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k3m0N9G0xSwL4wln7Cjgy2xukewtoyf9 | maths | permutations-and-combinations | number-of-permutations | If the letters of the word 'MOTHER' be permuted
and all the words so formed (with or without
meaning) be listed as in a dictionary, then the
position of the word 'MOTHER' is ______. | [] | null | 309 | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267803/exam_images/xbqcgksmphluv6sci9ir.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265109/exam_images/kdpljpefm6kve0qcetlt.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263903/exam_images/uepdoqg2xsu0wvhakoao.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264002/exam_images/e5meyec5crm1f4kzzfyx.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267595/exam_images/iqfugye1vg9tyuyciht6.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 2nd September Morning Slot Mathematics - Permutations and Combinations Question 124 English Explanation 1"></picture>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265950/exam_images/ckctc7z2vafuf0zlqhvf.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266728/exam_images/ebc7kqqjwjpkcofhvzew.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266454/exam_images/bupbnwexyxszyknef9g0.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263379/exam_images/np15dsek1lvmwsdns5g2.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265517/exam_images/k0f0rfyuhhvanqmcnfpu.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 2nd September Morning Slot Mathematics - Permutations and Combinations Question 124 English Explanation 2"></picture>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267076/exam_images/rvbxh42ujgpzuetksh4l.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263661/exam_images/mwobgckuzwokxsixbbmm.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263276/exam_images/duvsdkpymjbiwycmpd6z.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264367/exam_images/ajaugzxelk4kaywllcnh.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267333/exam_images/qyre0nrbhsgwy1cp8xgs.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265490/exam_images/ustv4otfdmz6sfaiqxez.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 2nd September Morning Slot Mathematics - Permutations and Combinations Question 124 English Explanation 3"></picture>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264631/exam_images/cqwosqpbqtnjalrosd1c.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267442/exam_images/bxfsjyajr5axhdg5rkqr.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266069/exam_images/yu6l4nbrqaxks8sqjoyq.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266879/exam_images/o6n3r8s38dmlipbdnapm.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263614/exam_images/ey8ucpg1jd1dowuj5uf6.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 2nd September Morning Slot Mathematics - Permutations and Combinations Question 124 English Explanation 4"></picture> | integer | jee-main-2020-online-2nd-september-morning-slot |
KcZ71werwgGvco0DDojgy2xukf0wa8ut | maths | permutations-and-combinations | number-of-permutations | The value of (2.<sup>1</sup>P<sub>0</sub>
β 3.<sup>2</sup>P<sub>1</sub> + 4.<sup>3</sup>P<sub>2</sub> .... up to
51<sup>th</sup> term)<br/> + (1! β 2! + 3! β ..... up to 51<sup>th</sup> term)
is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "1 + (51)!"}, {"identifier": "C", "content": "1 \u2013 51(51)!"}, {"identifier": "D", "content": "1 + (52)!"}] | ["D"] | null | (2.<sup>1</sup>P<sub>0</sub>
β 3.<sup>2</sup>P<sub>1</sub> + 4.<sup>3</sup>P<sub>2</sub> .... up to
51<sup>th</sup> term)<br> + (1! β 2! + 3! β ..... up to 51<sup>th</sup> term)
<br><br>= ( 2.1! - 3.2! + 4.3! - ....+ 52.51!)
<br>+ ( 1! β 2! + 3! β ..... + (51)! )
<br><br>= (2! - 3! + 4! .....+ 52! )
<br>+ ( 1! - 2! + 3! - 4! + .....+ 51! )
<br><br>= 1 + (52)! | mcq | jee-main-2020-online-3rd-september-morning-slot |
t2DoGkNqDTClpiz752jgy2xukf49ludb | maths | permutations-and-combinations | number-of-permutations | The total number of 3-digit numbers, whose
sum of digits is 10, is __________. | [] | null | 54 | Let xyz is 3 digits number.<br><br>Given that sum of digits = 10<br><br>$$ \therefore $$ x + y + z = 10 ......(1)<br><br>Also x can't be 0 as if x = 0 then it will become 2 digits number.<br><br>So, x $$ \ge $$ 1, y $$ \ge $$ 0, z $$ \ge $$ 0<br><br>As x $$ \ge $$ 1<br><br>$$ \Rightarrow $$ x $$-$$ 1 $$ \ge $$ 0<br><br>Let x $$-$$ 1 = t<br><br>$$ \therefore $$ t $$ \ge $$ 0<br><br>From equation (1)<br><br>(x $$-$$ 1) + y + z = 9<br><br>$$ \Rightarrow $$ t + y + z = 9<br><br>Now this problem becomes, distributing 9 things among 3 people t, y, z.<br><br>Number of ways we can do that<br><br>= $${}^{9 + 3 - 1}{C_{3 - 1}} = {}^{11}{C_2} = 55$$<br><br>Now when 3 digit number is 900 then t = 9, y = 0, z = 0.<br><br>And when t = 9, then<br><br>x $$-$$ 1 = 9<br><br>$$ \Rightarrow $$ x = 10<br><br>But we can't take x = 10 in a 3 digits number. So, we have to remove this case.<br><br>$$ \therefore $$ Total number of 3 digit numbers = 55 $$-$$ 1 = 54. | integer | jee-main-2020-online-3rd-september-evening-slot |
HzI0OnXimst1pZw1d21kmllyfaz | maths | permutations-and-combinations | number-of-permutations | The missing value in the following figure is <br/><br/><img src="data:image/png;base64,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"/> | [] | null | 4 | Inside number $=\left(\right.$ difference)${ }^{(\text {difference}) !}$
<br/><br/>$=($Greater number - Smaller number)<sup>(Greater number - Smaller number)!</sup>
<br/><br/>i.e. $1=(2-1)^{(2-1) !}, $
<br/><br/>$4^{24}=(12-8)^{(12-8) !}, $
<br/><br/>$3^6=(7-4)^{(7-4) !}$
<br/><br/>$\therefore \quad ?=(5-3)^{(5-3) !}$
<br/><br/>$\therefore$ Required number $=2^{2 !}=2^{2 \times 1}=4$ | integer | jee-main-2021-online-18th-march-morning-shift |
IMSC1eZiT13CQApXI81kmm3wdw0 | maths | permutations-and-combinations | number-of-permutations | If $$\sum\limits_{r = 1}^{10} {r!({r^3} + 6{r^2} + 2r + 5) = \alpha (11!)} $$, then the value of $$\alpha$$ is equal to ___________. | [] | null | 160 | <p>$$\sum\limits_{r = 1}^{10} {r![(r + 1)(r + 2)(r + 3) - 9(r + 1) + 8]} $$</p>
<p>$$ = \sum\limits_{r = 1}^{10} {[\{ (r + 3)! - (r + 1)!\} - 8\{ (r + 1)! - r!\} ]} $$</p>
<p>$$ = (13! + 12! - 2! - 3!) - 8(11! - 1)$$</p>
<p>$$ = (12\,.\,13 + 12 - 8)\,.\,11! - 8 + 8 = (160)(11!)$$</p>
<p>Therefore, $$\alpha = 160$$</p> | integer | jee-main-2021-online-18th-march-evening-shift |
1ktbicc13 | maths | permutations-and-combinations | number-of-permutations | If $${}^1{P_1} + 2.{}^2{P_2} + 3.{}^3{P_3} + .... + 15.{}^{15}{P_{15}} = {}^q{P_r} - s,0 \le s \le 1$$, then $${}^{q + s}{C_{r - s}}$$ is equal to ______________. | [] | null | 136 | $${}^1{P_1} + 2.{}^2{P_2} + 3.{}^3{P_3} + .... + 15.{}^{15}{P_{15}}$$<br><br>= 1! + 2 . 2! + 3 . 3! + ..... 15 $$\times$$ 15!<br><br>$$ = \sum\limits_{r = 1}^{15} {(r + 1)! - (r)!} $$<br><br>= 16! $$-$$ 1<br><br>= $${}^{16}{P_{16}}$$ $$-$$ 1<br><br>$$\Rightarrow$$ q = r = 16, s = 1<br><br>$${}^{q + s}{C_{r - s}} = {}^{17}{C_{15}}$$ = 136 | integer | jee-main-2021-online-26th-august-morning-shift |
1l6dx41xx | maths | permutations-and-combinations | number-of-permutations | <p>The letters of the word 'MANKIND' are written in all possible orders and arranged in serial order as in an English dictionary. Then the serial number of the word 'MANKIND' is _____________.</p> | [] | null | 1492 | <p>Step 1 :</p>
<p>Write all the alphabets in alphabetical order.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l79egidm/a6690954-7aeb-421e-841b-1c4509718aab/ba67f6a0-24a6-11ed-8d2e-5f0df5271c2d/file-1l79egidn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l79egidm/a6690954-7aeb-421e-841b-1c4509718aab/ba67f6a0-24a6-11ed-8d2e-5f0df5271c2d/file-1l79egidn.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Permutations and Combinations Question 75 English Explanation 1"></p>
<p>Step 2 :</p>
<p>Give a number to each alphabet starting from 0 in alphabetical order.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l79eif4t/9c168fac-5b9a-40c7-962f-9e1a2098a816/ef849cd0-24a6-11ed-8d2e-5f0df5271c2d/file-1l79eif4u.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l79eif4t/9c168fac-5b9a-40c7-962f-9e1a2098a816/ef849cd0-24a6-11ed-8d2e-5f0df5271c2d/file-1l79eif4u.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Permutations and Combinations Question 75 English Explanation 2"></p>
<p>Step 3 :</p>
<p>In word "MANKIND" first alphabet is "M" which is present in the 4<sup>th</sup> position in the Alphabet Box. And after "M", in the word "MANKIND" there are 6 alphabets "ANKIND" which can be arrange in $${{6!} \over {2!}}$$ ways. Here $$2!$$ present because two N presents.</p>
<p>So for alphabet "M" we write $$4.{{6!} \over {2!}}$$</p>
<p>Step 4 :</p>
<p>As in word "MANKIND" we calculated "M", so delete "M" from alphabet box and check remaining alphabets got right numbers or not in alphabet box.</p>
<p>As you can see after deleting "M", 4<sup>th</sup> position in the alphabet box is missing so we create a new alphabet box.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l79em0gg/97d3fb7a-f64e-4c46-bfbe-808b4622d7eb/536acd00-24a7-11ed-8d2e-5f0df5271c2d/file-1l79em0gh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l79em0gg/97d3fb7a-f64e-4c46-bfbe-808b4622d7eb/536acd00-24a7-11ed-8d2e-5f0df5271c2d/file-1l79em0gh.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Permutations and Combinations Question 75 English Explanation 3"></p>
<p>Step 5 :</p>
<p>Now next alphabet after "M" in word "MANKIND" is "A" which is present in the 0<sup>th</sup> position in the new alphabet box. And after "A" in the word "MANKIND" there are 5 alphabets "NKIND" which can be arrange in $${{5!} \over {2!}}$$ ways.</p>
<p>So, for alphabet "A" we write $$0.{{5!} \over {2!}}$$</p>
<p>Step 6 :</p>
<p>Now in word "MANKIND" we calculated "A", so delete "A" from previous new alphabet box and check remaining alphabets got right numbers or not in alphabet box.</p>
<p>As you can see after deleting "A", 0<sup>th</sup> positioin in the alphabet box is missing, so we have to create a new alphabet box.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l79ep4sk/1aa80241-a0dc-436c-8af5-3fbdc96cf0a2/aa31c440-24a7-11ed-8d2e-5f0df5271c2d/file-1l79ep4sl.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l79ep4sk/1aa80241-a0dc-436c-8af5-3fbdc96cf0a2/aa31c440-24a7-11ed-8d2e-5f0df5271c2d/file-1l79ep4sl.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Permutations and Combinations Question 75 English Explanation 4"></p>
<p>Step 7 :</p>
