} \times \frac{3}{10}+\frac{2}{100} \times \frac{5}{10}} \\\\
& =\frac{10}{6+12+10}=\frac{10}{28}=\frac{5}{14}
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-morning-shift |
lsamdflq | maths | probability | conditional-probability-and-multiplication-theorem | Let Ajay will not appear in JEE exam with probability $\mathrm{p}=\frac{2}{7}$, while both Ajay and Vijay will appear in the exam with probability $\mathrm{q}=\frac{1}{5}$. Then the probability, that Ajay will appear in the exam and Vijay will not appear is : | [{"identifier": "A", "content": "$\\frac{9}{35}$"}, {"identifier": "B", "content": "$\\frac{3}{35}$"}, {"identifier": "C", "content": "$\\frac{24}{35}$"}, {"identifier": "D", "content": "$\\frac{18}{35}$"}] | ["D"] | null | <p>We are given that the probability of Ajay not appearing in the JEE exam is $\mathrm{p}=\frac{2}{7}$, and the probability that both Ajay and Vijay will appear in the exam is $\mathrm{q}=\frac{1}{5}$.</p>
<p>We are asked to find the probability that Ajay will appear in the exam and Vijay will not. Let's denote this probability as $\mathrm{r}$.</p>
<p>To find $\mathrm{r}$, we need to use the concept of complementary events. The probability that Ajay will appear in the exam is the complement of the probability that he will not appear. So,</p>
<p>$$ P(\text{Ajay appears}) = 1 - P(\text{Ajay does not appear}) = 1 - \mathrm{p} = 1 - \frac{2}{7} = \frac{5}{7}. $$</p>
<p>The event that both Ajay and Vijay appear in the exam is independent of the event that only Ajay appears (and Vijay does not). Therefore, we can express the probability that only Ajay will appear (and Vijay will not) as the difference of Ajay appearing minus both Ajay and Vijay appearing, because the probability of both appearing ($\mathrm{q}$) is included in the probability of Ajay appearing:</p>
<p>$$ \mathrm{r} = P(\text{Ajay appears}) - P(\text{Both Ajay and Vijay appear}) = \frac{5}{7} - \frac{1}{5}. $$</p>
<p>To subtract these two fractions, we need a common denominator, which would be $35$ in this case. So,</p>
<p>$$ \mathrm{r} = \frac{5}{7} \cdot \frac{5}{5} - \frac{1}{5} \cdot \frac{7}{7} = \frac{25}{35} - \frac{7}{35} = \frac{25 - 7}{35} = \frac{18}{35}. $$</p>
<p>Therefore, the probability that Ajay will appear in the exam and Vijay will not appear is $\mathrm{r} = \frac{18}{35}$, which corresponds to Option D.</p> | mcq | jee-main-2024-online-1st-february-evening-shift |
lsblkyym | maths | probability | conditional-probability-and-multiplication-theorem | A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let <br/><br/>$a=P(X=3), b=P(X \geqslant 3)$ and $c=P(X \geqslant 6 \mid X>3)$. Then $\frac{b+c}{a}$ is equal to __________. | [] | null | 12 | <p>To solve this problem, we need to compute the probabilities $a$, $b$, and $c$, and then plug those values into the expression $\frac{b+c}{a}$.</p>
<p>Let's begin by defining each of the variables:</p>
<ul>
<li>$a = P(X=3)$: This is the probability that the first six appears on the third toss.</li>
<li>$b = P(X \geqslant 3)$: This is the probability that the first six appears on the third toss or later.</li>
<li>$c = P(X \geqslant 6 \mid X>3)$: This is the probability that the first six appears on the sixth toss or later, given that it has not appeared in the first three tosses.</li>
</ul>
<p>Since we're dealing with a fair die, each side has an equal probability of $\frac{1}{6}$ of landing face up. Let's find the probabilities step by step:</p>
<p><strong>Calculating $a$:</strong></p>
<p>
<p>The probability of rolling anything other than a six is $\frac{5}{6}$. So for the first six to show up exactly on the third roll, the sequence of rolls must be NN6, where N is anything but a six (i.e., the results of the first two rolls). Thus,</p>
<p>$a = P(X=3) = \left(\frac{5}{6}\right) \cdot \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right)$</p>
</p>
<p><strong>Calculating $b$:</strong></p>
<p>
<p>For the first six to appear on the third roll or later, we can think of two cases: when the first six appears on the third roll (which we've already calculated, $a$), and when it appears after the third roll. To combine these probabilities, we can use the fact that $P(X \geqslant 3) = 1 - P(X < 3)$, where $P(X < 3)$ is the probability that the first six appears on either the first or the second roll. So we calculate the latter first:</p>
<p>$ P(X < 3) = P(X=1) + P(X=2) $</p>
<p>$ P(X < 3) = \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right) $</p>
<p>Thus,</p>
<p>$ b = P(X \geqslant 3) = 1 - P(X < 3) = 1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right) \cdot \left(\frac{1}{6}\right) \right] $</p>
</p>
<p><strong>Calculating $c$:</strong></p>
<p>
<p>This is the probability that the first six appears on or after the sixth roll, given that it hasn't appeared in the first three rolls. Since $X>3$, the first three outcomes must not be a six, which occurs with probability $\left(\frac{5}{6}\right)^3$. The subsequent outcomes until (and including) the fifth roll also must not be a six. So,</p>
<p>$c = P(X \geqslant 6 \mid X>3) = \left(\frac{5}{6}\right)^2$</p>
<p>Notice here, we did not include the probability of rolling a six, because we are looking for the probability that we have not yet rolled a six after the fifth roll.</p>
</p>
<p>Now we can calculate $a$, $b$, and $c$:</p>
<p>
<p>$a = \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}$</p>
<p>$b = 1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)\cdot\left(\frac{1}{6}\right) \right]$</p>
<p>$c = \left(\frac{5}{6}\right)^2$</p>
<p>Now we'll substitute to find $\frac{b+c}{a}$:</p>
<p>$\frac{b+c}{a} = \frac{1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)\cdot\left(\frac{1}{6}\right) \right] + \left(\frac{5}{6}\right)^2}{\left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}}$</p>
</p>
<p>Simplifying the numerator:</p>
<p>
<p>$1 - \left[ \left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)\cdot\left(\frac{1}{6}\right) \right] + \left(\frac{5}{6}\right)^2$</p>
<p>$= 1 - \left[\frac{1}{6} + \frac{5}{36}\right] + \frac{25}{36}$</p>
<p>$= 1 - \left[\frac{6}{36} + \frac{5}{36}\right] + \frac{25}{36}$</p>
<p>$= 1 - \frac{11}{36} + \frac{25}{36}$</p>
<p>$= \frac{36}{36} - \frac{11}{36} + \frac{25}{36}$</p>
<p>$= \frac{50}{36}$</p>
<p>Now, substitute this back into the expression and solve:</p>
<p>$\frac{b+c}{a} = \frac{\frac{50}{36}}{\left(\frac{5}{6}\right)^2 \cdot \frac{1}{6}}$</p>
<p>$\frac{b+c}{a} = \frac{50}{36} \cdot \frac{6}{\left(\frac{5}{6}\right)^2}$</p>
<p>$\frac{b+c}{a} = \frac{50 \cdot 6}{25}$</p>
<p>$\frac{b+c}{a} = \frac{300}{25}$</p>
<p>$\frac{b+c}{a} = 12$</p>
<p>Therefore, $\frac{b+c}{a} = 12$.</p></p>
<p></p> | integer | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lse58eei | maths | probability | conditional-probability-and-multiplication-theorem | <p>Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability, that first drawn marble is red and second drawn marble is white, is</p> | [{"identifier": "A", "content": "$$\\frac{4}{25}$$\n"}, {"identifier": "B", "content": "$$\\frac{2}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{2}{25}$$\n"}, {"identifier": "D", "content": "$$\\frac{4}{75}$$"}] | ["D"] | null | <p>To solve this problem, we need to calculate the probability of two independent events occurring in succession: the first marble drawn is red, and the second marble drawn is white. Since the drawing is with replacement, the number of marbles of each color remains the same for both draws.</p>
<p>The total number of marbles in the box is the sum of red, white, blue, and orange marbles:</p>
$$
\text{Total marbles} = 10 (\text{red}) + 30 (\text{white}) + 20 (\text{blue}) + 15 (\text{orange}) = 75.
$$
<p>The probability of drawing a red marble in the first draw is the number of red marbles divided by the total number of marbles:</p>
$$
P(\text{First is red}) = \frac{10}{75}.
$$
<p>Since the marble is replaced, the probability of drawing a white marble in the second draw remains as the number of white marbles divided by the total number of marbles:</p>
$$
P(\text{Second is white}) = \frac{30}{75}.
$$
<p>The probability of both independent events occurring in succession (drawing a red marble first and then a white marble) is the product of their individual probabilities:</p>
$$
P(\text{First is red and second is white}) = P(\text{First is red}) \times P(\text{Second is white}) = \frac{10}{75} \times \frac{30}{75}.
$$
<p>Now, let's calculate this probability:</p>
<p>$$
P(\text{First is red and second is white}) = \frac{10 \times 30}{75 \times 75} = \frac{300}{5625}= \frac{4}{75}.
$$</p>
<p>Therefore, the correct answer is</p>
Option D
$$\frac{4}{75}.$$ | mcq | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lseygjua | maths | probability | conditional-probability-and-multiplication-theorem | <p>A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is</p> | [{"identifier": "A", "content": "$$\\frac{5}{11}$$\n"}, {"identifier": "B", "content": "$$\\frac{5}{6}$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{6}$$\n"}, {"identifier": "D", "content": "$$\\frac{6}{11}$$"}] | ["A"] | null | <p>Required probability $$=$$</p>
<p>$$\begin{aligned}
& \frac{5}{6} \times \frac{1}{6}+\left(\frac{5}{6}\right)^3 \times \frac{1}{6}+\left(\frac{5}{6}\right)^5 \times \frac{1}{6}+\ldots . . \\
& =\frac{1}{6} \times \frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{11}
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
8pKDwjndu91hVdqw | maths | probability | permutation-and-combination-based-problem | An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is : | [{"identifier": "A", "content": "$${2 \\over 7}$$"}, {"identifier": "B", "content": "$${1 \\over 21}$$"}, {"identifier": "C", "content": "$${1 \\over 23}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["A"] | null | Out of nine balls three balls can be chosen = $${}^9{C_3}$$ ways
<br><br>$$\therefore$$ Sample space = $${}^9{C_3}$$ = $${{9!} \over {3!6!}}$$ = $${{9 \times 8 \times 7} \over 6}$$ = 84
<br><br>According to the question, all three ball should be different. So out of 3 red balls 1 is chosen and out of 4 blue 1 is chosen and out of 2 green 1 is chosen.
<br><br>$$\therefore$$ Total cases = $${}^3{C_1} \times {}^4{C_1} \times {}^2{C_1}$$ = 3 $$ \times $$ 4 $$ \times $$ 2 = 24
<br><br>$$\therefore$$ Probability = $${{24} \over {84}}$$ = $${{2} \over {7}}$$ | mcq | aieee-2010 |
m6SF5Xwnm5L4nJ2s | maths | probability | permutation-and-combination-based-problem | Four numbers are chosen at random (without replacement) from the set $$\left\{ {1,2,3,....20} \right\}.$$
<br/><br/><b>Statement - 1:</b> The probability that the chosen numbers when arranged in some order will form an AP is $${1 \over {85}}.$$
<p><b>Statement - 2:</b> If the four chosen numbers form an AP, then the set of all possible values of common difference is $$\left( { \pm 1, \pm 2, \pm 3, \pm 4, \pm 5} \right).$$</p> | [{"identifier": "A", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is <b>not</b> a correct explanation for Statement - 1."}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is false."}, {"identifier": "C", "content": "Statement - 1 is false, Statement -2 is true."}, {"identifier": "D", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1."}] | ["B"] | null | Four numbers can be chosen $${}^{20}{C_4}$$ ways.