<p>Now next alphabet after "A" in word "MANKIND" is "N" which is present in the 3<sup>rd</sup> and 4<sup>th</sup> position in the alphabet box. We have to choose minimum position for "N" in the alphabet box. So, we choose 3<sup>rd</sup> position "N". And after "N" in the word "MANKIND" there are 4 alphabets "KIND" which can be arrange $${{4!} \over {2!}}$$ ways. Here $$2!$$ used because of 2 N's (one "N" is that "N" which we are considering and other "N" which is present in the word "KIND")</p>
<p>So for alphabet "N" we write $$3\, \times \,{{4!} \over {2!}}$$</p>
<p>Step 8 :</p>
<p>Now in word "MANKIND" we calculated "N", so delete 3<sup>rd</sup> "N" from alphabet box and check remaining alphabets got right number or not in alphabet box. As you can see after deleting 3<sup>rd</sup> "N", 3<sup>rd</sup> position is missing in the alphabet box so we have to create a new alphabet box.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l79ewvv3/7e732731-8eaf-4f1e-91a2-d2f8288f1660/81c4fdf0-24a8-11ed-8d2e-5f0df5271c2d/file-1l79ewvv4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l79ewvv3/7e732731-8eaf-4f1e-91a2-d2f8288f1660/81c4fdf0-24a8-11ed-8d2e-5f0df5271c2d/file-1l79ewvv4.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Permutations and Combinations Question 75 English Explanation 5"></p>
<p>Step 9 :</p>
<p>Now next alphabet after "N" in word "MANKIND" is "K" which is present at the 2<sup>nd</sup> position in the new alphabet box. And after "K" in the word "MANKIND" there are 3 alphabets "IND" which we can arrange $$3!$$ ways.</p>
<p>So for alphabet "K" we write $$2\times3!$$</p>
<p>Step 10 :</p>
<p>Now, in word "MANKIND" we calculated "K", so delete "K" from alphabet box and check remaining alphabets got right number or not in alphabet box. As you can see after deleting "K", 2<sup>nd</sup> position in the alphabet box is missing so we have to create a new alphabet box.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l79f1t81/275df426-ebcd-42da-bccc-f9c7f09fddb4/0ac6b120-24a9-11ed-8d2e-5f0df5271c2d/file-1l79f1t82.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l79f1t81/275df426-ebcd-42da-bccc-f9c7f09fddb4/0ac6b120-24a9-11ed-8d2e-5f0df5271c2d/file-1l79f1t82.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Permutations and Combinations Question 75 English Explanation 6"></p>
<p>Step 11 :</p>
<p>Now next alphabet after "K" in word "MANKIND" is "I" which is present at 1<sup>st</sup> position in the new alphabet box. And after "I" in the word "MANKIND" there are 2 alphabets "ND" which we can arrange in $$2!$$ ways.</p>
<p>So for alphabet "I" we write $$1\times2!$$</p>
<p>Step 12 :</p>
<p>Now in word "MANKIND" we calculated "I", so delete "I" from alphabet box and check remaining alphabets got right numbers or not in alphabet box. As you can see after deleting "I", 1<sup>st</sup> position in the alphabet box is missing so we have to create a new alphabet box.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l79f4997/9a6ce487-3cbd-4e4f-822e-42b6ff66ac73/4ec74ab0-24a9-11ed-8d2e-5f0df5271c2d/file-1l79f4998.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l79f4997/9a6ce487-3cbd-4e4f-822e-42b6ff66ac73/4ec74ab0-24a9-11ed-8d2e-5f0df5271c2d/file-1l79f4998.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Permutations and Combinations Question 75 English Explanation 7"></p>
<p>Step 13 :</p>
<p>Now next alphabet after "I" in word "MANKIND" is "N" which is present at 1<sup>st</sup> position in the new alphabet box. And after "N" in the word "MANKIND" there is 1 alphabet "D" which we can arrange in $$1!$$ ways.</p>
<p>So for alphabet "N" we write $$1\times1!$$.</p>
<p>Step 14 :</p>
<p>Now in word "MANKIND" we calculated "N", so delete "N" from alphabet box and check remaining alphabets got right numbering or not in alphabet box. As you can see after deleting "N", remaining alphabet "D" got correct numbering so new alphabet box will be same.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l79f79sr/bdabc9fc-26b6-4f3a-9eaa-4b6346d15d96/a2a022b0-24a9-11ed-8d2e-5f0df5271c2d/file-1l79f79ss.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l79f79sr/bdabc9fc-26b6-4f3a-9eaa-4b6346d15d96/a2a022b0-24a9-11ed-8d2e-5f0df5271c2d/file-1l79f79ss.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Permutations and Combinations Question 75 English Explanation 8"></p>
<p>Step 15 :</p>
<p>Now next alphabet after "N" in word "MANKIND" is "D" which is present at the 0<sup>th</sup> position in the new alphabet box. And after "D" in the word "MANKIND" there is no alphabet so it is the last alphabet.</p>
<p>So for last alphabet we always write $$0!$$</p>
<p>$$\therefore$$ Position of the word "MANKIND" in the dictionary</p>
<p>$$ = 4\,.\,{{6!} \over {2!}} + 0\,.\,{{5!} \over {2!}} + 3\,.\,{{4!} \over {2!}} + 2\,.\,3!\, + 1\,.\,2!\, + \,1\,.\,1!\, + \,0!$$</p>
<p>$$ = 1440 + 0 + 36 + 12 + 2 + 1 + 1$$</p>
<p>$$ = 1492$$</p> | integer | jee-main-2022-online-25th-july-morning-shift |
1l6m6g453 | maths | permutations-and-combinations | number-of-permutations | <p>Let $$S$$ be the set of all passwords which are six to eight characters long, where each character is either an alphabet from $$\{A, B, C, D, E\}$$ or a number from $$\{1,2,3,4,5\}$$ with the repetition of characters allowed. If the number of passwords in $$S$$ whose at least one character is a number from $$\{1,2,3,4,5\}$$ is $$\alpha \times 5^{6}$$, then $$\alpha$$ is equal to ___________.</p> | [] | null | 7073 | <p>If password is 6 character long, then</p>
<p>Total number of ways having atleast one number $$ = {10^6} - {5^6}$$</p>
<p>Similarly, if 7 character long $$ = {10^7} - {5^7}$$</p>
<p>and if 8-character long $$ = {10^8} - {5^8}$$</p>
<p>Number of password $$ = ({10^6} + {10^7} + {10^8}) - ({5^6} + {5^7} + {5^8})$$</p>
<p>$$ = {5^6}({2^6} + {5.2^7} + {25.2^8} - 1 - 5 - 25)$$</p>
<p>$$ = {5^6}(64 + 640 + 6400 - 31)$$</p>
<p>$$ = 7073 \times {5^6}$$</p>
<p>$$\therefore$$ $$\alpha = 7073$$.</p> | integer | jee-main-2022-online-28th-july-morning-shift |
ldoan5f0 | maths | permutations-and-combinations | number-of-permutations | If ${ }^{2 n+1} \mathrm{P}_{n-1}:{ }^{2 n-1} \mathrm{P}_{n}=11: 21$,
<br/><br/>then $n^{2}+n+15$ is equal to : | [] | null | 45 | $\frac{\frac{(2 n+1) !}{(n+2) !}}{\frac{(2 n-1) !}{(n-1) !}}=\frac{11}{21}$
<br/><br/>$\frac{(2 n+1) 2 n}{(n+2)(n+1) n}=\frac{11}{21}$
<br/><br/>$84 n+42=11\left(n^{2}+3 n+2\right)$
<br/><br/>$11 n^{2}-51 n-20=0$
<br/><br/>$(n-5)(11 n+4)=0$
<br/><br/>$n=5, \frac{-4}{11}$ (Rejected $)$
<br/><br/>$n^{2}+n+15=45$ | integer | jee-main-2023-online-31st-january-evening-shift |
1ldom13ou | maths | permutations-and-combinations | number-of-permutations | <p>The value of $$\frac{1}{1 ! 50 !}+\frac{1}{3 ! 48 !}+\frac{1}{5 ! 46 !}+\ldots .+\frac{1}{49 ! 2 !}+\frac{1}{51 ! 1 !}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{2^{51}}{50 !}$$"}, {"identifier": "B", "content": "$$\\frac{2^{51}}{51 !}$$"}, {"identifier": "C", "content": "$$\\frac{2^{50}}{50 !}$$"}, {"identifier": "D", "content": "$$\\frac{2^{50}}{51 !}$$"}] | ["D"] | null | $$
\begin{aligned}
& \mathrm{S}=\frac{1}{1 ! 50 !}+\frac{1}{3 ! 48 !}+\frac{1}{5 ! 46 !}+\ldots \ldots+\frac{1}{49 ! 2 !}+\frac{1}{51 ! 1 !} \\\\
& =\frac{1}{51 !}\left(\frac{51 !}{1 ! 50 !}+\frac{51 !}{3 ! 48 !}+\frac{51 !}{5 ! 46 !}+\ldots . .+\frac{51 !}{49 ! 2 !}+\frac{51 !}{51 ! 0 !}\right) \\\\
& =\frac{1}{51 !}\left({ }^{51} C_{50}+{ }^{51} C_{48}+{ }^{51} C_{46}+\ldots \ldots .+{ }^{51} C_2+{ }^{51} C_0\right) \\\\
& \because{ }^n C_0+{ }^n C_2+{ }^n C_4+\ldots \ldots=2^{n-1} \\\\
& \therefore S=\frac{2^{50}}{51 !}
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-morning-shift |
1ldoo8du5 | maths | permutations-and-combinations | number-of-permutations | <p>The number of 3-digit numbers, that are divisible by either 2 or 3 but not divisible by 7, is ____________.</p> | [] | null | 514 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le9xbqbh/6a03e2bb-e7c5-4885-abb8-b77d1c53361f/0699bad0-af86-11ed-8a80-c79017866e24/file-1le9xbqbi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le9xbqbh/6a03e2bb-e7c5-4885-abb8-b77d1c53361f/0699bad0-af86-11ed-8a80-c79017866e24/file-1le9xbqbi.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 1st February Morning Shift Mathematics - Permutations and Combinations Question 62 English Explanation">
<br><br>$A=$ Numbers divisible by 2
<br><br>$B=$ Numbers divisible by 3
<br><br>$C=$ Numbers divisible by 7
<br><br>$n(A \cup B)=n(A)+n(B)-n(A \cap B)$
<br><br>$=n(2)+n(3)-n(6)$
<br><br>$n(A)=n(2)=100,102 \ldots ., 998=450$
<br><br>$n(B)=n(3)=102,105, \ldots ., 999=30$
<br><br>$n(A \cap B)=n(6)=102,108, \ldots ., 996=150$
<br><br>$n(2$ or 3$)=450+300-150=600$
<br><br>Now,
<br><br>$n(\mathrm{~A} \cap C)=n(14)=112,126, \ldots ., 994=64$
<br><br>$n(A \cap B \cap C)=n(42)=126,168, \ldots ., 966=21$
<br><br>$n(B \cap C)=n(21)=105,126, \ldots \ldots, 987,=43$
<br><br>$n(2$ or 3 not by 7$)=600-[64+43-21]$
<br><br>$=514$ | integer | jee-main-2023-online-1st-february-morning-shift |
1ldpthzaf | maths | permutations-and-combinations | number-of-permutations | <p>Let 5 digit numbers be constructed using the digits $$0,2,3,4,7,9$$ with repetition allowed, and are arranged in ascending order with serial numbers. Then the serial number of the number 42923 is __________.</p> | [] | null | 2997 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lelo42n9/53ce0458-9680-4b44-8466-08f8e9ec8d51/6747a240-b5fb-11ed-9b52-9b6c0357d10e/file-1lelo42na.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lelo42n9/53ce0458-9680-4b44-8466-08f8e9ec8d51/6747a240-b5fb-11ed-9b52-9b6c0357d10e/file-1lelo42na.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 31st January Morning Shift Mathematics - Permutations and Combinations Question 61 English Explanation"> | integer | jee-main-2023-online-31st-january-morning-shift |
1ldseml3w | maths | permutations-and-combinations | number-of-permutations | <p>The letters of the word OUGHT are written in all possible ways and these words are arranged as in a dictionary, in a series. Then the serial number of the word TOUGH is :</p> | [{"identifier": "A", "content": "79"}, {"identifier": "B", "content": "84"}, {"identifier": "C", "content": "89"}, {"identifier": "D", "content": "86"}] | ["C"] | null | <p>Lets arrange the letters of OUGHT in alphabetical
order.
<br/><br/>G, H, O, T, U</p>
<br/>Words starting with
<br/><br/>$$
\begin{aligned}
& \mathrm{G}----\rightarrow 4 ! \\\\
& \mathrm{H}----\rightarrow 4 ! \\\\
& \mathrm{O}----\rightarrow 4 !
\end{aligned}
$$
<br/><br/>$\mathrm{T} \,\mathrm{G}---\rightarrow 3$ !
<br/><br/>T H $---\rightarrow 3$ !
<br/><br/>T O $\mathrm{G}--\rightarrow 2$ !
<br/><br/>T O H $--\rightarrow 2$ !
<br/><br/>T O U G H $\rightarrow 1$ !