<br><br>When common difference d = 1 then the possible sets are (1, 2, 3, 4) (2, 3, 4, 5) (3, 4, 5, 6) ................. (17, 18, 19, 20) = 17 sets
<br>So when d = 1 then 17 different AP's are possible with 4 numbers.
<br><br>Now let's create a table of all possible sets -
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<table class="tg">
<tbody><tr>
<th class="tg-wyy1">Common <br>Difference<br>(d)</th>
<th class="tg-wyy1">Possible Sets </th>
<th class="tg-wyy1">No of AP</th>
</tr>
<tr>
<td class="tg-s6z2">d = 1</td>
<td class="tg-s6z2">(1, 2, 3, 4) (2, 3, 4, 5) (3, 4, 5, 6) ................. (17, 18, 19, 20)</td>
<td class="tg-s6z2">17</td>
</tr>
<tr>
<td class="tg-s6z2">d = 2</td>
<td class="tg-s6z2">(1, 3, 5, 7) (2, 4, 6, 8) (3, 5, 7, 9) ................. (14, 16, 18, 20)</td>
<td class="tg-s6z2">14</td>
</tr>
<tr>
<td class="tg-s6z2">d = 3</td>
<td class="tg-s6z2">(1, 4, 7, 10) (2, 5, 8, 11) (3, 6, 9, 12) ................. (11, 14, 17, 20)</td>
<td class="tg-s6z2">11</td>
</tr>
<tr>
<td class="tg-baqh">d = 4</td>
<td class="tg-baqh">(1, 5, 9, 13) (2, 6, 10, 14) (3, 7, 11, 15) ................. (8, 12, 16, 20)</td>
<td class="tg-baqh">8</td>
</tr>
<tr>
<td class="tg-baqh">d = 5</td>
<td class="tg-baqh">(1, 6, 11, 16) (2, 7, 12, 17) (3, 8, 13, 18) (4, 9, 14, 19) (5, 10, 15, 20)</td>
<td class="tg-baqh">5</td>
</tr>
<tr>
<td class="tg-baqh">d = 6</td>
<td class="tg-baqh">(1, 7, 13, 19) (2, 8, 14, 20)</td>
<td class="tg-baqh">2</td>
</tr>
</tbody></table>
<br><br>$$\therefore$$ Total no of AP = 17 + 14 + 11 + 8 + 5 + 2 = 57
<br><br>$$\therefore$$ Required probability = $${{57} \over {{}^{20}{C_4}}}$$ = $${1 \over {85}}$$
<br><br>$$\therefore$$ <b>Statement - 1:</b> is true.
<br><br>$$\therefore$$ <b>Statement - 2:</b> is false as common difference can also be $$ \pm 6$$. | mcq | aieee-2010 |
v0MDihlczrEKxer8 | maths | probability | permutation-and-combination-based-problem | If $$12$$ different balls are to be placed in $$3$$ identical boxes, then the probability that one of the boxes contains exactly $$3$$ balls is : | [{"identifier": "A", "content": "$$220{\\left( {{1 \\over 3}} \\right)^{12}}$$ "}, {"identifier": "B", "content": "$$22{\\left( {{1 \\over 3}} \\right)^{11}}$$"}, {"identifier": "C", "content": "$${{55} \\over 3}{\\left( {{2 \\over 3}} \\right)^{11}}$$ "}, {"identifier": "D", "content": "$$55{\\left( {{2 \\over 3}} \\right)^{10}}$$"}] | ["C"] | null | 1<sup>st</sup> ball can go any of the 3 boxes. So total choices for 1<sup>st</sup> ball = 3
<br><br>2<sup>nd</sup> ball can also go any of the 3 boxes. So total choices for 2<sup>nd</sup> ball = 3
<br>.
<br>.
<br>.
<br>.
<br>12<sup>th</sup> ball can go any of the 3 boxes. So total choices for 12<sup>th</sup> ball = 3
<br><br>Total choices for all 12 balls = $$3 \times $$$$3 \times $$$$3 \times $$.................12 times = 3<sup>12</sup>.
<br><br>Now question says choose 3 balls from 12 balls. So no of ways = $${}^{12}{C_3}$$ ways.
<br>And then put it in a box. No of ways we can put = $${}^{12}{C_3} \times 1$$ ways.
<br><br>Now we have 9 balls left and we have to put those 9 balls in the remaining 2 boxes.
<br><br>Each ball can go to any of the 2 boxes, so for each ball there is 2 choices.
<br><br>$$\therefore$$ Total ways for 9 balls = 2<sup>9</sup>
<br><br>$$\therefore$$ Total ways we can put those 12 balls in the boxes = $${}^{12}{C_3} \times 1 \times {2^9}$$
<br><br>$$\therefore$$ Required probability = $${{{}^{12}{C_3} \times 1 \times {2^9}} \over {{3^{12}}}}$$ = $${{55} \over 3}{\left( {{2 \over 3}} \right)^{11}}$$
<br><br>So option (C) is correct. | mcq | jee-main-2015-offline |
hb12DaeYWQTsDI7quf0Na | maths | probability | permutation-and-combination-based-problem | From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is :
| [{"identifier": "A", "content": "$${{21} \\over {220}}$$ "}, {"identifier": "B", "content": "$${{3} \\over {11}}$$ "}, {"identifier": "C", "content": "$${{1} \\over {11}}$$ "}, {"identifier": "D", "content": "$${{2} \\over {23}}$$ "}] | ["C"] | null | <p>The number of ways to form a committee having at least one woman is</p>
<p>$$ = {}^5{C_1} \times {}^{10}{C_3} + {}^5{C_2} \times {}^{10}{C_2} + {}^5{C_3} \times {}^{10}{C_1} + {}^5{C_4}$$</p>
<p>$$ = {{5!} \over {4!}} \times {{10!} \over {7! \times 3!}} + {{5!} \over {2!3!}} \times {{10!} \over {8!2!}} + {{5!} \over {3!2!}} \times {{10!} \over {9!1!}} + {{5!} \over {4!}}$$</p>
<p>$$ = 5 \times {{10 \times 9 \times 8} \over {3 \times 2}} + {{5 \times 4} \over {2 \times 1}} \times {{10 \times 9} \over 2} + {{5 \times 4} \over {2 \times 1}} \times 10 + 5$$</p>
<p>$$ = 600 + 450 + 100 + 5 = 1155$$</p>
<p>The number of ways to form a committee having more women than men is</p>
<p>$${}^5{C_2} \times {}^{10}{C_2} + {}^5{C_4} = {{5!} \over {6!2!}} \times {{10!} \over {9!}} + {{5!} \over {4!}}$$</p>
<p>$$ = 10 \times 10 + 5 = 105$$</p>
<p>Therefore, the probability for the committees to have more women than men is</p>
<p>$${{105} \over {1155}} = {1 \over {11}}$$</p> | mcq | jee-main-2017-online-9th-april-morning-slot |
NAV5RgX2pFLPU0hYcWpBO | maths | probability | permutation-and-combination-based-problem | Two different families A and B are blessed with equal numbe of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family B is $${1 \over {12}},$$ then the number of children in each family is : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}] | ["C"] | null | Let the number of children in each family be x.
<br><br>Thus the total number of children in both the families are 2x
<br><br>Now, it is given that 3 tickets are distributed amongst the children of these two families.
<br><br>Thus, the probability that all the three tickets go to the children in family B
<br><br>= $${{{}^x{C_3}} \over {{}^{2x}{C_3}}}$$ = $${1 \over {12}}$$
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ $${{x\left( {x - 1} \right)\left( {x - 2} \right)} \over {2x\left( {2x - 1} \right)\left( {2x - 2} \right)}}$$ = $${1 \over {12}}$$
<br><br>$$ \Rightarrow $$ $${{\left( {x - 2} \right)} \over {\left( {2x - 1} \right)}}$$ = $${1 \over 6}$$
<br><br>Thus, the number of children in each family is 5. | mcq | jee-main-2018-online-16th-april-morning-slot |
miWACwbOxOgzdWPOMnhck | maths | probability | permutation-and-combination-based-problem | Let S = {1, 2, . . . . . ., 20}. A subset B of S is said to be "nice", if the sum of the elements of B is 203. Then the probability that a randonly chosen subset of S is "nice" is :
| [{"identifier": "A", "content": "$${5 \\over {{2^{20}}}}$$"}, {"identifier": "B", "content": "$${7 \\over {{2^{20}}}}$$"}, {"identifier": "C", "content": "$${4 \\over {{2^{20}}}}$$"}, {"identifier": "D", "content": "$${6 \\over {{2^{20}}}}$$"}] | ["A"] | null | <p>We can solve this problem by counting the number of "nice" subsets in the set S = {1, 2, $\ldots$, 20 }, and then dividing that number by the total number of possible subsets of S.</p>
<p>The sum of all elements in S is :</p>
<p>1 + 2 + $\ldots$ + 20 = $\frac{{20 \times 21}}{2}$ = 210</p>
<p>Since a "nice" subset must sum to 203, the elements not in the subset must sum to 210 - 203 = 7.</p>
<p>Now we need to find the ways to make the sum of 7 using the elements of S. The combinations are :</p>
<ol>
<li>1. 7</li>
<li>2. 1 + 6</li>
<li>3. 2 + 5</li>
<li>4. 3 + 4</li>
<li>5. 1 + 2 + 4</li>
<li>6. 1 + 3 + 3(This doesn't work since 3 is repeated)</li>
<li>7. 2 + 2 + 3(This doesn't work since 2 is repeated)</li>
</ol>
<p>So, there are 5 "nice" subsets.</p>
<p>Since the set S has 20 elements, there are $2^{20}$ possible subsets (including the empty set and the set itself). The probability of randomly choosing a "nice" subset is therefore :
<br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263858/exam_images/lnhcuunfwlwjtih9updz.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 11th January Evening Slot Mathematics - Probability Question 150 English Explanation"></p> | mcq | jee-main-2019-online-11th-january-evening-slot |
ZjWD3HbNe8roavEsKs3rsa0w2w9jx6gohcx | maths | probability | permutation-and-combination-based-problem | If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle
formed with these chosen vertices is equilateral is : | [{"identifier": "A", "content": "$${1 \\over {10}}$$"}, {"identifier": "B", "content": "$${3 \\over {10}}$$"}, {"identifier": "C", "content": "$${3 \\over {20}}$$"}, {"identifier": "D", "content": "$${1 \\over {5}}$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263722/exam_images/n4gew2oz06txcsbjoxw5.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Morning Slot Mathematics - Probability Question 139 English Explanation"><br>
Choosing vertices of a regular hexagon alternate, here A<sub>1</sub>, A<sub>3</sub>, A<sub>5</sub> or A<sub>2</sub>, A<sub>4</sub>, A<sub>6</sub> will result in an equilateral triangle.<br><br>
Hence the required probability = $${2 \over {{}^6{C_3}}}$$ = $${1 \over {10}}$$ | mcq | jee-main-2019-online-12th-april-morning-slot |
izhPmlaiTmaj8SvcuP7k9k2k5khcyom | maths | probability | permutation-and-combination-based-problem | If 10 different balls are to be placed in 4 distinct
boxes at random, then the probability that two
of these boxes contain exactly 2 and 3 balls is : | [{"identifier": "A", "content": "$${{965} \\over {{2^{11}}}}$$"}, {"identifier": "B", "content": "$${{965} \\over {{2^{10}}}}$$"}, {"identifier": "C", "content": "$${{945} \\over {{2^{11}}}}$$"}, {"identifier": "D", "content": "$${{945} \\over {{2^{10}}}}$$"}] | ["D"] | null | Total ways of distribution = 4<sup>10</sup> = 2<sup>20</sup>
<br><br>Number of ways selecting two boxes out of four = <sup>4</sup>C<sub>2</sub>
<br><br>Then number of ways selecting 5 balls out of 10 = <sup>10</sup>C<sub>5</sub>
<br><br>Then no of ways of distributing 5 balls into two groups of 2 balls and 3 balls = <sup>5</sup>C<sub>3</sub>.2!