<br/>____________________________
<br/>Total = 89 | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldswzrar | maths | permutations-and-combinations | number-of-permutations | <p>Five digit numbers are formed using the digits 1, 2, 3, 5, 7 with repetitions and are written in descending order with serial numbers. For example, the number 77777 has serial number 1. Then the serial number of 35337 is ____________.</p> | [] | null | 1436 | <p>Descending order means from highest to lowest.</p>
<p>$$\therefore$$ Descending order of digits 1, 2, 3, 5, 7 is = 7, 5, 3, 2, 1.</p>
<p>(1) Number starts with 7 :</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le2g13sg/30377b8c-aded-4085-8900-acce8af1dc3b/e75d6700-ab68-11ed-b566-111c81fc645a/file-1le2g13sh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le2g13sg/30377b8c-aded-4085-8900-acce8af1dc3b/e75d6700-ab68-11ed-b566-111c81fc645a/file-1le2g13sh.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Permutations and Combinations Question 52 English Explanation 1"></p>
<p>$$\therefore$$ Total possible numbers start's with 7 are = <sup>5</sup>C<sub>1</sub> $$\times$$ <sup>5</sup>C<sub>1</sub> $$\times$$ <sup>5</sup>C<sub>1</sub> $$\times$$ <sup>5</sup>C<sub>1</sub> = 5<sup>4</sup></p>
<p>(2) Number of starts with 5 :</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le2g1tke/43c45f16-3db2-4d80-a59f-b27d342fc2b4/fb46c0e0-ab68-11ed-b88f-0b6f40cc9ae6/file-1le2g1tkf.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le2g1tke/43c45f16-3db2-4d80-a59f-b27d342fc2b4/fb46c0e0-ab68-11ed-b88f-0b6f40cc9ae6/file-1le2g1tkf.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Permutations and Combinations Question 52 English Explanation 2"></p>
<p>$$\therefore$$ Total possible numbers start's with 5 are = <sup>5</sup>C<sub>1</sub> $$\times$$ <sup>5</sup>C<sub>1</sub> $$\times$$ <sup>5</sup>C<sub>1</sub> $$\times$$ <sup>5</sup>C<sub>1</sub> = 5<sup>4</sup></p>
<p>(3) Number starts with 37 :</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le2g2ax4/3d119479-b1ac-4838-9b4e-8989ddd3f6de/08ae7480-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g2ax5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le2g2ax4/3d119479-b1ac-4838-9b4e-8989ddd3f6de/08ae7480-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g2ax5.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Permutations and Combinations Question 52 English Explanation 3"></p>
<p>$$\therefore$$ Total possible numbers starts with 3 and 7 are = <sup>5</sup>C<sub>1</sub> $$\times$$ <sup>5</sup>C<sub>1</sub> $$\times$$ <sup>5</sup>C<sub>1</sub> = 5<sup>3</sup></p>
<p>(4) Number starts with 357 :</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le2g2w2h/254b12e3-a7f2-4f3b-a4c0-bba7e60d20fc/1904bc90-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g2w2i.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le2g2w2h/254b12e3-a7f2-4f3b-a4c0-bba7e60d20fc/1904bc90-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g2w2i.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Permutations and Combinations Question 52 English Explanation 4"></p>
<p>$$\therefore$$ Total possible numbers starts with 357 are = <sup>5</sup>C<sub>1</sub> $$\times$$ <sup>5</sup>C<sub>1</sub> = 5<sup>2</sup></p>
<p>(5) Number starts with 355 :</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le2g3fcm/c842699b-7c36-45e1-856a-a43d02492a13/27e9c160-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g3fcn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le2g3fcm/c842699b-7c36-45e1-856a-a43d02492a13/27e9c160-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g3fcn.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Permutations and Combinations Question 52 English Explanation 5"></p>
<p>$$\therefore$$ Total possible numbers starts with 355 are = <sup>5</sup>C<sub>1</sub> $$\times$$ <sup>5</sup>C<sub>1</sub> = 5<sup>2</sup></p>
<p>(6) Number starts with 3537 :</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le2g48bm/5e3286ff-1f06-4186-a954-c8121f8cf64f/3e4b4730-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g48bn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le2g48bm/5e3286ff-1f06-4186-a954-c8121f8cf64f/3e4b4730-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g48bn.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Permutations and Combinations Question 52 English Explanation 6"></p>
<p>$$\therefore$$ Total possible numbers starts with 3537 are = <sup>5</sup>C<sub>1</sub> = 5</p>
<p>(7) Number starts with 3535 :</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le2g4qkb/9fafb3cc-ba1e-4c03-a575-be2b1b45a714/4c6274b0-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g4qkc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le2g4qkb/9fafb3cc-ba1e-4c03-a575-be2b1b45a714/4c6274b0-ab69-11ed-b88f-0b6f40cc9ae6/file-1le2g4qkc.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Permutations and Combinations Question 52 English Explanation 7"></p>
<p>$$\therefore$$ Total possible numbers starts with 3535 are = <sup>5</sup>C<sub>1</sub> = 5</p>
<p>(8) Next number is 35337 which is the required number.</p>
<p>$$\therefore$$ Position of the number 35337 is = 5<sup>4</sup> + 5<sup>4</sup> + 5<sup>3</sup> + 5<sup>2</sup> + 5<sup>2</sup> + 5 + 5 + 1 = 1436</p> | integer | jee-main-2023-online-29th-january-morning-shift |
1ldww37nt | maths | permutations-and-combinations | number-of-permutations | <p>The number of integers, greater than 7000 that can be formed, using the digits 3, 5, 6, 7, 8 without repetition is :</p> | [{"identifier": "A", "content": "48"}, {"identifier": "B", "content": "120"}, {"identifier": "C", "content": "168"}, {"identifier": "D", "content": "220"}] | ["C"] | null | Four digit numbers greater than 7000 $=2 \times 4 \times 3 \times 2=48$<br/><br/>
Five digit number $=5 !=120$<br/><br/>
Total number greater than 7000 $=120+48=168$ | mcq | jee-main-2023-online-24th-january-evening-shift |
1lgow5dwx | maths | permutations-and-combinations | number-of-permutations | <p>All words, with or without meaning, are made using all the letters of the word MONDAY. These words are written as in a dictionary with serial numbers. The serial number of the word MONDAY is :</p> | [{"identifier": "A", "content": "324"}, {"identifier": "B", "content": "328"}, {"identifier": "C", "content": "326"}, {"identifier": "D", "content": "327"}] | ["D"] | null |
<p>The letters of "MONDAY" arranged in alphabetical order are: A, D, M, N, O, Y.</p>
<ol>
<li>Fix A as the first letter, we can arrange the remaining 5 letters in 5! ways = 120 ways.</li>
<li>Fix D as the first letter, we can arrange the remaining 5 letters in 5! ways = 120 ways.</li>
</ol>
<p>So far, we have 120 (for A) + 120 (for D) = 240 words.</p>
<p>Now, let's proceed with words starting with M:</p>
<ol>
<li>Fix MA as the first two letters, we can arrange the remaining 4 letters in 4! ways = 24 ways.</li>
<li>Fix MD as the first two letters, we can arrange the remaining 4 letters in 4! ways = 24 ways.</li>
<li>Fix MN as the first two letters, we can arrange the remaining 4 letters in 4! ways = 24 ways.</li>
</ol>
<p>Now, we have 240 (for A and D) + 24 (for MA) + 24 (for MD) + 24 (for MN) = 312 words.</p>
<p>After MN, we consider words that start with MO:</p>
<ol>
<li>Fix MOA as the first three letters, we can arrange the remaining 3 letters in 3! ways = 6 ways.</li>
<li>Fix MOD as the first three letters, we can arrange the remaining 3 letters in 3! ways = 6 ways.</li>
</ol>
<p>Adding these to the total, we have 312 (previous total) + 6 (for MOA) + 6 (for MOD) = 324 words.</p>
<p>Next, we consider words that start with 'MON'. The word 'MONDAY' comes after 'MONAD' in dictionary order, so:</p>
<ol>
<li>Fix 'MONA' as the first four letters. The remaining 2 letters ('D' and 'Y') can be arranged in 2! ways, which gives us 2 more words: 'MONADY' and 'MONAYD'.</li>
</ol>
<p>So, we have 324 (previous total) + 2 (for 'MONADY' and 'MONAYD') = 326 words.</p>
<p>Finally, we have the word 'MONDAY' itself, which is the 327th word. </p> | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgsugvol | maths | permutations-and-combinations | number-of-permutations | <p>If the letters of the word MATHS are permuted and all possible words so formed are arranged as in a dictionary with serial numbers, then the serial number of the word THAMS is :</p> | [{"identifier": "A", "content": "103"}, {"identifier": "B", "content": "104"}, {"identifier": "C", "content": "102"}, {"identifier": "D", "content": "101"}] | ["A"] | null | To solve this problem, we start by finding the number of permutations before we reach a word that begins with T.
<br/><br/>We have 5 letters in the word MATHS. If we fix the first letter, there are 4! (=4$$ \times $$3$$ \times $$2$$ \times $$1=24) ways to arrange the remaining letters.
<br/><br/>The letters before T in alphabetical order are A, H, M, and S. For each of these 4 letters, there are 4! ways to arrange the rest of the word, giving 4$$ \times $$4! = 96 words that come before any words starting with T.
<br/><br/>Then, in the section of words that start with 'T', we start with 'TA'. There are 3 remaining letters after 'TA', hence the permutations for words that start with 'TA' are (5-2)!=3!=6.
<br/><br/>So, the total permutations for all words that come before 'TH' is 96 + 6 = 102.
<br/><br/>Now, we need to count the words that come before 'THAMS'. After 'TH', the letters left are 'AMS'. Arranged in dictionary order, they would be 'AMS', 'ASM', 'MAS', 'MSA', 'SAM', 'SMA'. So, 'THAMS' is the first word in the 'TH' category, so it is the 103<sup>rd</sup> word overall when counting from the beginning.
<br/><br/>So, the correct answer is 103, which corresponds to Option A. | mcq | jee-main-2023-online-11th-april-evening-shift |
1lgvqlzes | maths | permutations-and-combinations | number-of-permutations | <p>The sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is equal to __________.</p> | [] | null | 26664 | We have, four digits are $2,1,2,3$.
<br/><br/>Total numbers when 1 is at unit digit $=\frac{3 !}{2 !}=3$
<br/><br/>Total number when 2 is at unit digit $=3 !=6$
<br/><br/>Total numbers when 3 is at unit digit $=\frac{3 !}{2 !}=3$
<br/><br/>Sum of digits at unit place $=3 \times 1+6 \times 2+3 \times 3=24$
<br/><br/>$\therefore$ Required sum $=24 \times 1000+24 \times 100+24 \times 10+24 \times 1$
<br/><br/>$=24 \times 1111=26664$ | integer | jee-main-2023-online-10th-april-evening-shift |
1lh2xtxhh | maths | permutations-and-combinations | number-of-permutations | <p> All the letters of the word PUBLIC are written in all possible orders and these words are written as in a dictionary with serial numbers. Then the serial number of the word PUBLIC is :</p> | [{"identifier": "A", "content": "578"}, {"identifier": "B", "content": "576"}, {"identifier": "C", "content": "580"}, {"identifier": "D", "content": "582"}] | ["D"] | null | Given word is PUBLIC.
<br/><br/>Alphabetical order of letters is BCILPU. So, number of words start with letter.
<br/><br/>B - - - - is $5 !=120$
<br/><br/>C ---- is $5 !=120$
<br/><br/>I ----- is $5 !=120$
<br/><br/>$\mathrm{L}-----$ is $5 !=120$
<br/><br/>$\mathrm{PB}----$ is $4 !=24$
<br/><br/>$\mathrm{PC}----$ is $4 !=24$
<br/><br/>PI ---- is $4 !=24$
<br/><br/>$\mathrm{PL}----$ is $4 !=24$
<br/><br/>PUBC -- is $2 !=2$
<br/><br/>PUBI - is $2 !=2$
<br/><br/>PUBLC - is $1 !=1$
<br/><br/>PUBLIC is $0 !=1$
<br/><br/>$\therefore$ Serial number of the word PUBLIC
<br/><br/>$=4 \times 120+4 \times 24+2 \times 2+2 \times 1=582$ | mcq | jee-main-2023-online-6th-april-evening-shift |
jaoe38c1lse5y7d4 | maths | permutations-and-combinations | number-of-permutations | <p>The total number of words (with or without meaning) that can be formed out of the letters of the word '<b>DISTRIBUTION</b>' taken four at a time, is equal to __________.</p> | [] | null | 3734 | <p>We have III, TT, D, S, R, B, U, O, N</p>
<p>Number of words with selection (a, a, a, b)</p>
<p>$$={ }^8 \mathrm{C}_1 \times \frac{4 !}{3 !}=32$$</p>
<p>Number of words with selection (a, a, b, b)</p>
<p>$$=\frac{4 !}{2 ! 2 !}=6$$</p>
<p>Number of words with selection (a, a, b, c)</p>
<p>$$={ }^2 \mathrm{C}_1 \times{ }^8 \mathrm{C}_2 \times \frac{4 !}{2 !}=672$$</p>
<p>Number of words with selection (a, b, c, d)</p>
<p>$$\begin{aligned}
& ={ }^9 \mathrm{C}_4 \times 4 !=3024 \\\\
& \therefore \text {Total }=3024+672+6+32 \\\\
& =3734
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lsf0oh50 | maths | permutations-and-combinations | number-of-permutations | <p>All the letters of the word "GTWENTY" are written in all possible ways with or without meaning and these words are written as in a dictionary. The serial number of the word "GTWENTY" is _________.</p> | [] | null | 553 | <p>Words starting with $$\mathrm{E}=360$$</p>
<p>Words starting with $$\mathrm{GE=60}$$</p>
<p>Words starting with $$\mathrm{GN=60}$$</p>
<p>Words starting with $$\mathrm{GTE=24}$$</p>
<p>Words starting with $$\mathrm{GTN=24}$$</p>
<p>Words starting with $$\mathrm{GTT=24}$$</p>
<p>GTWENTY $$=1$$</p>
<p>Total $$=553$$</p> | integer | jee-main-2024-online-29th-january-morning-shift |
lv5gs8ga | maths | permutations-and-combinations | number-of-permutations | <p>Let $$[t]$$ be the greatest integer less than or equal to $$t$$. Let $$A$$ be the set of all prime factors of 2310 and $$f: A \rightarrow \mathbb{Z}$$ be the function $$f(x)=\left[\log _2\left(x^2+\left[\frac{x^3}{5}\right]\right)\right]$$. The number of one-to-one functions from $$A$$ to the range of $$f$$ is</p> | [{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "120"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "24"}] | ["B"] | null | <p>$$\begin{aligned}
& A=\{2,3,5,7,11\} \\
& f(x)=\left[\log _2\left(x^2+\left[\frac{x^3}{5}\right]\right)\right] \\
& \text { Ranges } f(x)=\{2,3,5,6,8\}
\end{aligned}$$</p>
<p>$$\text { Number of one-one } A \rightarrow R_f$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw8oycbm/cc620a6d-aa22-48e0-ad06-9f4595901b72/65685c20-1334-11ef-9f8d-838c388c326d/file-1lw8oycbn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw8oycbm/cc620a6d-aa22-48e0-ad06-9f4595901b72/65685c20-1334-11ef-9f8d-838c388c326d/file-1lw8oycbn.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 - Permutations and Combinations Question 8 English Explanation"></p>
<p>$$5 \times 4 \times 3 \times 2 \times 1=120$$</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
lv5gsk30 | maths | permutations-and-combinations | number-of-permutations | <p>The number of 3-digit numbers, formed using the digits 2, 3, 4, 5 and 7, when the repetition of digits is not allowed, and which are not divisible by 3 , is equal to ________.</p> | [] | null | 36 | <p>To solve this problem, we need to find the number of 3-digit numbers formed using the digits 2, 3, 4, 5, and 7, with no repetition of digits allowed, and these numbers should not be divisible by 3. Let's break down the solution step-by-step:</p>
<p>1. <strong>Calculating the total number of 3-digit numbers without repetition:</strong></p>
<p>The number of ways to form a 3-digit number from 5 unique digits (2, 3, 4, 5, 7) without repetition can be calculated using permutations:</p>
<p>The total number of permutations for choosing 3 digits out of 5 and arranging them is given by:</p>
<p>$$5P3 = \frac{5!}{(5-3)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3 = 60$$</p>
<p>So, there are 60 possible 3-digit numbers that can be formed from the digits 2, 3, 4, 5, and 7 without repetition.</p>
<p>2. <strong>Finding the 3-digit numbers divisible by 3:</strong></p>
<p>A number is divisible by 3 if the sum of its digits is divisible by 3. Let's consider the sums of every combination of these three digits to find out which sums are divisible by 3:</p>
<p>Possible sums of combinations:</p>
<ul>
<li>2 + 3 + 4 = 9 (divisible by 3)</li>
<li>2 + 4 + 7 = 13</li>
<li>2 + 5 + 7 = 14</li>
<li>2 + 3 + 5 = 10</li>
<li>2 + 3 + 7 = 12 (divisible by 3)</li>
<li>3 + 4 + 5 = 12 (divisible by 3)</li>
<li>3 + 4 + 7 = 14</li>
<li>3 + 5 + 7 = 15 (divisible by 3)</li>
<li>4 + 5 + 7 = 16</li>
</ul>
<p>The combinations whose sums are divisible by 3 are:</p>
<ul>
<li>2, 3, 4</li>
<li>2, 3, 7</li>
<li>3, 4, 5</li>
<li>3, 5, 7</li>
</ul>
<p>Since the sum of the digits is divisible by 3 for these combinations, any permutation of these sets will yield a number divisible by 3:</p>
<p>The number of 3-digit numbers that can be formed from each set of 3 digits is:</p>
<p>$$3! = 6$$</p>
<p>So, the total number of 3-digit numbers divisible by 3 is:</p>
<p>$$4 \text{ sets} \times 6 \text{ permutations per set} = 24$$</p>
<p>3. <strong>Calculating the 3-digit numbers not divisible by 3:</strong></p>
<p>To find the 3-digit numbers not divisible by 3, we subtract the number of those divisible by 3 from the total number of 3-digit numbers:</p>
<p>$$60 - 24 = 36$$</p>
<p>Therefore, the number of 3-digit numbers that can be formed using the digits 2, 3, 4, 5, and 7 without repetition, and which are not divisible by 3, is equal to 36.</p> | integer | jee-main-2024-online-8th-april-morning-shift |
lv9s20f0 | maths | permutations-and-combinations | number-of-permutations | <p>60 words can be made using all the letters of the word $$\mathrm{BHBJO}$$, with or without meaning. If these words are written as in a dictionary, then the $$50^{\text {th }}$$ word is:</p> | [{"identifier": "A", "content": "OBBJH"}, {"identifier": "B", "content": "HBBJO"}, {"identifier": "C", "content": "OBBHJ"}, {"identifier": "D", "content": "JBBOH"}] | ["A"] | null | <p>To find the $$50^{\text{th}}$$ word formed by the letters of "BHBJO" as if listed in a dictionary, let's analyze the arrangement methodically. Given the letters are B, H, B, J, O, there are some repetitions with the letter B appearing twice.</p>
<p>First, calculate the total number of permutations of these letters:</p>
<p>$$ \frac{5!}{2!} = 60 $$</p>
<p>Let's arrange the letters in alphabetical order first: B, B, H, J, O.</p>
<p>We need to systematically count the words while following dictionary order:</p>
<p>1. Words starting with B:</p>
<ul>
<li>Next position letters: B, H, J, O</li>
<li>Number of permutations: $$\frac{4!}{1!} = 24$$ words</li>
</ul>
<p>Since 24 words starting with 'B' exist and are less than 50, Move to next starting letter alphabetically.</p>
<p>2. Words starting with H:</p>
<ul>
<li>Next position letters: B, B, J, O</li>
<li>Number of permutations: $$\frac{4!}{2!} = 12$$ words</li>
</ul>
<p>After 'B', tally becomes 24 (B-words) + 12 (H-words) = 36, needs more.</p>
<p>3. Words starting with J:</p>
<ul>
<li>Next position letters: B, B, H, O</li>
<li>Number of permutations: $$\frac{4!}{2!} = 12$$ words</li>
</ul>
<p>Now tally is 36 + 12 = 48 words.</p>
<p>Still 2 more to reach 50.</p>
<p>4. Words starting with O:</p>
<ul>
<li>Next position letters: B, B, H, J</li>
<li>Number of permutations: $$\frac{4!}{2!} = 12$$ words</li>
<li>Our answer must be here since 48 + 2 more = 50 total.</li>
</ul>
<p>First permutation: $$OBB H J$$</p>
<p>Second permutation (50th word): $$OBB J H$$</p>
<p>So, the $$50^{\text{th}}$$ word is: <b>OBBJH</b></p>
<p>Thus, the correct answer is <strong>Option A: OBBJH</strong>.</p> | mcq | jee-main-2024-online-5th-april-evening-shift |
lvb294sf | maths | permutations-and-combinations | number-of-permutations | <p>If all the words with or without meaning made using all the letters of the word "NAGPUR" are arranged as in a dictionary, then the word at $$315^{\text {th }}$$ position in this arrangement is :</p> | [{"identifier": "A", "content": "NRAPUG\n"}, {"identifier": "B", "content": "NRAGUP\n"}, {"identifier": "C", "content": "NRAPGU\n"}, {"identifier": "D", "content": "NRAGPU"}] | ["C"] | null | <p>NAGPUR</p>
<p>Word at $$315^{\text {th }}$$ position</p>
<p>$$\begin{aligned}
& \text { A...... }=5!=120 \\
& \text { G....... }=5!=120 \\
& \text { NA..... }=4!=24 \\
& \text { NG..... }=4!=24 \\
& \text { NP..... }=4!=24
\end{aligned}$$</p>
<p>..... Till 312 words</p>
<p>$$313^{\text {th }} \text { word }=\text { NRAGPU }$$</p>
<p>$$314^{\text {th }}$$ word $$=$$ $$\mathrm{NRAGUP}$$</p>
<p>$$315^{\text {th }} \text { word }=\text { NRAPGU }$$</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
ys2abeQH99za49tLBzsHL | maths | probability | bayes-theorem | A box 'A' contains $$2$$ white, $$3$$ red and $$2$$ black balls. Another box 'B' contains $$4$$ white, $$2$$ red and $$3$$ black balls. If two balls are drawn at random, without eplacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box 'B' is : | [{"identifier": "A", "content": "$${9 \\over {16}}$$ "}, {"identifier": "B", "content": "$${7 \\over {16}}$$"}, {"identifier": "C", "content": "$${9 \\over {32}}$$"}, {"identifier": "D", "content": "$${7 \\over {8}}$$"}] | ["B"] | null | Probability of drawing a white ball and then a red ball
<br><br>from bag B is given by
$${{{}^4{C_1} \times {}^2{C_1}} \over {{}^9{C_2}}}$$ = $${2 \over 9}$$
<br><br>Probability of drawing a white ball and then a red ball
<br><br>from bag A is given by $${{{}^2{C_1} \times {}^3{C_1}} \over {{}^7{C_2}}}$$ = $${2 \over 7}$$
<br><br>Hence, the probability of drawing a white ball and then
<br><br>a red ball from bag B = $${{{2 \over 9}} \over {{2 \over 7} + {2 \over 9}}}$$ = $${{2 \times 7} \over {18 + 14}}$$ = $${7 \over {16}}$$ | mcq | jee-main-2018-online-15th-april-morning-slot |
ddcEu6sJsoUFV2yTjBjgy2xukewmw2ol | maths | probability | bayes-theorem | Box I contains 30 cards numbered 1 to 30 and
Box II contains 20 cards numbered 31 to 50. A
box is selected at random and a card is drawn
from it. The number on the card is found to be
a non-prime number. The probability that the
card was drawn from Box I is : | [{"identifier": "A", "content": "$${8 \\over {17}}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${2 \\over 5}$$"}, {"identifier": "D", "content": "$${4 \\over {17}}$$"}] | ["A"] | null | Let B<sub>1</sub> be the event where Box-I is selected.
<br><br>And B<sub>2</sub> be the event where Box-II is selected.
<br><br>P(B<sub>1</sub>) = P(B<sub>2</sub>) = $${1 \over 2}$$
<br><br>Let E be the event where selected card is non prime.