<br><br>Then number of ways to distributing remaining balls into two boxes = 2<sup>5</sup>
<br><br>Number of ways placing exactly 2 and 3 balls in two of these boxes
<br><br>= <sup>4</sup>C<sub>2</sub> $$ \times $$ <sup>10</sup>C<sub>5</sub> $$ \times $$ <sup>5</sup>C<sub>3</sub>.2! $$ \times $$ 2<sup>5</sup>
<br><br>= $${{{{6.252.10.2.2}^5}} \over {{2^{20}}}}$$
<br><br>= $${{945} \over {{2^{10}}}}$$ | mcq | jee-main-2020-online-9th-january-evening-slot |
D78VhXLXjBEt1DRB79jgy2xukf49fh2z | maths | probability | permutation-and-combination-based-problem | The probability that a randomly chosen 5-digit
number is made from exactly two digits is : | [{"identifier": "A", "content": "$${{150} \\over {{{10}^4}}}$$"}, {"identifier": "B", "content": "$${{134} \\over {{{10}^4}}}$$"}, {"identifier": "C", "content": "$${{121} \\over {{{10}^4}}}$$"}, {"identifier": "D", "content": "$${{135} \\over {{{10}^4}}}$$"}] | ["D"] | null | Sample space = 9 $$ \times $$ 10<sup>4</sup><br><br>Case - I<br><br>Out of exactly two digits selected one is zero then favourable cases = $${}^9{C_1}({2^4} - 1)$$<br><br>Case - II<br><br>Both selected digits are non-zero then favourable cases = $${}^9{C_2}({2^5} - 2)$$<br><br>Probability = $${{9({2^4} - 1) + {{9.8} \over 2}({2^5} - 2)} \over {9 \times {{10}^4}}}$$<br><br>$$ = {{15 + 120} \over {{{10}^4}}} = {{135} \over {{{10}^4}}}$$ | mcq | jee-main-2020-online-3rd-september-evening-slot |
2gXd4NRSYBvLj2D1oajgy2xukfuvqwpi | maths | probability | permutation-and-combination-based-problem | Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is : | [{"identifier": "A", "content": "$${{10} \\over {99}}$$"}, {"identifier": "B", "content": "$${{5} \\over {33}}$$"}, {"identifier": "C", "content": "$${{15} \\over {101}}$$"}, {"identifier": "D", "content": "$${{5} \\over {101}}$$"}] | ["B"] | null | Out of 11 consecutive natural numbers either
6 even and 5 odd numbers or 5 even and 6
odd numbers.
<br><br>Let, E = Even
<br>O = Odd
<br><br><b>Case-1 :</b>
<br><br>E, O, E, O, E, O, E, O, E, O, E
<br><br>2b = a + c $$ \Rightarrow $$ Even
<br><br>$$ \Rightarrow $$ Both a and c should be either even or odd.
<br><br>P = $${{{}^6{C_2} + {}^5{C_2}} \over {{}^{11}{C_3}}}$$ = $${5 \over {33}}$$
<br><br><b>Case -2 :</b>
<br><br>O, E, O, E, O, E, O, E, O, E, O
<br><br>P = $${{{}^5{C_2} + {}^6{C_2}} \over {{}^{11}{C_3}}}$$ = $${5 \over {33}}$$
<br><br>Total probability = $${1 \over 2} \times {5 \over {33}}$$ + $${1 \over 2} \times {5 \over {33}}$$ = $${5 \over {33}}$$ | mcq | jee-main-2020-online-6th-september-morning-slot |
JvSQmAt1Odx2wCs08v1klrlyrob | maths | probability | permutation-and-combination-based-problem | The probability that two randomly selected subsets of the set {1, 2, 3, 4, 5} have exactly two elements in their intersection, is : | [{"identifier": "A", "content": "$${{135} \\over {{2^9}}}$$"}, {"identifier": "B", "content": "$${{65} \\over {{2^8}}}$$"}, {"identifier": "C", "content": "$${{65} \\over {{2^7}}}$$"}, {"identifier": "D", "content": "$${{35} \\over {{2^7}}}$$"}] | ["A"] | null | Given, set P = {1, 2, 3, 4, 5}<br/><br/>Let the two subsets be A and B<br/><br/>Then, n (A $$\cap$$ B) = 2 (as given in question)
<br/><br/>We can choose two elements from set P in <sup>5</sup>C<sub>2</sub> ways.
<br/><br/>After choosing two common elements for set A and B, each of remaining three elements from set P have three choice (1) It can go to set A (2) It can go to set B (3) It don't go to any sets it stays at set P.
<br/><br/>$$ \therefore $$ Total ways for the three elements = 3 $$ \times $$ 3 $$ \times $$ 3 = 3<sup>3</sup>
<br/><br/>$$\therefore$$ Required probability = $${{{}^5{C_2} \times {3^3}} \over {\left( {{2^5}} \right)\left( {{2^5}} \right)}}$$ = $${{{}^5{C_2} \times {3^3}} \over {{4^5}}} = {{10 \times 27} \over {{2^{10}}}} = {{135} \over {{2^9}}}$$ | mcq | jee-main-2021-online-24th-february-evening-slot |
BKJKsH0G7if3cZd50d1klt7nltg | maths | probability | permutation-and-combination-based-problem | Let A be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of A leaves remainder 2 when divided by 5 is : | [{"identifier": "A", "content": "$${2 \\over 9}$$"}, {"identifier": "B", "content": "$${1 \\over 5}$$"}, {"identifier": "C", "content": "$${122 \\over 297}$$"}, {"identifier": "D", "content": "$${97 \\over 297}$$"}] | ["D"] | null | n(s) = n(when 7 appears on thousands place)<br><br>+ n (7 does not appear on thousands place)<br><br>= 9 $$\times$$ 9 $$\times$$ 9 + 8 $$\times$$ 9 $$\times$$ 9 $$\times$$ 3<br><br>= 33 $$\times$$ 9 $$\times$$ 9<br><br>n(E) = n(last digit 7 & 7 appears once)<br><br>+n(last digit 2 when 7 appears once)<br><br>= 8 $$\times$$ 9 $$\times$$ 9 + (3 $$\times$$ 9 $$\times$$ 9 - 2 $$\times$$ 9)<br><br>$$\therefore$$ $$P(E) = {{8 \times 9 \times 9 + 9 \times 25} \over {33 \times 9 \times 9}} = {{97} \over {297}}$$ | mcq | jee-main-2021-online-25th-february-evening-slot |
GJ1zE8RkCmDVd30zIh1kluxtvqr | maths | probability | permutation-and-combination-based-problem | A seven digit number is formed using digits 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is : | [{"identifier": "A", "content": "$${1 \\over 7}$$"}, {"identifier": "B", "content": "$${4 \\over 7}$$"}, {"identifier": "C", "content": "$${6 \\over 7}$$"}, {"identifier": "D", "content": "$${3 \\over 7}$$"}] | ["D"] | null | Digits = 3, 3, 4, 4, 4, 5, 5<br><br>Total 7 digit numbers = $${{7!} \over {2!2!3!}}$$<br><br>Number of 7 digit number divisible by 2 $$ \Rightarrow $$ last digit = 4<br><br> <picture><source media="(max-width: 2210px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263557/exam_images/xuvtn4gwslaitxll4yp9.webp"><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265492/exam_images/au3xqinyxld1l4hjeaso.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265171/exam_images/eeu2lwysmlvu3odhfdrs.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263836/exam_images/y9lzr2tj3abwsbodp1cx.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264537/exam_images/vvir1mxtzvslxg81ywvk.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266451/exam_images/q2kwdgywrif048t1trat.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264077/exam_images/yc1gwgxjzvxqymummtib.webp"><source media="(max-width: 1400px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263429/exam_images/b0bzozhagttkkdhav7tl.webp"><source media="(max-width: 1580px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264423/exam_images/ousq7k6cfagspu8jzhgh.webp"><source media="(max-width: 1760px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264587/exam_images/gsxhokydxfh9u8o35bsr.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266109/exam_images/qg9czmccpl9q9jom4wds.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Evening Shift Mathematics - Probability Question 112 English Explanation"></picture> <br><br>Now 7 digit numbers which are divisible by 2 = $${{6!} \over {2!2!2!}}$$<br><br>Required probability = $${{{{6!} \over {2!2!2!}}} \over {{{7!} \over {3!2!2!}}}} = {3 \over 7}$$ | mcq | jee-main-2021-online-26th-february-evening-slot |
1krpzjgtr | maths | probability | permutation-and-combination-based-problem | Words with or without meaning are to be formed using all the letters of the word EXAMINATION. The probability that the letter M appears at the fourth position in any such word is : | [{"identifier": "A", "content": "$${1 \\over {66}}$$"}, {"identifier": "B", "content": "$${1 \\over {11}}$$"}, {"identifier": "C", "content": "$${1 \\over {9}}$$"}, {"identifier": "D", "content": "$${2 \\over {11}}$$"}] | ["B"] | null | AAEIIMNNOTX<br><br>----------------M----------------<br><br>Total words with M at fourth Place = $${{10!} \over {2!2!2!}}$$<br><br>Total words = $${{11!} \over {2!2!2!}}$$<br><br>Required probability = $${{10!} \over {11!}}$$ = $${{1} \over {11}}$$ | mcq | jee-main-2021-online-20th-july-morning-shift |
1krtcgdaz | maths | probability | permutation-and-combination-based-problem | Four dice are thrown simultaneously and the numbers shown on these dice are recorded in 2 $$\times$$ 2 matrices. The probability that such formed matrix have all different entries and are non-singular, is : | [{"identifier": "A", "content": "$${{45} \\over {162}}$$"}, {"identifier": "B", "content": "$${{21} \\over {81}}$$"}, {"identifier": "C", "content": "$${{22} \\over {81}}$$"}, {"identifier": "D", "content": "$${{43} \\over {162}}$$"}] | ["D"] | null | $$A = \left| {\matrix{
a & b \cr
c & d \cr
} } \right|$$<br><br>| A | = ad $$-$$ bc<br><br>Total case = 6<sup>4</sup><br><br>For non-singular matrix | A | $$\ne$$ 0 $$\Rightarrow$$ ad $$-$$ bc $$\ne$$ 0<br><br>$$\Rightarrow$$ ad $$\ne$$ bc<br><br>And a, b, c, d are all different numbers in the set {1, 2, 3, 4, 5, 6}<br><br>Now for ad = bc<br><br>(i) 6 $$\times$$ 1 = 2 $$\times$$ 3<br><br>$$\Rightarrow$$ $$\left. \matrix{
a = 6,b = 2,c = 3,d = 1 \hfill \cr
or\,a = 1,b = 2,c = 3,d = 6 \hfill \cr
: \hfill \cr
: \hfill \cr} \right\}$$ 8 each cases<br><br>(ii) 6 $$\times$$ 2 = 3 $$\times$$ 4<br><br>$$\Rightarrow$$ $$\left. \matrix{