<br><br>For B<sub>1</sub> : Prime numbers: {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
<br><br>For B<sub>2</sub> : Prime numbers: {31, 37, 41, 43, 47}
<br><br>P(E) = P(B<sub>1</sub>) $$ \times $$ $$P\left( {{E \over {{B_1}}}} \right)$$ + P(B<sub>2</sub>) $$ \times $$ $$P\left( {{E \over {{B_2}}}} \right)$$
<br><br>= $${1 \over 2} \times {{20} \over {30}}$$ + $${1 \over 2} \times {{15} \over {20}}$$
<br><br>Required probability :
<br><br>$$P\left( {{{{B_1}} \over E}} \right)$$ = $${{P\left( {{B_2}} \right).P\left( {{E \over {{B_1}}}} \right)} \over {P\left( E \right)}}$$
<br><br>= $${{{1 \over 2} \times {{20} \over {30}}} \over {{1 \over 2} \times {{20} \over {30}} + {1 \over 2}{{15} \over {20}}}}$$
<br><br>= $${{{2 \over 3}} \over {{2 \over 3} + {3 \over 4}}}$$
<br><br>= $${8 \over {17}}$$ | mcq | jee-main-2020-online-2nd-september-morning-slot |
RTeG7N1NMdtZ3ITnFI1klt88cer | maths | probability | bayes-theorem | In a group of 400 people, 160 are smokers and non-vegetarian; 100 are smokers and vegetarian and the remaining 140 are non-smokers and vegetarian. Their chances of getting a particular chest disorder are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the chest disorder. The probability that the selected person is a smoker and non-vegetarian is : | [{"identifier": "A", "content": "$${{14} \\over {45}}$$"}, {"identifier": "B", "content": "$${{8} \\over {45}}$$"}, {"identifier": "C", "content": "$${{7} \\over {45}}$$"}, {"identifier": "D", "content": "$${{28} \\over {45}}$$"}] | ["D"] | null | Consider following events<br><br>A : Person chosen is a smoker and non vegetarian.<br><br>B : Person chosen in a smoker and vegetarian.<br><br>C : Person chosen is a non-smoker and vegetarian.<br><br>E : Person chosen has a chest disorder<br><br>Given <br><br>$$P(A) = {{160} \over {400}}$$
<br><br>$$P(B) = {{100} \over {400}}$$
<br><br>$$P(C) = {{140} \over {400}}$$<br><br>$$P\left( {{E \over A}} \right) = {{35} \over {100}}$$
<br><br>$$P\left( {{E \over B}} \right) = {{20} \over {100}}$$
<br><br>$$P\left( {{E \over C}} \right) = {{10} \over {100}}$$<br><br>To find<br><br>$$P\left( {{A \over E}} \right) = {{P(A)P\left( {{E \over A}} \right)} \over {P(A).P\left( {{E \over A}} \right) + P(B).P\left( {{E \over B}} \right) + P(C).P\left( {{E \over C}} \right)}}$$<br><br>$$ = {{{{160} \over {400}} \times {{35} \over {100}}} \over {{{160} \over {400}} \times {{35} \over {100}} \times {{100} \over {400}} \times {{20} \over {200}} + {{140} \over {400}} \times {{10} \over {100}}}}$$<br><br>$$ = {{28} \over {45}}$$ | mcq | jee-main-2021-online-25th-february-evening-slot |
1l5c0v06l | maths | probability | bayes-theorem | <p>Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is $${6 \over {11}}$$, then n is equal to __________.</p> | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l8la6rik/8b350f4d-948a-4095-a004-5c8116776062/a60740c0-3efb-11ed-8e30-11b9f1cb84fb/file-1l8la6ril.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l8la6rik/8b350f4d-948a-4095-a004-5c8116776062/a60740c0-3efb-11ed-8e30-11b9f1cb84fb/file-1l8la6ril.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 24th June Morning Shift Mathematics - Probability Question 71 English Explanation"><br>$$
\begin{aligned}
&P(1 R \text { and } 1 B)=P(A) \cdot P\left(\frac{1 R 1 B}{A}\right)+P(B) \cdot P\left(\frac{1 R 1 B}{B}\right) \\\\
&=\frac{1}{2} \cdot \frac{{ }^{3} C_{1} \cdot{ }^{1} C_{1}}{{ }^{6} C_{2}}+\frac{1}{2} \cdot \frac{{ }^{2} C_{1} \cdot{ }^{3} C_{1}}{{ }^{n+5} C_{2}} \\\\
&P\left(\frac{1 R 1 B}{A}\right)=\frac{\frac{1}{2} \cdot \frac{3}{15}}{\frac{1}{2} \cdot \frac{3}{15}+\frac{1}{2} \cdot \frac{6 \cdot 2}{(n+5)(n+4)}}=\frac{6}{11} \\\\
&\Rightarrow \frac{\frac{1}{10}}{\frac{1}{10}+\frac{6}{(n+5)(n+4)}}=\frac{6}{11}
\end{aligned}
$$
<br><br>
$$
\begin{aligned}
&\Rightarrow \frac{11}{10}=\frac{6}{10}+\frac{36}{(n+5)(n+4)} \\\\
&\Rightarrow \frac{5}{10 \times 36}=\frac{1}{(n+5)(n+4)} \\\\
&\Rightarrow n^{2}+9 n-52=0 \\\\
&\Rightarrow n=4 \text { is only possible value }
\end{aligned}
$$ | mcq | jee-main-2022-online-24th-june-morning-shift |
1l6rev2p3 | maths | probability | bayes-theorem | <p>Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is :</p> | [{"identifier": "A", "content": "$$\\frac{4}{9}$$"}, {"identifier": "B", "content": "$$\\frac{5}{18}$$"}, {"identifier": "C", "content": "$$\\frac{1}{6}$$"}, {"identifier": "D", "content": "$$\\frac{3}{10}$$"}] | ["B"] | null | Let $E \rightarrow$ Ball drawn from Bag II is black.
<br/><br/>
$E_{R} \rightarrow$ Bag I to Bag II red ball transferred.
<br/><br/>
$E_{B} \rightarrow$ Bag I to Bag II black ball transferred.
<br/><br/>
$E_{w} \rightarrow$ Bag I to Bag II white ball transferred.
<br/><br/>
$P\left(E_{R} / E\right)=\frac{P\left(E / E_{R}\right) \cdot P\left(E_{R}\right)}{P\left(E / E_{R}\right) P\left(E_{R}\right)+P\left(E / E_{B}\right) P\left(E_{B}\right)+P\left(E / E_{W}\right) P\left(E_{W}\right)}$
<br/><br/>
Here,
<br/><br/>
$P\left(E_{R}\right)=3 / 10, \quad P\left(E_{B}\right)=4 / 10, \quad P\left(E_{W}\right)=3 / 10$
<br/><br/>
and
<br/><br/>
$$
\begin{aligned}
& P\left(E / E_{R}\right)=5 / 10, \quad P\left(E / E_{B}\right)=6 / 10, \quad P\left(E / E_{W}\right)=5 / 10 \\\\
& \therefore \quad P\left(E_{R} / E\right)=\frac{15 / 100}{15 / 100+24 / 100+15 / 100} \\\\
& =\frac{15}{54}=\frac{5}{18}
\end{aligned}
$$ | mcq | jee-main-2022-online-29th-july-evening-shift |
1ldprftih | maths | probability | bayes-theorem | <p>A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is :</p> | [{"identifier": "A", "content": "$$\\frac{3}{7}$$"}, {"identifier": "B", "content": "$$\\frac{5}{6}$$"}, {"identifier": "C", "content": "$$\\frac{5}{7}$$"}, {"identifier": "D", "content": "$$\\frac{2}{7}$$"}] | ["C"] | null | Let $E_{i} \rightarrow$ Bag have at least $i$ black balls
<br/><br/>$E \rightarrow 2$ balls are drawn & both black
<br/><br/>$$
\begin{aligned}
& \therefore P\left(\frac{E_{5} \text { or } E_{6}}{E}\right)=\frac{P\left(\frac{E}{E_{5}}\right)+P\left(\frac{E}{E_{6}}\right)}{\sum\limits_{i=1}^{6} P\left(\frac{E}{E_{i}}\right)} \\\\
& =\frac{\frac{{ }^{5} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{6} C_{2}}{{ }^{6} C_{2}}}{0+\frac{{ }^{2} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{3} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{4} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{5} C_{2}}{{ }^{6} C_{2}}+\frac{{ }^{6} C_{2}}{{ }^{6} C_{2}}} \\\\
& =\frac{10+15}{1+3+6+10+15}=\frac{25}{35}=\frac{5}{7}
\end{aligned}
$$ | mcq | jee-main-2023-online-31st-january-morning-shift |
1ldwxvdwq | maths | probability | bayes-theorem | <p>Three urns A, B and C contain 4 red, 6 black; 5 red, 5 black; and $$\lambda$$ red, 4 black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn C is 0.4 then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola $$y^2=\lambda x$$ with one vertex at the vertex of the parabola, is :</p> | [] | null | 432 | $E_{1}$ : Ball is drawn from urn $A(4 R+6 B)$
<br><br>
$E_{2}:$ Ball is drawn from urn $B(5 R+5 B)$
<br><br>
$E_{3}$ : Ball is drawn from urn $C(\lambda R+4 B)$
<br><br>
$A \rightarrow$ Ball drawn is red.
<br><br>
Required probability $=P\left(\frac{E_{3}}{A}\right)$
<br><br>
$=\frac{\frac{1}{3} \times \frac{\lambda}{\lambda+4}}{\frac{1}{3} \times \frac{4}{10}+\frac{1}{3} \times \frac{5}{10}+\frac{1}{3} \times \frac{\lambda}{\lambda+4}}=\frac{2}{5}$
<br><br>
$\Rightarrow\frac{10 \lambda}{19 \lambda+36}=\frac{2}{5}$
<br><br>
$\Rightarrow \lambda=6$
<br><br>So, parabola $y^2=6 x$
<br><br>Let side length of the triangle be $l$.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lfq9svjl/bc19241f-0f31-4930-9f1d-f97a67d8717c/92b63210-cc4f-11ed-8d1a-d560c3385a80/file-1lfq9svjm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lfq9svjl/bc19241f-0f31-4930-9f1d-f97a67d8717c/92b63210-cc4f-11ed-8d1a-d560c3385a80/file-1lfq9svjm.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 24th January Evening Shift Mathematics - Probability Question 42 English Explanation">
<br>$$
\begin{aligned}
& \tan 30^{\circ}=\frac{3 t} {\frac{3}{2} t^2} \\\\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{2}{t} \\\\
& \therefore t=2 \sqrt{3} \\\\
& \text { So, }\left(\frac{3}{2} t^2, 3 t\right) \\\\
& =(18,6 \sqrt{3})
\end{aligned}
$$
<br><br>$$
\text { Now, } l^2=18^2+(6 \sqrt{3})^2=324+108=432
$$ | integer | jee-main-2023-online-24th-january-evening-shift |
1lsg4n319 | maths | probability | bayes-theorem | <p>Bag A contains 3 white, 7 red balls and Bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn is white, is</p> | [{"identifier": "A", "content": "1/4"}, {"identifier": "B", "content": "1/3"}, {"identifier": "C", "content": "3/10"}, {"identifier": "D", "content": "1/9"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxmoxg/bddbd042-adc7-47f2-8492-6edfb8c4f19c/00793740-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxmoxh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsoxmoxg/bddbd042-adc7-47f2-8492-6edfb8c4f19c/00793740-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxmoxh.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) 30th January Evening Shift Mathematics - Probability Question 17 English Explanation"></p>
<p>$$\mathrm{E}_1: \mathrm{A}$$ is selected</p>
<p>$$\mathrm{E}_2: \mathrm{B}$$ is selected</p>
<p>$$\mathrm{E}$$ : white ball is drawn</p>
<p>$$\mathrm{P}\left(\mathrm{E}_1 / \mathrm{E}\right)=$$</p>
<p>$$\begin{aligned}
& \frac{P(E) \cdot P\left(E / E_1\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)}=\frac{\frac{1}{2} \times \frac{3}{10}}{\frac{1}{2} \times \frac{3}{10}+\frac{1}{2} \times \frac{3}{5}} \\
& =\frac{3}{3+6}=\frac{1}{3}
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-evening-shift |
lv0vxct6 | maths | probability | bayes-theorem | <p>Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn $$\mathrm{A}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{4}{17}$$\n"}, {"identifier": "B", "content": "$$\\frac{5}{16}$$\n"}, {"identifier": "C", "content": "$$\\frac{5}{18}$$\n"}, {"identifier": "D", "content": "$$\\frac{7}{18}$$"}] | ["C"] | null | <p>Let's denote the events as follows :</p>
<p>Let $$ A_1, A_2, $$ and $$ A_3 $$ be the events that urns A, B, and C are chosen, respectively.</p>
<p>Let $$ B $$ be the event that a black ball is drawn.</p>
<p>We need to find the probability that the chosen urn is A given that a black ball is drawn, which is $$ P(A_1|B) $$.</p>
<p>Using Bayes' theorem, we have :</p>
<p>$$ P(A_1|B) = \frac{P(B|A_1)P(A_1)}{P(B)} $$</p>
<p>First, we calculate each term individually :</p>
<p>1. The probability of choosing any urn, since they are chosen at random :</p>
<p>$$ P(A_1) = P(A_2) = P(A_3) = \frac{1}{3} $$</p>
<p>2. The probability of drawing a black ball from each urn :</p>
<p>Urn A: $$ P(B|A_1) = \frac{5}{12} $$</p>
<p>Urn B: $$ P(B|A_2) = \frac{7}{12} $$</p>
<p>Urn C: $$ P(B|A_3) = \frac{6}{12} = \frac{1}{2} $$</p>
<p>3. The total probability of drawing a black ball, $$ P(B) $$ :</p>
<p>$$ P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + P(B|A_3)P(A_3) $$</p>
<p>$$ P(B) = \frac{5}{12} \cdot \frac{1}{3} + \frac{7}{12} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} $$</p>
<p>$$ P(B) = \frac{5}{36} + \frac{7}{36} + \frac{6}{36} $$</p>
<p>$$ P(B) = \frac{18}{36} = \frac{1}{2} $$</p>