a = 6,b = 3,c = 4,d = 2 \hfill \cr
or\,a = 2,b = 3,c = 4,d = 6 \hfill \cr
: \hfill \cr
: \hfill \cr} \right\}$$ 8 such cases<br><br>favourable cases<br><br>= $$^6C_4 \times 4! - 16$$<br><br>required probability <br><br>$$ = {{{^6C_4 \times 4!} - 16} \over {{6^4}}} = {{43} \over {162}}$$ | mcq | jee-main-2021-online-22th-july-evening-shift |
1krw0yts1 | maths | probability | permutation-and-combination-based-problem | Let 9 distinct balls be distributed among 4 boxes, B<sub>1</sub>, B<sub>2</sub>, B<sub>3</sub> and B<sub>4</sub>. If the probability than B<sub>3</sub> contains exactly 3 balls is $$k{\left( {{3 \over 4}} \right)^9}$$ then k lies in the set : | [{"identifier": "A", "content": "{x $$\\in$$ R : |x $$-$$ 3| < 1}"}, {"identifier": "B", "content": "{x $$\\in$$ R : |x $$-$$ 2| $$\\le$$ 1}"}, {"identifier": "C", "content": "{x $$\\in$$ R : |x $$-$$ 1| < 1}"}, {"identifier": "D", "content": "{x $$\\in$$ R : |x $$-$$ 5| $$\\le$$ 1}"}] | ["A"] | null | Required probability = $${{{}^9{C_3}{{.3}^6}} \over {{4^9}}}$$<br><br>$$ = {{{}^9{C_3}} \over {27}}.{\left( {{3 \over 4}} \right)^9}$$<br><br>$$ = {{28} \over 9}.{\left( {{3 \over 4}} \right)^9} \Rightarrow k = {{28} \over 9}$$<br><br>Which satisfies $$\left| {x - 3} \right| < 1$$ | mcq | jee-main-2021-online-25th-july-morning-shift |
1ktk4uzqz | maths | probability | permutation-and-combination-based-problem | Let S = {1, 2, 3, 4, 5, 6}. Then the probability that a randomly chosen onto function g from S to S satisfies g(3) = 2g(1) is : | [{"identifier": "A", "content": "$${1 \\over {10}}$$"}, {"identifier": "B", "content": "$${1 \\over {15}}$$"}, {"identifier": "C", "content": "$${1 \\over {5}}$$"}, {"identifier": "D", "content": "$${1 \\over {30}}$$"}] | ["A"] | null | g(3) = 2g(1) can be defined in 3 ways<br><br>number of onto functions in this condition = 3 $$\times$$ 4!<br><br>Total number of onto functions = 6!<br><br>Required probability = $${{3 \times 4!} \over {6!}} = {1 \over {10}}$$ | mcq | jee-main-2021-online-31st-august-evening-shift |
1kto2k0k0 | maths | probability | permutation-and-combination-based-problem | Two squares are chosen at random on a chessboard (see figure). The probability that they have a side in common is :<br/><br/><img src="data:image/png;base64,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"/> | [{"identifier": "A", "content": "$${2 \\over 7}$$"}, {"identifier": "B", "content": "$${1 \\over 18}$$"}, {"identifier": "C", "content": "$${1 \\over 7}$$"}, {"identifier": "D", "content": "$${1 \\over 9}$$"}] | ["B"] | null | Total ways of choosing square = $${}^{64}{C_2}$$<br><br>$$ = {{64 \times 63} \over {2 \times 1}} = 32 \times 63$$<br><br>ways of choosing two squares having common side = 2 (7 $$\times$$ 8) = 112<br><br>Required probability $$ = {{112} \over {32 \times 63}} = {{16} \over {32 \times 9}} = {1 \over {18}}$$.<br><br>Ans. (b) | mcq | jee-main-2021-online-1st-september-evening-shift |
1l544hp5b | maths | probability | permutation-and-combination-based-problem | <p>The probability that a randomly chosen 2 $$\times$$ 2 matrix with all the entries from the set of first 10 primes, is singular, is equal to :</p> | [{"identifier": "A", "content": "$${{133} \\over {{{10}^4}}}$$"}, {"identifier": "B", "content": "$${{18} \\over {{{10}^3}}}$$"}, {"identifier": "C", "content": "$${{19} \\over {{{10}^3}}}$$"}, {"identifier": "D", "content": "$${{271} \\over {{{10}^4}}}$$"}] | ["C"] | null | <p>First 10 prime numbers are</p>
<p>={2, 3, 5, 7, 11, 13, 17, 19, 23, 29}</p>
<p>Let A is a 2 $$\times$$ 2 matrix,</p>
<p>$$A = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right]$$</p>
<p>Given that matrix A is singular.</p>
<p>$$\therefore$$ | A | = 0</p>
<p>$$ \Rightarrow \left| {\matrix{
a & b \cr
c & d \cr
} } \right| = 0$$</p>
<p>$$ \Rightarrow ad = bc$$</p>
<p>Case I :</p>
<p>ad = bc condition satisfy when a = b = c = d.</p>
<p>For ex when a = 2, b = 2, c = 2, d = 2, then ad = bc satisfy.</p>
<p>Now there are 10 prime numbers.</p>
<p>We can choose any one of the 10 prime number in $${}^{10}C_{1}$$ = 10 ways and put them in the four positions of the matrix and matrix will be singular.</p>
<p>$$\therefore$$ In this case, total favorable case = 10</p>
<p>Case 2 :</p>
<p>ad = bc condition satisfies when</p>
<p>(1)</p>
<p>a = 2, d - 3 then</p>
<p>(a) b = 2, c = 3</p>
<p>(b) b = 3, c = 2</p>
<p>or</p>
<p>a = 3, d = 2 then</p>
<p>(a) b = 2, c = 3</p>
<p>(b) b = 3, c = 2</p>
<p>So you can see for two different prime number for a and d there are 4 possible value of b and c which satisfy ad = bc condition.</p>
<p>Two different values of a and d can be chosen from 10 prime numbers = $${}^{10}C_{2}$$ ways</p>
<p>And for each combination of a and d there are 4 possible values of b and c.</p>
<p>$$\therefore$$ Total possible values = $${}^{10}C_{2}$$ $$\times$$ 4</p>
<p>From case I and case II total possible values of 10 prime numbers which satisfy ad = bc condition</p>
<p>= 10 + $${}^{10}C_{2}$$ $$\times$$ 4</p>
<p>For sample space,</p>
<p>Number of ways to fill element a of matrix A = chose any prime number among 10 available prime number = $${}^{10}C_{1}$$ ways</p>
<p>Similarly,</p>
<p>For element b of matrix A = $${}^{10}C_{1}$$ ways</p>
<p>For element c of matrix A = $${}^{10}C_{1}$$ ways</p>
<p>For element d of matrix A = $${}^{10}C_{1}$$ ways</p>
<p>$$\therefore$$ Sample space = $${}^{10}C_{1}$$ $$\times$$ $${}^{10}C_{1}$$ $$\times$$ $${}^{10}C_{1}$$ $$\times$$ $${}^{10}C_{1}$$ = 10<sup>4</sup></p>
<p>$$\therefore$$ Probability $$ = {{10 + {{}^{10}C_{2}} \times 4} \over {{{10}^4}}}$$</p>
<p>$$ = {{10 + 180} \over {{{10}^4}}}$$</p>
<p>$$ = {{190} \over {{{10}^4}}}$$</p>
<p>$$ = {{19} \over {{{10}^3}}}$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift |
1l55iov0n | maths | probability | permutation-and-combination-based-problem | <p>The probability that a randomly chosen one-one function from the set {a, b, c, d} to the set {1, 2, 3, 4, 5} satisfies f(a) + 2f(b) $$-$$ f(c) = f(d) is :</p> | [{"identifier": "A", "content": "$${1 \\over {24}}$$"}, {"identifier": "B", "content": "$${1 \\over {40}}$$"}, {"identifier": "C", "content": "$${1 \\over {30}}$$"}, {"identifier": "D", "content": "$${1 \\over {20}}$$"}] | ["D"] | null | Number of one-one function from $\{a, b, c, d\}$ to set $\{1,2,3,4,5\}$ is ${ }^{5} P_{4}=120 n(s)$.
<br/><br/>
The required possible set of value (f(a), $f(b), f(c), f(d))$ such that $f(a)+2 f(b)-f(c)=f(d)$ are $(5,3,2,1),(5,1,2,3),(4,1,3,5),(3,1,4,5)$, $(5,4,3,2)$ and $(3,4,5,2)$
<br/><br/>
$\therefore n(E)=6$
<br/><br/>
$\therefore $ Required probability $=\frac{n(E)}{n(S)}=\frac{6}{120}=\frac{1}{20}$ | mcq | jee-main-2022-online-28th-june-evening-shift |
1l57onrrd | maths | probability | permutation-and-combination-based-problem | <p>Five numbers $${x_1},{x_2},{x_3},{x_4},{x_5}$$ are randomly selected from the numbers 1, 2, 3, ......., 18 and are arranged in the increasing order $$({x_1} < {x_2} < {x_3} < {x_4} < {x_5})$$. The probability that $${x_2} = 7$$ and $${x_4} = 11$$ is :</p> | [{"identifier": "A", "content": "$${1 \\over {136}}$$"}, {"identifier": "B", "content": "$${1 \\over {72}}$$"}, {"identifier": "C", "content": "$${1 \\over {68}}$$"}, {"identifier": "D", "content": "$${1 \\over {34}}$$"}] | ["C"] | null | No. of ways to select and arrange $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4, \mathrm{x}_5$ from $1,2,3$.......18<br/><br/>
$$
\begin{aligned}
& \mathrm{n}(\mathrm{s})={ }^{18} \mathrm{C}_5 \\\\
& \begin{array}{lllll}
x_1 & \underset{7}{\left(x_2\right)} & x_3 & \underset{11}{\left(x_4\right)} & x_5
\end{array} \\\\
& \mathrm{n}(\mathrm{E})={ }^6 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1 \times{ }^7 \mathrm{C}_1 \\\\
& P(E)=\frac{6 \times 3 \times 7}{{ }^{18} C_5} \\\\
& \frac{1}{17 \times 4}=\frac{1}{68} \\\\
\end{aligned}
$$ | mcq | jee-main-2022-online-27th-june-morning-shift |
1l58h8xsg | maths | probability | permutation-and-combination-based-problem | <p>If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is p, then 96 p is equal to _______________.</p> | [] | null | 33 | <p>Total number of numbers from given</p>
<p>Condition = n(s) = 2<sup>6</sup>.</p>
<p>Every required number is of the form</p>
<p>A = 7 . (10<sup>a<sub>1</sub></sup> + 10<sup>a<sub>2</sub></sup> + 10<sup>a<sub>3</sub></sup> + .......) + 111111</p>
<p>Here 111111 is always divisible by 21.</p>
<p>$$\therefore$$ If A is divisible by 21 then</p>
<p>10<sup>a<sub>1</sub></sup> + 10<sup>a<sub>2</sub></sup> + 10<sup>a<sub>3</sub></sup> + ....... must be divisible by 3.</p>
<p>For this we have <sup>6</sup>C<sub>0</sub> + <sup>6</sup>C<sub>3</sub> + <sup>6</sup>C<sub>6</sub> cases are there</p>
<p>$$\therefore$$ n(E) = <sup>6</sup>C<sub>0</sub> + <sup>6</sup>C<sub>3</sub> + <sup>6</sup>C<sub>6</sub> = 22</p>
<p>$$\therefore$$ Required probability = $${{22} \over {{2^6}}} = p$$</p>
<p>$$\therefore$$ $${{11} \over {32}} = p$$</p>
<p>$$\therefore$$ $$96p = 33$$</p> | integer | jee-main-2022-online-26th-june-evening-shift |
1ldsuemqy | maths | probability | permutation-and-combination-based-problem | <p>Fifteen football players of a club-team are given 15 T-shirts with their names written on the backside. If the players pick up the T-shirts randomly, then the probability that at least 3 players pick the correct T-shirt is :</p> | [{"identifier": "A", "content": "$$\\frac{1}{6}$$"}, {"identifier": "B", "content": "$$\\frac{2}{15}$$"}, {"identifier": "C", "content": "$$\\frac{5}{24}$$"}, {"identifier": "D", "content": "0.08"}] | ["D"] | null | Required probability $=1-\frac{D_{(15)}+{ }^{15} C_1 \cdot D_{(14)}+{ }^{15} C_2 D_{(13)}}{15 !}$
<br/><br/>Taking $\mathrm{D}_{(15)}$ as $\frac{15 !}{e}$
<br/><br/>$\mathrm{D}_{(14)}$ as $\frac{14 \text { ! }}{e}$
<br/><br/>$\mathrm{D}_{(13)}$ as $\frac{13 \text { ! }}{e}$
<br/><br/>$$
\text { We get, } \text { Required probability } = 1-\left(\frac{\frac{15 !}{e}+15 \cdot \frac{14 !}{e}+\frac{15 \times 14}{2} \times \frac{13 !}{e}}{15 !}\right)
$$
<br/><br/>$$
=1-\left(\frac{1}{e}+\frac{1}{e}+\frac{1}{2 e}\right)=1-\frac{5}{2 e} \approx .08
$$ | mcq | jee-main-2023-online-29th-january-morning-shift |
jaoe38c1lscn159w | maths | probability | permutation-and-combination-based-problem | <p>An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability, that the first draw gives all white balls and the second draw gives all black balls, is :</p> | [{"identifier": "A", "content": "$$\\frac{3}{256}$$"}, {"identifier": "B", "content": "$$\\frac{5}{256}$$"}, {"identifier": "C", "content": "$$\\frac{3}{715}$$"}, {"identifier": "D", "content": "$$\\frac{5}{715}$$"}] | ["C"] | null | <p>$$\frac{{ }^6 \mathrm{C}_4}{{ }^{15} \mathrm{C}_4} \times \frac{{ }^9 \mathrm{C}_4}{{ }^{11} \mathrm{C}_4}=\frac{3}{715}$$</p>
<p>Hence option (3) is correct.</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsd3atnw | maths | probability | permutation-and-combination-based-problem | <p>A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is</p> | [{"identifier": "A", "content": "$$\\frac{1}{9}$$"}, {"identifier": "B", "content": "$$\\frac{2}{9}$$"}, {"identifier": "C", "content": "$$\\frac{1}{27}$$"}, {"identifier": "D", "content": "$$\\frac{2}{27}$$"}] | ["B"] | null | <p>To solve this problem, we need to first determine the probability of getting a head (H) and the probability of getting a tail (T).</p>
<p>Since a head is twice as likely to occur as a tail, we can denote the probability of getting a tail as $$ P(T) = p $$ and the probability of getting a head as $$ P(H) = 2p $$.</p>
<p>These probabilities must sum to 1 because those are the only two possible outcomes for each coin toss :
<br/><br/>$$ P(H) + P(T) = 1 $$
<br/><br/>$$ 2p + p = 1 $$
<br/><br/>$$ 3p = 1 $$
<br/><br/>$$ p = \frac{1}{3} $$</p>
<p>Therefore, the probability of getting a tail (T) is $$ P(T) = \frac{1}{3} $$ and the probability of getting a head (H) is $$ P(H) = 2 \times \frac{1}{3} = \frac{2}{3} $$.</p>
<p>Now to find the probability of getting two tails and one head, we need to consider the different sequences in which this can occur. There are three unique sequences: TTH, THT, and HTT.</p>
<p>The probability of each sequence is found by multiplying the probabilities of each individual event since each coin toss is independent:
<br/><br/>$$ P(TTH) = P(T) \times P(T) \times P(H) = \left(\frac{1}{3}\right)^2 \times \frac{2}{3} = \frac{1}{9} \times \frac{2}{3} = \frac{2}{27} $$
<br/><br/>$$ P(THT) = P(T) \times P(H) \times P(T) = \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{2}{27} $$
<br/><br/>$$ P(HTT) = P(H) \times P(T) \times P(T) = \frac{2}{3} \times \left(\frac{1}{3}\right)^2 = \frac{2}{27} $$</p>
<p>The overall probability of getting two tails and one head in any order is the sum of these individual probabilities :
<br/><br/>$$ P(2T1H) = P(TTH) + P(THT) + P(HTT) = \frac{2}{27} + \frac{2}{27} + \frac{2}{27} = \frac{6}{27} $$</p>
<p>Simplifying this expression gives us:
$$ P(2T1H) = \frac{6}{27} = \frac{2}{9} $$</p>
<p>Therefore, the correct answer is :
<br/><br/>Option B :
$$\frac{2}{9}$$</p>
| mcq | jee-main-2024-online-31st-january-evening-shift |
lv9s209z | maths | probability | permutation-and-combination-based-problem | <p>The coefficients $$\mathrm{a}, \mathrm{b}, \mathrm{c}$$ in the quadratic equation $$\mathrm{a} x^2+\mathrm{bx}+\mathrm{c}=0$$ are from the set $$\{1,2,3,4,5,6\}$$. If the probability of this equation having one real root bigger than the other is p, then 216p equals :</p> | [{"identifier": "A", "content": "38"}, {"identifier": "B", "content": "76"}, {"identifier": "C", "content": "57"}, {"identifier": "D", "content": "19"}] | ["A"] | null | <p>Equation is $$a x^2+b x+c=0$$</p>
<p>$$\mathrm{D}>0$$ [for roots to be real & distinct]</p>
<p>$$\Rightarrow b^2-4 a c>0$$</p>
<p>For $$b<2$$ no value of $$a$$ & $$c$$ are possible</p>
<p>$$\begin{aligned}
& \text { For } b=3 \Rightarrow a c<\frac{9}{4} \\
& (a, c) \in\{(1,1),(1,2),(2,1)\} \Rightarrow 3 \text { cases }
\end{aligned}$$</p>
<p>For $$b=4 \Rightarrow a c<4$$</p>
<p>$$(a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3)\} \Rightarrow 5 \text { cases }$$</p>
<p>For $$b=5 \Rightarrow a c<\frac{25}{4}$$</p>
<p>$$\begin{aligned}
& (a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3),(2,2), \\
& (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6), \\
& (6,1)\}=14 \text { cases }
\end{aligned}$$</p>
<p>For $$b=6 \Rightarrow a c<9$$</p>
<p>$$\begin{aligned}
& (a, c) \in\{(1,1),(1,2),(2,1),(3,1),(1,3),(2,2), \\
& (4,1),(1,4),(3,2),(2,3),(5,1),(1,5),(1,6), \\
& (6,1),(2,4),(4,2)\}=16 \text { cases }
\end{aligned}$$</p>
<p>Total cases $$=3+5+14+16=38$$ cases</p>
<p>$$\Rightarrow$$ Probability, $$p=\frac{38}{216}$$</p>
<p>$$\Rightarrow 216 p=38$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift |
uiR8NU8AS1zaGHd3 | maths | probability | probability-distribution-of-a-random-variable | A random variable $$X$$ has Poisson distribution with mean $$2$$.
<br/>Then $$P\left( {X > 1.5} \right)$$ equals : | [{"identifier": "A", "content": "$${2 \\over {{e^2}}}$$ "}, {"identifier": "B", "content": "$$0$$"}, {"identifier": "C", "content": "$$1 - {3 \\over {{e^2}}}$$ "}, {"identifier": "D", "content": "$${3 \\over {{e^2}}}$$ "}] | ["C"] | null | <p>In a position distribution,</p>
<p>$$P(X = r) = {{{e^{ - \lambda }}{\lambda ^r}} \over {r!}}$$ ($$\lambda$$ = mean).</p>
<p>Now, $$P(X = r > 1.5) = P(2) + P(3) + $$ ..... $$\infty$$</p>
<p>$$ = 1 - \{ P(0) + P(1)\} $$</p>
<p>$$ = 1 - \left( {{e^{ - 2}} + {{{e^{ - 2}} \times 2} \over 1}} \right) = 1 - {3 \over {{e^2}}}$$</p> | mcq | aieee-2005 |
ZjpgMVecKoOHSSxa | maths | probability | probability-distribution-of-a-random-variable | At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of $$5$$ phone calls during $$10$$ minute time intervals. The probability that there is at the most one phone call during a $$10$$-minute time period is : | [{"identifier": "A", "content": "$${6 \\over {{5^e}}}$$ "}, {"identifier": "B", "content": "$${5 \\over 6}$$ "}, {"identifier": "C", "content": "$${6 \\over 55}$$"}, {"identifier": "D", "content": "$${6 \\over {{e^5}}}$$ "}] | ["D"] | null | <p>Required probability</p>
<p>$$ = P(X = 0) + P(X = 1)$$</p>
<p>$$ = {{{e^{ - 5}}} \over {0!}}{5^0} + {{{e^{ - 5}}} \over {1!}}{5^1}$$</p>
<p>$$ = {e^{ - 5}} + 5{e^{ - 5}}$$</p>
<p>$$ = {6 \over {{e^5}}}$$</p> | mcq | aieee-2006 |
P6xO9hto9apIDYTTsa3rsa0w2w9jxayoap3 | maths | probability | probability-distribution-of-a-random-variable | A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins
Rs. 12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the
expected gain/loss (in Rs.) of the person is : | [{"identifier": "A", "content": "$${1 \\over 4}$$ loss"}, {"identifier": "B", "content": "$${1 \\over 2}$$ gain"}, {"identifier": "C", "content": "$${1 \\over 2}$$ loss"}, {"identifier": "D", "content": "2 gain"}] | ["C"] | null | When two dice are thrown then sample space will {(1, 1), (2, 2) ....... (6, 6)} contain total 36 elements number of cases.<br><br>
Then the expectation will be $${6 \over {36}} \times 15 \times {4 \over {36}} \times 12 - {{26} \over {36}} \times 6$$<br><br>
$${{90 + 48 - 156} \over {36}} = - {1 \over 2}$$ = $$ {1 \over 2}$$ loss
| mcq | jee-main-2019-online-12th-april-evening-slot |
FxEzwiF0BbebQgJ2yf7k9k2k5e2z1s8 | maths | probability | probability-distribution-of-a-random-variable | An unbiased coin is tossed 5 times. Suppose that a variable X is assigned the value of k when k
consecutive heads are obtained for k = 3, 4, 5, otherwise X takes the value -1. Then the expected
value of X, is :
| [{"identifier": "A", "content": "$$ - {3 \\over {16}}$$"}, {"identifier": "B", "content": "$$ - {1 \\over 8}$$"}, {"identifier": "C", "content": "$${1 \\over 8}$$"}, {"identifier": "D", "content": "$${3 \\over {16}}$$"}] | ["C"] | null | Number of ways 3 consecutive heads can appers
<br><br>(1) HHHT_
<br><br>(2) _THHH
<br><br>(3) THHHT
<br><br>$$ \therefore $$ Probablity of getting 3 consecutive heads
<br><br>= $${2 \over {32}}$$ + $${2 \over {32}}$$ + $${1 \over {32}}$$ = $${5 \over {32}}$$