<p>Now, substitute these into Bayes' theorem :</p>
<p>$$ P(A_1|B)= \frac{P(B|A_1)P(A_1)}{P(B)} $$</p>
<p>$$ P(A_1|B) = \frac{\frac{5}{12} \cdot \frac{1}{3}}{\frac{1}{2}} $$</p>
<p>$$ P(A_1|B) = \frac{\frac{5}{36}}{\frac{1}{2}} $$</p>
<p>$$ P(A_1|B) = \frac{5}{18} $$</p>
<p>Therefore, the probability that the black ball drawn is from urn A is option C :</p>
<p>$$ \frac{5}{18} $$</p> | mcq | jee-main-2024-online-4th-april-morning-shift |
lv3ve61j | maths | probability | bayes-theorem | <p>There are three bags $$X, Y$$ and $$Z$$. Bag $$X$$ contains 5 one-rupee coins and 4 five-rupee coins; Bag $$Y$$ contains 4 one-rupee coins and 5 five-rupee coins and Bag $$Z$$ contains 3 one-rupee coins and 6 five-rupee coins. A bag is selected at random and a coin drawn from it at random is found to be a one-rupee coin. Then the probability, that it came from bag $$\mathrm{Y}$$, is :</p> | [{"identifier": "A", "content": "$$\\frac{1}{2}$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{5}{12}$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{4}$$"}] | ["B"] | null | <p>To solve this problem, we use Bayes' theorem. Let's define the events:</p>
<ul>
<li>$$X$$: Selecting bag $$X$$</li>
<li>$$Y$$: Selecting bag $$Y$$</li>
<li>$$Z$$: Selecting bag $$Z$$</li>
<li>$$A$$: Drawing a one-rupee coin</li>
</ul>
<p>We are given that a bag is selected at random, so the probabilities for choosing any of the bags are:</p>
<p>
<p>$$P(X) = P(Y) = P(Z) = \frac{1}{3}$$</p>
</p>
<p>Next, we need the probability of drawing a one-rupee coin from each bag:</p>
<ul>
<li>From bag $$X$$: $$P(A|X) = \frac{5}{5+4} = \frac{5}{9}$$</li>
<li>From bag $$Y$$: $$P(A|Y) = \frac{4}{4+5} = \frac{4}{9}$$</li>
<li>From bag $$Z$$: $$P(A|Z) = \frac{3}{3+6} = \frac{3}{9} = \frac{1}{3}$$</li>
</ul>
<p>We need to find the probability that the coin came from bag $$Y$$ given that a one-rupee coin was drawn, i.e., we need $$P(Y|A)$$. Using Bayes' theorem:</p>
<p>
<p>$$P(Y|A) = \frac{P(A|Y) \cdot P(Y)}{P(A)}$$</p>
</p>
<p>To find $$P(A)$$, the total probability of drawing a one-rupee coin can be calculated as follows:</p>
<p>
<p>$$P(A) = P(A|X) \cdot P(X) + P(A|Y) \cdot P(Y) + P(A|Z) \cdot P(Z)$$</p>
</p>
<p>Substituting the values:</p>
<p>
<p>$$P(A) = \left(\frac{5}{9} \cdot \frac{1}{3}\right) + \left(\frac{4}{9} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right)$$</p>
</p>
<p>Calculating the above, we get:</p>
<p>
<p>$$P(A) = \frac{5}{27} + \frac{4}{27} + \frac{1}{9}$$</p>
</p>
<p>Note that $$\frac{1}{9} = \frac{3}{27}$$, so:</p>
<p>
<p>$$P(A) = \frac{5}{27} + \frac{4}{27} + \frac{3}{27} = \frac{12}{27} = \frac{4}{9}$$</p>
</p>
<p>Now, substituting back into Bayes' theorem:</p>
<p>
<p>$$P(Y|A) = \frac{\left(\frac{4}{9}\right) \cdot \left(\frac{1}{3}\right)}{\frac{4}{9}} = \frac{4}{9} \cdot \frac{1}{3} \cdot \frac{9}{4} = \frac{1}{3}$$</p>
</p>
<p>Hence, the probability that the coin came from bag $$Y$$ is:</p>
<p>Option B: $$\frac{1}{3}$$</p> | mcq | jee-main-2024-online-8th-april-evening-shift |
lvc586ak | maths | probability | bayes-theorem | <p>A company has two plants $$A$$ and $$B$$ to manufacture motorcycles. $$60 \%$$ motorcycles are manufactured at plant $$A$$ and the remaining are manufactured at plant $$B .80 \%$$ of the motorcycles manufactured at plant $$A$$ are rated of the standard quality, while $$90 \%$$ of the motorcycles manufactured at plant $$B$$ are rated of the standard quality. A motorcycle picked up randomly from the total production is found to be of the standard quality. If $$p$$ is the probability that it was manufactured at plant $$B$$, then $$126 p$$ is</p> | [{"identifier": "A", "content": "54"}, {"identifier": "B", "content": "66"}, {"identifier": "C", "content": "56"}, {"identifier": "D", "content": "64"}] | ["A"] | null | <p>$$\begin{aligned}
& P(\text { standard automobile from } A)=\frac{6}{10} \times \frac{8}{10}=\frac{12}{25} \\
& P(\text { standard automobile from } B)=\frac{4}{10} \times \frac{9}{10}=\frac{9}{25}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { Required Probability } \frac{\frac{9}{25}}{\frac{12}{25}+\frac{9}{25}} \\
& P=\frac{9}{21}=\frac{3}{7}
\end{aligned}$$</p>
<p>So, $$126 P=126 \times \frac{3}{7}=54$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
byM0olMAPi99wdnA | maths | probability | binomial-distribution | A dice is tossed $$5$$ times. Getting an odd number is considered a success. Then the variance of distribution of success is : | [{"identifier": "A", "content": "$$8/3$$"}, {"identifier": "B", "content": "$$3/8$$"}, {"identifier": "C", "content": "$$4/5$$ "}, {"identifier": "D", "content": "$$5/4$$"}] | ["D"] | null | Total no of trials = 5
<br>$$\therefore$$ n = 5
<br>Odd no possibe are = {1, 3, 5} =3
<br>Sample space = {1, 2, 3, 4, 5, 6} = 6
<br><br>Probablity of getting odd number = $${3 \over 6}$$ = $${1 \over 2}$$
<br>$$\therefore$$ p = $${1 \over 2}$$
<br><br>Formula for variance = npq
<br>where q = 1 - p = 1 - $${1 \over 2}$$ = $${1 \over 2}$$
<br>$$\therefore$$ Variance = $$5 \times {1 \over 2} \times {1 \over 2}$$ = $${5 \over 4}$$ | mcq | aieee-2002 |
KSCoYtIeFC6mdWAN | maths | probability | binomial-distribution | The mean and variance of a random variable $$X$$ having binomial distribution are $$4$$ and $$2$$ respectively, then $$P(X=1)$$ is : | [{"identifier": "A", "content": "$${1 \\over 4}$$ "}, {"identifier": "B", "content": "$${1 \\over 32}$$"}, {"identifier": "C", "content": "$${1 \\over 16}$$"}, {"identifier": "D", "content": "$${1 \\over 8}$$"}] | ["B"] | null | Mean = $$np$$ = 4
<br>Variance = $$npq$$ = 2
<br><br>$$ \Rightarrow 4 \times q$$ = 2
<br>$$ \Rightarrow$$ $$q$$ = $${1 \over 2}$$
<br>p = 1 - q = 1 - $${1 \over 2}$$ = $${1 \over 2}$$
<br>n = 8
<br><br>Now P(X = 1) = $${}^8{C_1}{\left( {{1 \over 2}} \right)^1}{\left( {{1 \over 2}} \right)^{8 - 1}}$$
<br> = $$8 \times {1 \over {{2^8}}}$$
<br> = $${1 \over {32}}$$ | mcq | aieee-2003 |
WwUXVwpKwqJWZr3N | maths | probability | binomial-distribution | The mean and the variance of a binomial distribution are $$4$$ and $$2$$ respectively. Then the probability of $$2$$ successes is : | [{"identifier": "A", "content": "$${28 \\over 256}$$"}, {"identifier": "B", "content": "$${219 \\over 256}$$"}, {"identifier": "C", "content": "$${128 \\over 256}$$"}, {"identifier": "D", "content": "$${37 \\over 256}$$"}] | ["A"] | null | <p>Given that mean = 4 = np = 4 and variance = 2</p>
<p>npq = 2 $$\Rightarrow$$ 4q = 2</p>
<p>$$ \Rightarrow q = {1 \over 2}$$</p>
<p>$$\therefore$$ $$p = 1 - q = 1 - {1 \over 2} = {1 \over 2}$$</p>
<p>Also, n = 8</p>
<p>Probability of 2 successes $$ = P(X = 2) = {}^8{C_2}{p^2}{q^6}$$</p>
<p>$$ = {{8!} \over {2! \times 6!}} \times {\left( {{1 \over 2}} \right)^2} \times {\left( {{1 \over 2}} \right)^6} = 28 \times {1 \over {{2^8}}}$$</p>
<p>$$ = {{28} \over {256}}$$</p> | mcq | aieee-2004 |
R9zlWlJjkftCAb8Q | maths | probability | binomial-distribution | A pair of fair dice is thrown independently three times. The probability of getting a score of exactly $$9$$ twice is : | [{"identifier": "A", "content": "$$8/729$$"}, {"identifier": "B", "content": "$$8/243$$"}, {"identifier": "C", "content": "$$1/729$$"}, {"identifier": "D", "content": "$$8/9.$$"}] | ["B"] | null | <p>Probability of getting score 9 in a single throw</p>
<p>$$ = {4 \over {36}} = {1 \over 9}.$$</p>
<p>Probability of getting score 9 exactly twice</p>
<p>$$ = {}^3{C_2} \times {\left( {{1 \over 9}} \right)^2} \times {8 \over 9} = {8 \over {243}}.$$</p> | mcq | aieee-2007 |
uWKlDoLk0I45E7uG | maths | probability | binomial-distribution | In a binomial distribution $$B\left( {n,p = {1 \over 4}} \right),$$ if the probability of at least one success is greater than or equal to $${9 \over {10}},$$ then $$n$$ is greater than : | [{"identifier": "A", "content": "$${1 \\over {\\log _{10}^4 + \\log _{10}^3}}$$"}, {"identifier": "B", "content": "$${9 \\over {\\log _{10}^4 - \\log _{10}^3}}$$"}, {"identifier": "C", "content": "$${4 \\over {\\log _{10}^4 - \\log _{10}^3}}$$"}, {"identifier": "D", "content": "$${1 \\over {\\log _{10}^4 - \\log _{10}^3}}$$"}] | ["D"] | null | Given that, no of trials = n
<br>Probability of success (p) = $${{1 \over 4}}$$
<br>$$\therefore$$ Probability of no success = 1 - $${{1 \over 4}}$$ = $${{3 \over 4}}$$
<br><br>As we know, probability of at least one success = 1 - probability of no success
<br><br>$$\therefore$$ According to the question,
<br><br>1 - (Probability of no success in n trials) $$\ge$$ $${9 \over {10}}$$
<br><br>$$ \Rightarrow $$ 1 - P(x = 0) $$\ge$$ $${9 \over {10}}$$
<br><br>$$ \Rightarrow $$ 1 - $${}^n{C_0}$$$${\left( {{1 \over 4}} \right)^0}$$$${\left( {{3 \over 4}} \right)^n}$$ $$\ge$$ $${9 \over {10}}$$
<br><br>$$ \Rightarrow $$ 1 - $${\left( {{3 \over 4}} \right)^n}$$ $$\ge$$ $${9 \over {10}}$$
<br><br>$$ \Rightarrow $$ $${\left( {{3 \over 4}} \right)^n}$$ $$\le$$ $${1 \over {10}}$$
<br><br>$$ \Rightarrow $$ $${\left( {{4 \over 3}} \right)^n}$$ $$\ge$$ $$10$$
<br><br>[Taking log on both sides]
<br><br>$$ \Rightarrow $$ $$n\left( {\log _{10}^4 - \log _{10}^3} \right)$$ $$\ge$$ $${\log _{10}^{10}}$$
<br><br>$$ \Rightarrow $$ $$n\left( {\log _{10}^4 - \log _{10}^3} \right)$$ $$\ge$$ $$1$$
<br><br>$$ \Rightarrow $$ $$n$$ $$\ge$$ $${1 \over {\left( {\log _{10}^4 - \log _{10}^3} \right)}}$$
<br><br>$$\therefore$$ Option (D) is correct. | mcq | aieee-2009 |
OE21asjIebiZi6xf | maths | probability | binomial-distribution | Consider $$5$$ independent Bernoulli's trials each with probability of success $$p.$$ If the probability of at least one failure is greater than or equal to $${{31} \over 32},$$ then $$p$$ lies in the interval : | [{"identifier": "A", "content": "$$\\left( {{3 \\over 4},{{11} \\over {12}}} \\right]$$ "}, {"identifier": "B", "content": "$$\\left[ {0,{1 \\over 2}} \\right]$$ "}, {"identifier": "C", "content": "$$\\left( {{11 \\over 12},1} \\right]$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 2},{{3} \\over {4}}} \\right]$$ "}] | ["B"] | null | Here is 5 trials.
<br>So according to Bernoulli trial n = 5
<br><br>P( at least one failure) = 1 - P( no failure)
<br><br>According to question,
<br>1 - P( no failure) $$ \ge {{31} \over {32}}$$
<br><br>$$ \Rightarrow $$ 1 - P( x = 5 ) $$ \ge {{31} \over {32}}$$
<br><br>[<b>Note:</b> no failure = all success]
<br><br>$$ \Rightarrow $$ 1 - $${}^5{C_5}\,{p^5}$$ $$ \ge {{31} \over {32}}$$
<br><br>$$ \Rightarrow $$ 1 - $${p^5}$$ $$ \ge {{31} \over {32}}$$
<br><br>$$ \Rightarrow $$ $${p^5}$$ $$ \le $$ $${1 \over {32}}$$
<br><br>$$ \Rightarrow $$ $$p$$ $$ \le $$ $${1 \over {2}}$$
<br><br>$$\therefore$$ $$p$$ $$ \in $$ $$\left[ {0,{1 \over 2}} \right]$$
<br><br><b>Note:</b> $$p$$ can not be less than 0 as probability is always $$\ge$$ 0 and $$\le$$ 1. | mcq | aieee-2011 |
6W0qksJnJVRp3xfN | maths | probability | binomial-distribution | A multiple choice examination has $$5$$ questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get $$4$$ or more correct answers just by guessing is : | [{"identifier": "A", "content": "$${{17} \\over {{3^5}}}$$ "}, {"identifier": "B", "content": "$${{13} \\over {{3^5}}}$$"}, {"identifier": "C", "content": "$${{11} \\over {{3^5}}}$$"}, {"identifier": "D", "content": "$${{10} \\over {{3^5}}}$$"}] | ["C"] | null | Each question has 3 alternative and exactly one is correct.
<br><br>$$\therefore$$ Probability of giving correct answer P(correct) = $${1 \over 3}$$
<br><br>$$\therefore$$ Probability of giving wrong answer P(wrong) = $${2 \over 3}$$
<br><br>Here student give 4 or more correct answer.
<br><br>$$\therefore$$ Student give either 4 correct answer or 5 correct answer.
<br><br>Using Binomial Distribution,
<br><br>Required probability = $${}^5{C_4}{\left( {{1 \over 3}} \right)^4}{\left( {{2 \over 3}} \right)^1}$$ + $${}^5{C_5}{\left( {{1 \over 3}} \right)^5}{\left( {{2 \over 3}} \right)^0}$$ = $${{11} \over {{3^5}}}$$ | mcq | jee-main-2013-offline |
fPr4yNo79AsCQj24OgUaE | maths | probability | binomial-distribution | An experiment succeeds twice as often as it fails. The probability of at least 5 successes in the six trials of this experiment is : | [{"identifier": "A", "content": "$${{240} \\over {729}}$$ "}, {"identifier": "B", "content": "$${{192} \\over {729}}$$"}, {"identifier": "C", "content": "$${{256} \\over {729}}$$"}, {"identifier": "D", "content": "$${{496} \\over {729}}$$"}] | ["C"] | null | Probability of fail, P(F) = p
<br><br>$$ \therefore $$ Probability of success, P(S) = 2p
<br><br>We know,
<br><br>p + 2Pp = 1
<br><br>$$ \Rightarrow $$ p = $${1 \over 3}$$
<br><br>$$ \therefore $$ Now, probability of at least 5 success in 6 trials,
<br><br>P(x $$ \ge $$ 5)
<br><br>= P(x = 5) + P (x = 6)
<br><br>= <sup>6</sup>C<sub>5</sub> $${\left( {{2 \over 3}} \right)^5}{\left( {{1 \over 3}} \right)^1}{ + ^6}{C_6}{\left( {{2 \over 3}} \right)^6}{\left( {{1 \over 3}} \right)^o}$$
<br><br>= $${\left( {{2 \over 3}} \right)^5}\left( {{6 \over 3} + {2 \over 3}} \right)$$
<br><br>= $${{256} \over {729}}$$ | mcq | jee-main-2016-online-10th-april-morning-slot |
TNLBuSQYBYqruPcC | maths | probability | binomial-distribution | A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with
replacement, then the variance of the number of green balls drawn is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$${6 \\over {25}}$$"}, {"identifier": "D", "content": "$${{12} \\over 5}$$"}] | ["D"] | null | We can apply binomial probability distribution
<br><br>n = 10
<br><br>p = Probability of drawing a green ball = $${{15} \over {25}}$$ = $${3 \over 5}$$
<br><br>Also q = 1 - $${3 \over 5}$$ = $${2 \over 5}$$
<br><br>Variance = npq
<br><br>= $$10 \times {3 \over 5} \times {2 \over 5}$$ = $${{12} \over 5}$$
| mcq | jee-main-2017-offline |
PQxlp1mj9JtBpAOqST5uh | maths | probability | binomial-distribution | If the probability of hitting a target by a shooter, in any shot, is $${1 \over 3}$$, then the minimum number of independent
shots at the target required by him so that the probability of hitting the target atleast once is greater than $${5 \over 6}$$ is :
| [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "3"}] | ["C"] | null | $$1 - {}^n{C_0}{\left( {{1 \over 3}} \right)^0}{\left( {{2 \over 3}} \right)^n} > {5 \over 6}$$
<br><br>$${1 \over 6} > {\left( {{2 \over 3}} \right)^n}\,\, \Rightarrow \,\,0.1666 > {\left( {{2 \over 3}} \right)^n}$$
<br><br>$${n_{\min }} = 5$$ | mcq | jee-main-2019-online-10th-january-evening-slot |
dNyf7SQZWaNvnOFNKi3rsa0w2w9jxb0q6m7 | maths | probability | binomial-distribution | For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability
that the candidate solve any problem is $${4 \over 5}$$
, then the probability that he is unable to solve less than two
problems is : | [{"identifier": "A", "content": "$${{164} \\over {25}}{\\left( {{1 \\over 5}} \\right)^{48}}$$"}, {"identifier": "B", "content": "$${{316} \\over {25}}{\\left( {{4 \\over 5}} \\right)^{48}}$$"}, {"identifier": "C", "content": "$${{201} \\over 5}{\\left( {{1 \\over 5}} \\right)^{49}}$$"}, {"identifier": "D", "content": "$${{54} \\over 5}{\\left( {{4 \\over 5}} \\right)^{49}}$$"}] | ["D"] | null | There are 50 questions in an exam.<br><br>
So Probability of each question to be correct is p = $${4 \over 5}$$<br><br>
and also probability of each question to be incorrect is q = $${1 \over 5}$$<br><br>
Let Y is the number of correct question in 50 questions, hence required probability is <br><br>
$${\left( {{4 \over 5}} \right)^{50}} + {}^{50}{C_1}\left( {{1 \over 5}} \right){\left( {{4 \over 5}} \right)^{49}}$$<br><br>
= $${\left( {{4 \over 5}} \right)^{49}} + \left( {{4 \over 5} + 10} \right)\left( {{4 \over 5}} \right)$$<br><br>
= $${{54} \over 5}{\left( {{4 \over 5}} \right)^{49}}$$ | mcq | jee-main-2019-online-12th-april-evening-slot |
y42Fg1dEZL6zPE14iO3rsa0w2w9jx66ep25 | maths | probability | binomial-distribution | Let a random variable X have a binomial distribution with mean 8 and variance 4. If $$P\left( {X \le 2} \right) = {k \over {{2^{16}}}}$$, then k
is equal to : | [{"identifier": "A", "content": "17"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "137"}, {"identifier": "D", "content": "121"}] | ["C"] | null | Let number of trials be n and probability of success = p, probability of failure = q<br><br>
Given np = 8, npq = 4<br><br>
$$ \Rightarrow $$ q = $${1 \over 2}$$, p = $${1 \over 2}$$, n = 16 (as p + q = 1)<br><br>
p$$\left( {x \le 2} \right)$$ = $${{{}^{16}{C_0} + {}^{16}{C_1} + {}^{16}{C_2}} \over {{2^{16}}}} = {{1 + 16 + 120} \over {{2^{16}}}} = {{137} \over {{2^{16}}}}$$ | mcq | jee-main-2019-online-12th-april-morning-slot |
8LM5BOab51jC57AsuYgDh | maths | probability | binomial-distribution | A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then $$\left( {{{mean\,\,of\,X} \over {s\tan dard\,\,deviation\,\,of\,X}}} \right)$$ is equal to : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "$$3\\sqrt 2 $$"}, {"identifier": "C", "content": "$${{4\\sqrt 3 } \\over 3}$$"}, {"identifier": "D", "content": "$$4\\sqrt 3 $$"}] | ["D"] | null | p (probability of getting white ball) = $${{30} \over {40}}$$
<br><br>q = $${1 \over 4}$$ and n = 16
<br><br>mean = np = 16.$${3 \over 4}$$ = 12
<br><br>and standard diviation
<br><br>= $$\sqrt {npq} $$ = $$\sqrt {16.{3 \over 4}.{1 \over 4}} = \sqrt 3 $$<br><br>
$$ \because $$ $$mean \over standard\,deviation$$ = $$12\over{\sqrt3}$$ = $$4{\sqrt3}$$ | mcq | jee-main-2019-online-11th-january-evening-slot |
sLnGz2MXngSYJtUcmwWgT | maths | probability | binomial-distribution | Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then P(X = 1) + P (X = 2) equals : | [{"identifier": "A", "content": "$$25 \\over 169$$"}, {"identifier": "B", "content": "$$49\\over 169$$"}, {"identifier": "C", "content": "$$24 \\over 169$$"}, {"identifier": "D", "content": "$$52 \\over 169$$"}] | ["A"] | null | P (X = 1) means out of two drawn cards one card is ace.
<br><br>and P(X = 2) means both the drawn cards are ace.
<br><br>$$ \therefore $$ P(X = 1) = first card is ace or 2nd card is ace.
<br><br>= A _ $$+$$ _ A
<br><br>= $${4 \over {52}} \times {{48} \over {52}} + {{48} \over {52}} \times {4 \over {52}}$$
<br><br>= $$2 \times {4 \over {52}} \times {{48} \over {52}}$$
<br><br>P(X = 2) = First and second both cards arc ace.
<br><br>= A A
<br><br>= $${4 \over {52}} \times {4 \over {52}}$$
<br><br>$$ \therefore $$ P(X = 1) + P(X = 2)
<br><br>= $$2 \times {4 \over {52}} \times {{48} \over {52}} + {4 \over {52}} \times {4 \over {52}}$$
<br><br>= $${{25} \over {169}}$$ | mcq | jee-main-2019-online-9th-january-morning-slot |
2nkSMMiE7kPpEjjIP47k9k2k5filgzm | maths | probability | binomial-distribution | In a workshop, there are five machines and the probability of any one of them to be out of service on a day is $${{1 \over 4}}$$
. If the probability that at most two machines will be out of service on the same day is $${\left( {{3 \over 4}} \right)^3}k$$, then k is equal to : | [{"identifier": "A", "content": "$${{{17} \\over 4}}$$"}, {"identifier": "B", "content": "$${{{17} \\over 2}}$$"}, {"identifier": "C", "content": "$${{{17} \\over 8}}$$"}, {"identifier": "D", "content": "4"}] | ["C"] | null | Probablity of at most two machines will be out of service = $${\left( {{3 \over 4}} \right)^3}k$$
<br><br>$$ \Rightarrow $$ <sup>5</sup>C<sub>0</sub>$${\left( {{3 \over 4}} \right)^5}$$ + <sup>5</sup>C<sub>1</sub>$$\left( {{1 \over 4}} \right){\left( {{3 \over 4}} \right)^5}$$ + <sup>5</sup>C<sub>2</sub>$${\left( {{1 \over 4}} \right)^2}{\left( {{3 \over 4}} \right)^5}$$ = $${\left( {{3 \over 4}} \right)^3}k$$
<br><br>$$ \Rightarrow $$ $${{17} \over 8}{\left( {{3 \over 4}} \right)^3}$$ = $${\left( {{3 \over 4}} \right)^3}k$$
<br><br>$$ \Rightarrow $$ k = $${{17} \over 8}$$ | mcq | jee-main-2020-online-7th-january-evening-slot |
ypdvxtIllDnLKrAoI3jgy2xukfqfwl3g | maths | probability | binomial-distribution | In a bombing attack, there is 50% chance that
a bomb will hit the target. Atleast two
independent hits are required to destroy the
target completely. Then the minimum number
of bombs, that must be dropped to ensure that
there is at least 99% chance of completely
destroying the target, is __________. | [] | null | 11 | Let n is total no. of bombs being dropped
<br><br>at least 2 bombs should hit.