<br><br>Number of ways 4 consecutive heads can appers
<br><br>(1) HHHHT
<br><br>(2) THHHH
<br><br>$$ \therefore $$ Probablity of getting 4 consecutive heads
<br><br>= $${1 \over {32}}$$ + $${1 \over {32}}$$ = $${2 \over {32}}$$
<br><br>Number of ways 5 consecutive heads can appers
<br><br>(1) HHHHH
<br><br>$$ \therefore $$ Probablity of getting 5 consecutive heads
<br><br>= $${1 \over {32}}$$
<br><br>Now Probablity of getting 0, 1, and 2 consecutive heads
<br><br>= 1 - $$\left( {{5 \over {32}} + {2 \over {32}} + {1 \over {32}}} \right)$$ = $${{{24} \over {32}}}$$
<br><br>Now, Expectation
<br><br>= (-1) $$ \times $$ $${{{24} \over {32}}}$$ + 3 $$ \times $$ $${{{5} \over {32}}}$$ + 4 $$ \times $$ $${{{2} \over {32}}}$$ + 5 $$ \times $$ $${{{1} \over {32}}}$$
<br><br>= $${1 \over 8}$$ | mcq | jee-main-2020-online-7th-january-morning-slot |
F1meBFGPVyGyNy1nKL7k9k2k5kh288t | maths | probability | probability-distribution-of-a-random-variable | A random variable X has the following
probability distribution :<br/><br/>
<style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;width:100%}
.tg td{font-family:Arial, sans-serif;font-size:14px;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black;}
.tg th{font-family:Arial, sans-serif;font-size:14px;font-weight:normal;padding:10px 5px;border-style:solid;border-width:1px;overflow:hidden;word-break:normal;border-color:black;}
.tg .tg-baqh{text-align:center;vertical-align:top}
.tg .tg-wa1i{font-weight:bold;text-align:center;vertical-align:middle}
.tg .tg-nrix{text-align:center;vertical-align:middle}
</style>
<table class="tg">
<tbody><tr>
<th class="tg-wa1i">X:</th>
<th class="tg-nrix">1</th>
<th class="tg-baqh">2</th>
<th class="tg-baqh">3</th>
<th class="tg-baqh">4</th>
<th class="tg-baqh">5</th>
</tr>
<tr>
<td class="tg-wa1i">P(X):</td>
<td class="tg-nrix">K<sup>2</sup></td>
<td class="tg-baqh">2K</td>
<td class="tg-baqh">K</td>
<td class="tg-baqh">2K</td>
<td class="tg-baqh">5K<sup>2</sup></td>
</tr>
</tbody></table><br/>
Then P(X > 2) is equal to : | [{"identifier": "A", "content": "$${1 \\over {6}}$$"}, {"identifier": "B", "content": "$${7 \\over {12}}$$"}, {"identifier": "C", "content": "$${1 \\over {36}}$$"}, {"identifier": "D", "content": "$${23 \\over {36}}$$"}] | ["D"] | null | $$\sum\limits_{i = 1}^5 {P(X)} $$ = 1
<br><br>$$ \Rightarrow $$ K<sup>2</sup>
+ 2K + K + 2K + 5K<sup>2</sup>
= 1
<br><br>$$ \Rightarrow $$ 6K<sup>2</sup>
+ 5K – 1 = 0
<br><br>$$ \Rightarrow $$ (6K - 1)(k + 1) = 0
<br><br>$$ \Rightarrow $$ K = $${1 \over 6}$$ and K = -1(rejected)
<br><br>$$ \therefore $$ P(X $$ > $$ 2)
<br><br>= K + 2K + 5K<sup>2</sup>
<br><br>= $${1 \over 6} + {2 \over 6} + {5 \over {36}}$$
<br><br>= $${{23} \over {36}}$$ | mcq | jee-main-2020-online-9th-january-evening-slot |
1krzncxyp | maths | probability | probability-distribution-of-a-random-variable | Let X be a random variable such that the probability function of a distribution is given by $$P(X = 0) = {1 \over 2},P(X = j) = {1 \over {{3^j}}}(j = 1,2,3,...,\infty )$$. Then the mean of the distribution and P(X is positive and even) respectively are : | [{"identifier": "A", "content": "$${3 \\over 8}$$ and $${1 \\over 8}$$"}, {"identifier": "B", "content": "$${3 \\over 4}$$ and $${1 \\over 8}$$"}, {"identifier": "C", "content": "$${3 \\over 4}$$ and $${1 \\over 9}$$"}, {"identifier": "D", "content": "$${3 \\over 4}$$ and $${1 \\over 16}$$"}] | ["B"] | null | Mean = $$\sum {{X_i}{P_i}} = \sum\limits_{r = 0}^\infty {r.{1 \over {{3^r}}} = {3 \over 4}} $$<br><br>P(X is even) $$ = {1 \over {{3^2}}} + {1 \over {{3^4}}} + ...\infty $$<br><br>$$ = {{{1 \over 9}} \over {1 - {1 \over 9}}} = {{1/9} \over {8/9}} = {1 \over 8}$$ | mcq | jee-main-2021-online-25th-july-evening-shift |
1ktgohohi | maths | probability | probability-distribution-of-a-random-variable | The probability distribution of random variable X is given by :<br/><br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-0lax{text-align:left;vertical-align:top}
</style>
<table class="tg">
<thead>
<tr>
<th class="tg-0lax">X</th>
<th class="tg-0lax">1</th>
<th class="tg-0lax">2</th>
<th class="tg-0lax">3</th>
<th class="tg-0lax">4</th>
<th class="tg-0lax">5</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0lax">P(X)</td>
<td class="tg-0lax">K</td>
<td class="tg-0lax">2K</td>
<td class="tg-0lax">2K</td>
<td class="tg-0lax">3K</td>
<td class="tg-0lax">K</td>
</tr>
</tbody>
</table><br/><br/>Let p = P(1 < X < 4 | X < 3). If 5p = $$\lambda$$K, then $$\lambda$$ equal to ___________. | [] | null | 30 | $$\sum {P(X) = 1 \Rightarrow k + 2k + 3} k + k = 1$$<br><br>$$ \Rightarrow k = {1 \over 9}$$<br><br>Now, $$p = P\left( {{{kx < 4} \over {X < 3}}} \right) = {{P(X = 2)} \over {P(X < 3)}} = {{{{2k} \over {9k}}} \over {{k \over {9k}} + {{2k} \over {9k}}}} = {2 \over 3}$$<br><br>$$ \Rightarrow p = {2 \over 3}$$<br><br>Now, $$5p = \lambda k$$<br><br>$$ \Rightarrow (5)\left( {{2 \over 3}} \right) = \lambda (1/9)$$<br><br>$$ \Rightarrow \lambda = 30$$ | integer | jee-main-2021-online-27th-august-evening-shift |
1ktoapaky | maths | probability | probability-distribution-of-a-random-variable | Let X be a random variable with distribution.<br/><br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg">
<thead>
<tr>
<th class="tg-baqh">x</th>
<th class="tg-baqh">$$ - $$2</th>
<th class="tg-baqh">$$ - $$1</th>
<th class="tg-baqh">3</th>
<th class="tg-baqh">4</th>
<th class="tg-baqh">6</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">P(X = x)</td>
<td class="tg-baqh">$${1 \over 5}$$</td>
<td class="tg-baqh">a</td>
<td class="tg-baqh">$${1 \over 3}$$</td>
<td class="tg-baqh">$${1 \over 5}$$</td>
<td class="tg-baqh">b</td>
</tr>
</tbody>
</table><br/><br/>If the mean of X is 2.3 and variance of X is $$\sigma$$<sup>2</sup>, then 100 $$\sigma$$<sup>2</sup> is equal to : | [] | null | 781 | <table class="tg">
<thead>
<tr>
<th class="tg-baqh">x</th>
<th class="tg-baqh">$$ - $$2</th>
<th class="tg-baqh">$$ - $$1</th>
<th class="tg-baqh">3</th>
<th class="tg-baqh">4</th>
<th class="tg-baqh">6</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">P(X = x)</td>
<td class="tg-baqh">$${1 \over 5}$$</td>
<td class="tg-baqh">a</td>
<td class="tg-baqh">$${1 \over 3}$$</td>
<td class="tg-baqh">$${1 \over 5}$$</td>
<td class="tg-baqh">b</td>
</tr>
</tbody>
</table><br><br>$$\overline X $$ = 2.3<br><br>$$-$$a + 6b = $${9 \over {10}}$$ ..... (1)<br><br>$$\sum {{P_i} = {1 \over 5} + a + {1 \over 3} + {1 \over 5} + b = 1} $$<br><br>$$a + b = {4 \over {15}}$$ .... (2)<br><br>From equation (1) and (2)<br><br>$$a = {1 \over {10}},b = {1 \over 6}$$<br><br>$${\sigma ^2} = \sum {{p_i}x_i^2 - {{(\overline X )}^2}} $$<br><br>$${1 \over 5}(4) + a(1) + {1 \over 3}(9) + {1 \over 5}(16) + b(36) - {(2.3)^2}$$<br><br>$$ = {4 \over 5} + a + 3 + {{16} \over 5} + 36b - {(2.3)^2}$$<br><br>$$ = 4 + a + 3 + 36b - {(2.3)^2}$$<br><br>$$ = 7 + a + 36b - {(2.3)^2}$$<br><br>$$ = 7 + {1 \over {10}} + 6 - {(2.3)^2}$$<br><br>$$ = 13 + {1 \over {10}} - {\left( {{{23} \over {10}}} \right)^2}$$<br><br>$$ = {{131} \over {10}} - {\left( {{{23} \over {10}}} \right)^2}$$<br><br>$$ = {{1310 - {{(23)}^2}} \over {100}}$$<br><br>$$ = {{1310 - 529} \over {100}}$$<br><br>$${\sigma ^2} = {{781} \over {100}}$$<br><br>$$100{\sigma ^2} = 781$$ | integer | jee-main-2021-online-1st-september-evening-shift |
1l59kq3cf | maths | probability | probability-distribution-of-a-random-variable | <p>A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its faces and the probability of getting a face with mark n is $${1 \over n}$$. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is :</p> | [{"identifier": "A", "content": "$${7 \\over {{2^{11}}}}$$"}, {"identifier": "B", "content": "$${7 \\over {{2^{12}}}}$$"}, {"identifier": "C", "content": "$${3 \\over {{2^{10}}}}$$"}, {"identifier": "D", "content": "$${{13} \\over {{2^{12}}}}$$"}] | ["D"] | null | <p>There are only two ways to get sum 48, which are (32, 8, 8) and (16, 16, 16)</p>
<p>So, required probability</p>
<p>$$ = 3\left( {{2 \over {32}}\,.\,{1 \over 8}\,.\,{1 \over 8}} \right) + \left( {{1 \over {16}}\,.\,{1 \over {16}}\,.\,{1 \over {16}}} \right)$$</p>
<p>$$ = {3 \over {{2^{10}}}} + {1 \over {{2^{12}}}}$$</p>
<p>$$ = {{13} \over {{2^{12}}}}$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5babc18 | maths | probability | probability-distribution-of-a-random-variable | <p>A random variable X has the following probability distribution :</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 374px">
<colgroup>
<col style="width: 96px"/>
<col style="width: 53px"/>
<col style="width: 54px"/>
<col style="width: 55px"/>
<col style="width: 57px"/>
<col style="width: 59px"/>
</colgroup>
<thead>
<tr>
<th class="tg-c3ow">X</th>
<th class="tg-c3ow">0</th>
<th class="tg-c3ow">1</th>
<th class="tg-c3ow">2</th>
<th class="tg-c3ow">3</th>
<th class="tg-c3ow">4</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-c3ow">P(X)</td>
<td class="tg-c3ow">k</td>
<td class="tg-c3ow">2k</td>
<td class="tg-c3ow">4k</td>
<td class="tg-c3ow">6k</td>
<td class="tg-c3ow">8k</td>
</tr>
</tbody>
</table></p>
<p>The value of P(1 < X < 4 | X $$\le$$ 2) is equal to :</p> | [{"identifier": "A", "content": "$${4 \\over 7}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${3 \\over 7}$$"}, {"identifier": "D", "content": "$${4 \\over 5}$$"}] | ["A"] | null | <p>$$\because$$ x is a random variable</p>
<p>$$\therefore$$ $$k + 2k + 4k + 6k + 8k = 1$$</p>
<p>$$\therefore$$ $$k = {1 \over {21}}$$</p>
<p>Now, $$P(1 < x < 4\,|\,x \le 2) = {{4k} \over {7k}} = {4 \over 7}$$</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
1l5w1mu8i | maths | probability | probability-distribution-of-a-random-variable | <p>The probability distribution of X is :</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 451px">
<colgroup>
<col style="width: 55px"/>
<col style="width: 89px"/>
<col style="width: 94px"/>
<col style="width: 104px"/>
<col style="width: 109px"/>
</colgroup>
<thead>
<tr>
<th class="tg-baqh">X</th>
<th class="tg-baqh">0</th>
<th class="tg-baqh">1</th>
<th class="tg-baqh">2</th>
<th class="tg-baqh">3</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">P(X)</td>
<td class="tg-baqh">$${{1 - d} \over 4}$$</td>
<td class="tg-baqh">$${{1 + 2d} \over 4}$$</td>
<td class="tg-baqh">$${{1 - 4d} \over 4}$$</td>
<td class="tg-baqh">$${{1 + 3d} \over 4}$$</td>
</tr>
</tbody>
</table></p>
<p>For the minimum possible value of d, sixty times the mean of X is equal to _______________.</p> | [] | null | 75 | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 298px">
<colgroup>
<col style="width: 48px">
<col style="width: 59px">
<col style="width: 60px">
<col style="width: 63px">
<col style="width: 68px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">X</th>
<th class="tg-baqh">0</th>
<th class="tg-baqh">1</th>
<th class="tg-baqh">2</th>
<th class="tg-baqh">3</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">$$P(x)$$</td>
<td class="tg-baqh">$${{1 - d} \over 4}$$</td>
<td class="tg-baqh">$${{1 + 2d} \over 4}$$</td>
<td class="tg-baqh">$${{1 - 4d} \over 4}$$</td>
<td class="tg-baqh">$${{1 + 3d} \over 4}$$</td>
</tr>
</tbody>
</table></p>
<p>We know, $$0 \le P(x) \le 1$$</p>
<p>$$\therefore$$ $$0 \le {{1 - d} \over 4} \le 1$$</p>
<p>$$ \Rightarrow 0 \le 1 - d \le 4$$</p>
<p>$$ \Rightarrow - 1 \le - d \le 3$$</p>
<p>$$ \Rightarrow 1 \ge d \ge - 3$$</p>
<p>Also,</p>
<p>$$0 \le {{1 + 2d} \over 4} \le 1$$</p>
<p>$$ \Rightarrow 0 \le 1 + 2d \le 4$$</p>
<p>$$ \Rightarrow - 1 \le 2d \le 3$$</p>
<p>$$ \Rightarrow - {1 \over 2} \le d \le {3 \over 2}$$</p>
<p>Also,</p>
<p>$$0 \le {{1 - 4d} \over 4} \le 1$$</p>
<p>$$ \Rightarrow 0 \le 1 - 4d \le 4$$</p>
<p>$$ \Rightarrow - 1 \le - 4d \le 3$$</p>
<p>$$ \Rightarrow 1 \ge 4d \ge - 3$$</p>
<p>$$ \Rightarrow {1 \over 4} \ge d \ge - {3 \over 4}$$</p>
<p>And,</p>
<p>$$0 \le {{1 + 3d} \over 4} \le 1$$</p>
<p>$$ \Rightarrow 0 \le 1 + 3d \le 4$$</p>
<p>$$ \Rightarrow - 1 \le 3d \le 3$$</p>
<p>$$ \Rightarrow - {1 \over 3} \le d \le 1$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l6dn1sg2/70ac6607-e35b-496d-816f-33268be95f9a/3e6d8930-132f-11ed-941a-4dd6502f33e3/file-1l6dn1sg3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l6dn1sg2/70ac6607-e35b-496d-816f-33268be95f9a/3e6d8930-132f-11ed-941a-4dd6502f33e3/file-1l6dn1sg3.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Probability Question 69 English Explanation"></p>
<p>Common range is $$ = - {1 \over 3}$$ to $${1 \over 4}$$</p>
<p>$$\therefore$$ $$d\, \in \left[ { - {1 \over 3},{1 \over 4}} \right]$$</p>
<p>$$\therefore$$ Minimum value of $$d = - {1 \over 3}$$</p>
<p>We know, mean</p>
<p>$$E(x) = \sum {x\,.\,P(x)} $$</p>
<p>$$ = 0 \times {{1 - d} \over 4} + 1 \times {{1 + 2d} \over 4} + 2 \times {{1 - 4d} \over 4} + 3 \times {{1 + 3d} \over 4}$$</p>
<p>$$ = {{1 + 2d + 2 - 8d + 3 + 9d} \over 4}$$</p>
<p>$$ = {{6 + 3d} \over 4}$$</p>
<p>For $$d = - {1 \over 3}$$, $$E(x) = {{6 + 3 \times - {1 \over 3}} \over 4} = {5 \over 4}$$</p>
<p>$$\therefore$$ $$60E(x) = 60 \times {5 \over 4} = 75$$</p> | integer | jee-main-2022-online-30th-june-morning-shift |
1l6kkv5he | maths | probability | probability-distribution-of-a-random-variable | <p>A six faced die is biased such that</p>
<p>$$3 \times \mathrm{P}($$a prime number$$)\,=6 \times \mathrm{P}($$a composite number$$)\,=2 \times \mathrm{P}(1)$$.</p>
<p>Let X be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of X is :</p> | [{"identifier": "A", "content": "$$\\frac{3}{11}$$"}, {"identifier": "B", "content": "$$\\frac{5}{11}$$"}, {"identifier": "C", "content": "$$\\frac{7}{11}$$"}, {"identifier": "D", "content": "$$\\frac{8}{11}$$"}] | ["D"] | null | <p>Let P(a prime number) = $$\alpha$$</p>
<p>P(a composite number) = $$\beta$$</p>
<p>and P(1) = $$\gamma$$</p>
<p>$$\because$$ $$3\alpha = 6\beta = 2\gamma = k$$ (say)</p>
<p>and $$3\alpha + 2\beta + \gamma = 1$$</p>
<p>$$ \Rightarrow k + {k \over 3} + {k \over 2} = 1 \Rightarrow k = {6 \over {11}}$$</p>
<p>Mean = np where n = 2</p>
<p>and p = probability of getting perfect square</p>
<p>$$ = P(1) + P(4) = {k \over 2} + {k \over 6} = {4 \over {11}}$$</p>
<p>So, mean $$ = 2\,.\,\left( {{4 \over {11}}} \right) = {8 \over {11}}$$</p> | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6npi0xr | maths | probability | probability-distribution-of-a-random-variable | <p>A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let $$\mathrm{X}$$ be the number of white balls, among the drawn balls. If $$\sigma^{2}$$ is the variance of $$\mathrm{X}$$, then $$100 \sigma^{2}$$ is equal to ________.</p> | [] | null | 57 | <p>$$X = $$ Number of white ball drawn</p>
<p>$$P(X = 0) = {{{}^6{C_3}} \over {{}^{10}{C_3}}} = {1 \over 6}$$</p>
<p>$$P(X = 1) = {{{}^6{C_2} \times {}^4{C_1}} \over {{}^{10}{C_3}}} = {1 \over 2},$$</p>
<p>$$P(X = 2) = {{{}^6{C_1} \times {}^4{C_2}} \over {{}^{10}{C_3}}} = {3 \over {10}}$$</p>
<p>and $$P(X = 3) = {{{}^6{C_0} \times {}^4{C_3}} \over {{}^{10}{C_3}}} = {1 \over {30}}$$</p>
<p>Variance $$ = {\sigma ^2} = \sum {{P_i}X_i^2 - {{\left( {\sum {{P_i}{X_i}} } \right)}^2}} $$</p>
<p>$${\sigma ^2} = {1 \over 2} + {{12} \over {10}} + {3 \over {10}} - {\left( {{1 \over 2} + {6 \over {10}} + {1 \over {10}}} \right)^2}$$</p>
<p>$$ = {{56} \over {100}}$$</p>
<p>$$100{\sigma ^2} = 56.$$</p> | integer | jee-main-2022-online-28th-july-evening-shift |
1ldsey88d | maths | probability | probability-distribution-of-a-random-variable | <p>Let $$\mathrm{S} = \{ {w_1},{w_2},......\} $$ be the sample space associated to a random experiment. Let $$P({w_n}) = {{P({w_{n - 1}})} \over 2},n \ge 2$$. Let $$A = \{ 2k + 3l:k,l \in N\} $$ and $$B = \{ {w_n}:n \in A\} $$. Then P(B) is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{3}{32}$$"}, {"identifier": "B", "content": "$$\\frac{1}{32}$$"}, {"identifier": "C", "content": "$$\\frac{1}{16}$$"}, {"identifier": "D", "content": "$$\\frac{3}{64}$$"}] | ["D"] | null | <p>$$P({w_1}) + {{P({w_1})} \over 2} + {{P({w_1})} \over {{2^2}}}\, + \,..... = 1$$</p>
<p>$$\therefore$$ $$P({w_1}) = {1 \over 2}$$</p>
<p>Hence, $$P({w_n}) = {1 \over {{2^n}}}$$</p>
<p>Every number except 1, 2, 3, 4, 6 is representable in the form</p>
<p>$$2k + 3l$$ where $$k,l \in N$$.</p>
<p>$$\therefore$$ $$P(B) = 1 - P({w_1}) - P({w_2}) - P({w_3}) - P({w_4}) - P({w_6})$$</p>
<p>$$ = {3 \over {64}}$$</p> | mcq | jee-main-2023-online-29th-january-evening-shift |
1lgpxw3jh | maths | probability | probability-distribution-of-a-random-variable | <p> A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If $$\mathrm{X}$$ denotes the number of tosses of the coin, then the mean of $$\mathrm{X}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{81}{64}$$"}, {"identifier": "B", "content": "$$\\frac{37}{16}$$"}, {"identifier": "C", "content": "$$\\frac{21}{16}$$"}, {"identifier": "D", "content": "$$\\frac{15}{16}$$"}] | ["C"] | null | The given probabilities for getting a head (H) and a tail (T) are as follows:
<br/><br/>$$ P(H) = \frac{3}{4}, \quad P(T) = \frac{1}{4} $$
<br/><br/>The random variable X can take the values 1, 2, or 3. These correspond to the following events:
<br/><br/>- X = 1 : A head is obtained on the first toss. This happens with probability $P(H) = \frac{3}{4}$.
<br/><br/>- X = 2 : A tail is obtained on the first toss and a head on the second. This happens with probability $P(T)P(H) = \frac{1}{4} \times \frac{3}{4}$.
<br/><br/>- X = 3 : Either two tails and then a head are obtained, or three tails are obtained.
<br/> This happens with probability $P(T)P(T)P(H) + P(T)P(T)P(T) = \left(\frac{1}{4}\right)^2 \times \frac{3}{4} + \left(\frac{1}{4}\right)^3$.
<br/><br/>Now, we calculate the mean (expected value) of X:
<br/><br/>$$
\begin{aligned}
E(X) & = 1 \cdot P(X = 1) + 2 \cdot P(X = 2) + 3 \cdot P(X = 3) \\\\
& = 1 \cdot \frac{3}{4} + 2 \cdot \left(\frac{1}{4} \times \frac{3}{4}\right) + 3 \cdot \left[\left(\frac{1}{4}\right)^2 \times \frac{3}{4} + \left(\frac{1}{4}\right)^3\right] \\\\
& = \frac{3}{4} + \frac{3}{8} + 3 \cdot \left(\frac{1}{64} + \frac{3}{64}\right) \\\\
& = \frac{3}{4} + \frac{3}{8} + \frac{3}{16} \\\\
& = 3 \cdot \left(\frac{7}{16}\right) \\\\
& = \frac{21}{16}.
\end{aligned}
$$ | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgre62ht | maths | probability | probability-distribution-of-a-random-variable | <p>Two dice A and B are rolled. Let the numbers obtained on A and B be $$\alpha$$ and $$\beta$$ respectively. If the variance of $$\alpha-\beta$$ is $$\frac{p}{q}$$, where $$p$$ and $$q$$ are co-prime, then the sum of the positive divisors of $$p$$ is equal to :</p> | [{"identifier": "A", "content": "48"}, {"identifier": "B", "content": "31"}, {"identifier": "C", "content": "72"}, {"identifier": "D", "content": "36"}] | ["A"] | null | $$
\begin{array}{|c|l|c|}
\hline \alpha-\beta & {\text { Case }} & \mathbf{P} \\
\hline 5 & (6,1) & 1 / 36 \\
\hline 4 & (6,2)(5,1) & 2 / 36 \\
\hline 3 & (6,3)(5,2)(4,1) & 3 / 36 \\
\hline 2 & (6,4)(5,3)(4,3)(3,1) & 4 / 36 \\
\hline 1 & (6,5)(5,4)(4,3)(3,2)(2,1) & 5 / 36 \\
\hline 0 & (6,6)(5,5) \ldots \ldots(1,1) & 6 / 36 \\
\hline-1 & ----- & 5 / 36 \\
\hline-2 & -----& 4 / 36 \\
\hline-3 & ----- & 3 / 36 \\
\hline-4 & (2,6)(1,5) & 2 / 36 \\
\hline-5 & (1,6) & 1 / 36 \\
\hline
\end{array}
$$
<br/><br/>
$$E[X^2] = \sum\limits_{i=-5}^{5} (x_i)^2 P(x_i)$$
<br/><br/>Substituting the values from table :
<br/><br/>$$E[X^2] = 2\left[\frac{25}{36}+\frac{32}{36}+\frac{27}{36}+\frac{16}{36}+\frac{5}{36}\right] = \frac{105}{18} = \frac{35}{6}$$
<br/><br/>Next, we calculate the expected value of the differences. The expected value is calculated as the sum of the products of each outcome and its corresponding probability. Given that the table is symmetric around 0, the expected value is 0.
<br/><br/>$$E[X] = \sum\limits_{i=-5}^{5} x_i P(x_i) = 0$$
<br/><br/>Now, we can calculate the variance, which is the expected value of the squared differences minus the square of the expected value of the differences :
<br/><br/>$$Var[X] = E[X^2] - (E[X])^2 = \frac{35}{6} - 0^2 = \frac{35}{6}$$
<br/><br/>Here, $$p = 35$$ and $$q = 6$$, and they are co-prime.
<br/><br/>The positive divisors of 35 are 1, 5, 7, and 35. The sum of these divisors is $$1 + 5 + 7 + 35 = 48$$. | mcq | jee-main-2023-online-12th-april-morning-shift |
1lgrgkm6x | maths | probability | probability-distribution-of-a-random-variable | <p>A fair $$n(n > 1)$$ faces die is rolled repeatedly until a number less than $$n$$ appears. If the mean of the number of tosses required is $$\frac{n}{9}$$, then $$n$$ is equal to ____________.</p> | [] | null | 10 | In this case, the success is defined as getting a number less than $n$ when rolling an $n$-sided die. The probability of success, $p$, in each roll is then $(n-1)/n$, and the probability of failure, $q = 1 - p$, is $1/n$.
<br/><br/>$$
\text { Mean }=\sum\limits_{i=1}^{\infty} p_i x_i=1 \cdot \frac{n-1}{n}+\frac{2}{n} \cdot\left(\frac{n-1}{n}\right)+\frac{3}{n^2}\left(\frac{n-1}{n}\right)+\ldots
$$
<br/><br/>$$
\frac{n}{9}=\left(1-\frac{1}{n}\right) S
$$ ......(1)
<br/><br/>where
<br/><br/>$$
\begin{aligned}
& S=1+\frac{2}{n}+\frac{3}{n^2}+\frac{4}{n^3}+\ldots \\\\
& \frac{1}{n} S=\frac{1}{n}+\frac{2}{n^2}+\frac{3}{n^3}+\ldots \\\\
& ----------------\\\\
& \left(1-\frac{1}{n}\right) S=1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\ldots
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \left(1-\frac{1}{n}\right) S=\frac{1}{1-\frac{1}{n}} \\\\
& \Rightarrow \frac{n}{9}=\left(1-\frac{1}{n}\right) \times \frac{1}{\left(1-\frac{1}{n}\right)^2}=\frac{n}{n-1}
\end{aligned}
$$
<br/><br/>$$ \Rightarrow $$ n = 10 | integer | jee-main-2023-online-12th-april-morning-shift |
1lgsw8nze | maths | probability | probability-distribution-of-a-random-variable | <p>Let the probability of getting head for a biased coin be $$\frac{1}{4}$$. It is tossed repeatedly until a head appears. Let $$\mathrm{N}$$ be the number of tosses required. If the probability that the equation $$64 \mathrm{x}^{2}+5 \mathrm{Nx}+1=0$$ has no real root is $$\frac{\mathrm{p}}{\mathrm{q}}$$, where $$\mathrm{p}$$ and $$\mathrm{q}$$ are coprime, then $$q-p$$ is equal to ________.</p> | [] | null | 27 | We have the quadratic equation $64x^2 + 5Nx + 1 = 0$. For it to have no real roots, the discriminant ($b^2 - 4ac$) should be less than 0. Here, $a = 64$, $b = 5N$, and $c = 1$.
<br/><br/>This gives us :
<br/><br/>$(5N)^2 - 4\times64\times1 < 0$
<br/><br/>$\Rightarrow 25N^2 < 256$
<br/><br/>$\Rightarrow N^2 < \frac{256}{25}$
<br/><br/>$\Rightarrow N < \sqrt{\frac{256}{25}} = \frac{16}{5}$
<br/><br/>Since $N$ must be an integer (as it represents the number of tosses), the possible values of $N$ are 1, 2, or 3.
<br/><br/>The probability of getting the first head on the $n$-th toss (given the probability of getting a head is $1/4$) is given by the geometric distribution formula, $(1 - p)^{n-1}\times p$.
<br/><br/>So, the probability for our specific values of $N$ is:
<br/><br/>$P(N=1) = (1 - 1/4)^{1-1}\times(1/4) = 1/4$
<br/><br/>$P(N=2) = (1 - 1/4)^{2-1}\times(1/4) = 3/4 \times 1/4 = 3/16$
<br/><br/>$P(N=3) = (1 - 1/4)^{3-1}\times(1/4) = (3/4)^2 \times 1/4 = 9/64$
<br/><br/>Therefore, the total probability (p/q) is :
<br/><br/>$p/q = P(N=1) + P(N=2) + P(N=3)$
<br/><br/>$= 1/4 + 3/16 + 9/64$
<br/><br/>$= 16/64 + 12/64 + 9/64$
<br/><br/>$= 37/64$
<br/><br/>So, $p = 37$, $q = 64$ and $q-p = 64 - 37 = 27$.
<br/><br/>Therefore, $q-p$ is equal to $27$.
| integer | jee-main-2023-online-11th-april-evening-shift |
1lgyl9ba2 | maths | probability | probability-distribution-of-a-random-variable | <p>If the probability that the random variable $$\mathrm{X}$$ takes values $$x$$ is given by $$\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1) 3^{-x}, x=0,1,2,3, \ldots$$, where $$\mathrm{k}$$ is a constant, then $$\mathrm{P}(\mathrm{X} \geq 2)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{7}{18}$$"}, {"identifier": "B", "content": "$$\\frac{20}{27}$$"}, {"identifier": "C", "content": "$$\\frac{7}{27}$$"}, {"identifier": "D", "content": "$$\\frac{11}{18}$$"}] | ["C"] | null | As, we know that sum of all the probabilities $=1$
<br/><br/>$$
\begin{aligned}
& \text { So, } \sum_{x=1}^{\infty} \mathrm{P}(\mathrm{X}=x)=1 \\\\
& \Rightarrow k\left[1+2 \cdot 3^{-1}+3 \cdot 3^{-2}+\ldots . \infty\right]=1
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Let } S=1+\frac{2}{3}+\frac{3}{3^2}+\ldots .+\infty \\\\
& \Rightarrow \frac{S}{3}=0+\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\ldots .+\infty
\end{aligned}
$$
<br/><br/>On subtracting, we get
<br/><br/>$$
\begin{aligned}
& \frac{2 S}{3}=1+\frac{1}{3}+\frac{1}{3^2}+\ldots .+\infty \\\\
& \Rightarrow \frac{2 S}{3}=\frac{1}{1-\frac{1}{3}}=\frac{1}{\frac{2}{3}} \\\\
& \Rightarrow \frac{2 S}{3}=\frac{3}{2} \\\\
& \Rightarrow S=\frac{9}{4}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { So, } k \times \frac{9}{4}=1 \Rightarrow k=\frac{4}{9} \\\\
& \text { Now, } \mathrm{P}(\mathrm{X} \geq 2)=1-\mathrm{P}(\mathrm{X}<2) \\\\
& =1-\mathrm{P}(\mathrm{X}=0)-\mathrm{P}(\mathrm{X}=1) \\\\
& =1-\frac{4}{9}(1)-\frac{4}{9} \times \frac{2}{3} \\\\
& =1-\frac{4}{9}-\frac{8}{27}=\frac{27-12-8}{27}=\frac{7}{27}
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
jaoe38c1lse553je | maths | probability | probability-distribution-of-a-random-variable | <p>Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable $$x$$ to be the number of rotten apples in a draw of two apples, the variance of $$x$$ is</p> | [{"identifier": "A", "content": "$$\\frac{57}{153}$$\n"}, {"identifier": "B", "content": "$$\\frac{40}{153}$$\n"}, {"identifier": "C", "content": "$$\\frac{37}{153}$$\n"}, {"identifier": "D", "content": "$$\\frac{47}{153}$$"}] | ["B"] | null | <p>3 bad apples, 15 good apples.</p>
<p>Let $$\mathrm{X}$$ be no of bad apples</p>
<p>$$\begin{aligned}
& \text { Then } \mathrm{P}(\mathrm{X}=0)=\frac{{ }^{15} \mathrm{C}_2}{{ }^{18} \mathrm{C}_2}=\frac{105}{153} \\
& \mathrm{P}(\mathrm{X}=1)=\frac{{ }^3 \mathrm{C}_1 \times{ }^{15} \mathrm{C}_1}{{ }^{18} \mathrm{C}_2}=\frac{45}{153} \\
& \mathrm{P}(\mathrm{X}=2)=\frac{{ }^3 \mathrm{C}_2}{{ }^{18} \mathrm{C}_2}=\frac{3}{153} \\
& \mathrm{E}(\mathrm{X})=0 \times \frac{105}{153}+1 \times \frac{45}{153}+2 \times \frac{3}{153}=\frac{51}{153} \\
& =\frac{1}{3} \\
& \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-(\mathrm{E}(\mathrm{X}))^2 \\
& =0 \times \frac{105}{153}+1 \times \frac{45}{153}+4 \times \frac{3}{153}-\left(\frac{1}{3}\right)^2 \\
& =\frac{57}{153}-\frac{1}{9}=\frac{40}{153}
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
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