<br><br>P(x > 2) $$ \ge $$ 0.99
<br><br>$$ \Rightarrow $$ 1 - p(x < 2) $$ \ge $$ 0.99
<br><br>$$ \Rightarrow $$ 1 - (p(x = 0) + p(x = 1)) $$ \ge $$ 0.99
<br><br>$$ \Rightarrow $$ 1 - <sup>n</sup>C<sub>0</sub>$${\left( {{1 \over 2}} \right)^0}{\left( {{1 \over 2}} \right)^n}$$ - <sup>n</sup>C<sub>1</sub>.$${\left( {{1 \over 2}} \right)^1}{\left( {{1 \over 2}} \right)^{n - 1}}$$ $$ \ge $$ 0.99
<br><br>$$ \Rightarrow $$ 1 - $${1 \over {{2^n}}}$$ - $${n \over {{2^n}}}$$ $$ \ge $$ $${{99} \over {100}}$$
<br><br>$$ \Rightarrow $$ $${1 \over {100}}$$ $$ \ge $$ $${{n + 1} \over {{2^n}}}$$
<br><br>$$ \Rightarrow $$ 2<sup>n</sup> $$ \ge $$ 100(n + 1)
<br><br>Now checking for value of n, we get
<br><br>n = 11 | integer | jee-main-2020-online-5th-september-evening-slot |
wcKeZHloIi9ghYWsTA1klrhaah4 | maths | probability | binomial-distribution | An ordinary dice is rolled for a certain number of times. If the probability of getting an odd
number 2 times is equal to the probability of getting an even number 3 times, then the
probability of getting an odd number for odd number of times is : | [{"identifier": "A", "content": "$${5 \\over {36}}$$"}, {"identifier": "B", "content": "$${3 \\over {16}}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over {32}}$$"}] | ["C"] | null | P(odd no. twice) = P(even no. thrice)<br><br>$$ \Rightarrow {}^n{C_2}{\left( {{1 \over 2}} \right)^n} = {}^n{C_3}{\left( {{1 \over 2}} \right)^n} \Rightarrow n = 5$$<br><br>Success is getting an odd number then P(odd successes) = P(1) + P(3) + P(5)<br><br>$$ = {}^5{C_1}{\left( {{1 \over 2}} \right)^5} + {}^5{C_3}{\left( {{1 \over 2}} \right)^5} + {}^5{C_5}{\left( {{1 \over 2}} \right)^5}$$<br><br>$$ = {{16} \over {{2^5}}} = {1 \over 2}$$ | mcq | jee-main-2021-online-24th-february-morning-slot |
jvIiWm7JdY4spvuhU21klugrqgz | maths | probability | binomial-distribution | A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is : | [{"identifier": "A", "content": "$${{15} \\over {{2^8}}}$$"}, {"identifier": "B", "content": "$${{15} \\over {{2^{12}}}}$$"}, {"identifier": "C", "content": "$${{15} \\over {{2^{13}}}}$$"}, {"identifier": "D", "content": "$${{15} \\over {{2^{14}}}}$$"}] | ["C"] | null | Let the coin be tossed n-times<br><br>$$P(H) = P(T) = {1 \over 2}$$<br><br>P(7 heads) = $${}^n{C_7}{\left( {{1 \over 2}} \right)^{n - 7}}{\left( {{1 \over 2}} \right)^7} = {{{}^n{C_7}} \over {{2^n}}}$$<br><br>P(9 heads) = $${}^n{C_9}{\left( {{1 \over 2}} \right)^{n - 9}}{\left( {{1 \over 2}} \right)^9} = {{{}^n{C_9}} \over {{2^n}}}$$<br><br>P(7 heads) = P(9 heads)<br><br>$${}^n{C_7} = {}^n{C_9} \Rightarrow n = 16$$<br><br>P(2 heads) = $${}^{16}{C_2}{\left( {{1 \over 2}} \right)^{14}}{\left( {{1 \over 2}} \right)^2} = {{15 \times 8} \over {{2^{16}}}}$$<br><br>P(2 heads) $$ = {{15} \over {{2^{13}}}}$$ | mcq | jee-main-2021-online-26th-february-morning-slot |
uamm4PJDXxn7SEVBRT1kmm2wyow | maths | probability | binomial-distribution | Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to : | [{"identifier": "A", "content": "$${{40} \\over {243}}$$"}, {"identifier": "B", "content": "$${{128} \\over {625}}$$"}, {"identifier": "C", "content": "$${{80} \\over {243}}$$"}, {"identifier": "D", "content": "$${{32} \\over {625}}$$"}] | ["D"] | null | $${}^5{C_1}{p^1}{q^4}$$ = 0.4096 ..... (1)<br><br>$${}^5{C_2}{p^2}{q^3}$$ = 0.2048 ..... (2)<br><br>$${{(1)} \over {(2)}} \Rightarrow {q \over {2p}} = 2 \Rightarrow q = 4p$$<br><br>$$p + q = 1 \Rightarrow P = {1 \over 5},q = {4 \over 5}$$<br><br>P (exactly 3) = $${}^5{C_3}{(p)^3}{(q)^2} = {}^5{C_3}{\left( {{1 \over 5}} \right)^3}{\left( {{4 \over 5}} \right)^2}$$<br><br>$$ = 10 \times {1 \over {125}} \times {{16} \over {25}} = {{32} \over {625}}$$ | mcq | jee-main-2021-online-18th-march-evening-shift |
1krxij6cm | maths | probability | binomial-distribution | A student appeared in an examination consisting of 8 true-false type questions. The student guesses the answers with equal probability.
the smallest value of n, so that the probability of guessing at least 'n' correct answers is less than $${1 \over 2}$$, is : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["A"] | null | $$P(E) < {1 \over 2}$$<br><br>$$ \Rightarrow \sum\limits_{r = n}^8 {{}^8{C_r}} {\left( {{1 \over 2}} \right)^{8 - r}}{\left( {{1 \over 2}} \right)^r} < {1 \over 2}$$<br><br>$$ \Rightarrow \sum\limits_{r = n}^8 {{}^8{C_r}} {\left( {{1 \over 2}} \right)^8} < {1 \over 2}$$<br><br>$$ \Rightarrow {}^8{C_n} + {}^8{C_{n + 1}} + .... + {}^8{C_8} < 128$$<br><br>$$ \Rightarrow 256 - \left( {{}^8{C_0} + {}^8{C_1} + ... + {}^8{C_{n - 1}}} \right) < 128$$<br><br>$$ \Rightarrow {}^8{C_0} + {}^8{C_1} + .... + {}^8{C_{n - 1}} > 128$$<br><br>$$ \Rightarrow n - 1 \ge 4$$<br><br>$$ \Rightarrow n \ge 5$$ | mcq | jee-main-2021-online-27th-july-evening-shift |
1ktfz7m08 | maths | probability | binomial-distribution | Each of the persons A and B independently tosses three fair coins. The probability that both of them get the same number of heads is : | [{"identifier": "A", "content": "$${1 \\over 8}$$"}, {"identifier": "B", "content": "$${5 \\over 8}$$"}, {"identifier": "C", "content": "$${5 \\over 16}$$"}, {"identifier": "D", "content": "1"}] | ["C"] | null | <p>Let x be the number of heads obtained by A, and y be the number of heads obtained by B.</p>
<p>Note that x and y are binomial variable with parameters n = 3 and p = $${1 \over 2}$$</p>
<p>$$\therefore$$ Probability that both A and B obtained the same number of heads is</p>
<p>$$ = P(x = 0)\,.\,P(y = 0) + P(x = 1)\,.\,P(y = 1) + P(x = 2)\,.\,P(y = 2) + P(x = 3)\,.\,P(y = 3)$$</p>
<p>$$ = {\left[ {{3_{{C_0}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2} + {\left[ {{3_{{C_1}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2} + {\left[ {{3_{{C_2}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2} + {\left[ {{3_{{C_3}}}{{\left( {{1 \over 2}} \right)}^3}} \right]^2}$$</p>
<p>$$ = {\left( {{1 \over 2}} \right)^6}\left[ {1 + 9 + 9 + 1} \right]$$</p>
<p>$$ = {{20} \over {64}}$$</p>
<p>$$ = {5 \over {16}}$$</p> | mcq | jee-main-2021-online-27th-august-evening-shift |
1l57op7rj | maths | probability | binomial-distribution | <p>Let X be a random variable having binomial distribution B(7, p). If P(X = 3) = 5P(x = 4), then the sum of the mean and the variance of X is :</p> | [{"identifier": "A", "content": "$${105 \\over {16}}$$"}, {"identifier": "B", "content": "$${7\\over {16}}$$"}, {"identifier": "C", "content": "$${77\\over {36}}$$"}, {"identifier": "D", "content": "$${49\\over {16}}$$"}] | ["C"] | null | <p>Given P(X = 3) = 5P(X = 4) and n = 7</p>
<p>$$ \Rightarrow {}^7{C_3}{p^3}{q^4} = 5\,.\, \Rightarrow {}^7{C_4}{p^4}{q^3}$$</p>
<p>$$ \Rightarrow q = 5p$$ and also $$p + q = 1$$</p>
<p>$$ \Rightarrow p = {1 \over 6}$$ and $$q = {5 \over 6}$$</p>
<p>Mean $$ = {7 \over 6}$$ and variance $$ = {{35} \over {36}}$$</p>
<p>Mean + Variance $$ = {7 \over 6} + {{35} \over {36}} = {{77} \over {36}}$$</p> | mcq | jee-main-2022-online-27th-june-morning-shift |
1l58a4o04 | maths | probability | binomial-distribution | <p>Let a biased coin be tossed 5 times. If the probability of getting 4 heads is equal to the probability of getting 5 heads, then the probability of getting atmost two heads is :</p> | [{"identifier": "A", "content": "$${{275} \\over {{6^5}}}$$"}, {"identifier": "B", "content": "$${{36} \\over {{5^4}}}$$"}, {"identifier": "C", "content": "$${{181} \\over {{5^5}}}$$"}, {"identifier": "D", "content": "$${{46} \\over {{6^4}}}$$"}] | ["D"] | null | <p>Coin is tossed 5 times, so n = 5</p>
<p>Let, p = probability of getting heads</p>
<p>q = probability of getting tails.</p>
<p>$$\therefore$$ p + q = 1 ...... (1)</p>
<p>$$\therefore$$ Probability of getting 4 heads</p>
<p>= <sup>5</sup>C<sub>4</sub> . p<sup>4</sup> . q</p>
<p>And probability of getting 5 heads</p>
<p>= <sup>5</sup>C<sub>5</sub> . p<sup>5</sup></p>
<p>Given, <sup>5</sup>C<sub>4</sub> . p<sup>4</sup> . q = <sup>5</sup>C<sub>5</sub> . p<sup>5</sup></p>
<p>$$\Rightarrow$$ 5q = p ....... (2)</p>
<p>From equation (1) and (2), we get,</p>
<p>5q + q = 1</p>
<p>$$\Rightarrow$$ 6q = 1</p>
<p>$$\Rightarrow$$ q = $${1 \over 6}$$</p>
<p>$$\therefore$$ p = 1 $$-$$ $${1 \over 6}$$ = $${5 \over 6}$$</p>
<p>Now, probability of getting atmost two heads</p>
<p>= p (x = 0) + p (x = 1) + p (x = 2)</p>
<p>p (x = 0) = Getting zero head in 5 trials</p>
<p>= <sup>5</sup>C<sub>0</sub> . p<sup>0</sup> . q<sup>5</sup></p>
<p>p (x = 1) = Getting one head in 5 trials</p>
<p>= <sup>5</sup>C<sub>1</sub> . p<sup>1</sup> . q<sup>4</sup></p>
<p>p (x = 2) = Getting two heads in 5 trials</p>
<p>= <sup>5</sup>C<sub>2</sub> . p<sup>2</sup> . q<sup>3</sup></p>
<p>= <sup>5</sup>C<sub>0</sub> . q<sup>5</sup> + <sup>5</sup>C<sub>1</sub> . pq<sup>4</sup> + <sup>5</sup>C<sub>2</sub> . p<sup>2</sup>q<sup>3</sup></p>
<p>$$ = {\left( {{1 \over 6}} \right)^5} + 5\,.\,{5 \over 6}\,.\,{\left( {{1 \over 6}} \right)^4} + 10\,.\,{\left( {{5 \over 6}} \right)^2}\,.\,{\left( {{1 \over 6}} \right)^3}$$</p>
<p>$$ = {{1 + 25 + 250} \over {{6^5}}} = {{276} \over {{6^5}}}$$ = $${{46} \over {{6^4}}}$$ | mcq | jee-main-2022-online-26th-june-morning-shift |
1l5bb8fiy | maths | probability | binomial-distribution | <p>In an examination, there are 10 true-false type questions. Out of 10, a student can guess the answer of 4 questions correctly with probability $${3 \over 4}$$ and the remaining 6 questions correctly with probability $${1 \over 4}$$. If the probability that the student guesses the answers of exactly 8 questions correctly out of 10 is $${{{{27}k}} \over {{4^{10}}}}$$, then k is equal to ___________.</p> | [] | null | 479 | <p>Student guesses only two wrong. So there are three possibilities.</p>
<p>(i) Student guesses both wrong from 1<sup>st</sup> section</p>
<p>(ii) Student guesses both wrong from 2<sup>nd</sup> section</p>
<p>(iii) Student guesses two wrong one from each section</p>
<p>Required probabilities</p>
<p>$$ = {}^4{C_2}{\left( {{3 \over 4}} \right)^2}{\left( {{1 \over 4}} \right)^2}{\left( {{1 \over 6}} \right)^6} + {}^6{C_2}{\left( {{3 \over 4}} \right)^2}{\left( {{1 \over 4}} \right)^4}{\left( {{3 \over 4}} \right)^4} + {}^4{C_1}\,.\,{}^6{C_1}\left( {{3 \over 4}} \right)\left( {{1 \over 4}} \right){\left( {{3 \over 4}} \right)^3}{\left( {{1 \over 4}} \right)^5}$$</p>
<p>$$ = {1 \over {{4^{10}}}}\left[ {6 \times 9 + 15 \times {9^4} + 24 \times {9^2}} \right]$$</p>
<p>$$ = {{27} \over {{4^{10}}}}\left[ {2 + 27 \times 15 + 72} \right]$$</p>
<p>$$ = {{27 \times 479} \over {{4^{10}}}}$$</p> | integer | jee-main-2022-online-24th-june-evening-shift |
1l5c1q77f | maths | probability | binomial-distribution | <p>If a random variable X follows the Binomial distribution B(33, p) such that <br/><br/>$$3P(X = 0) = P(X = 1)$$, then the value of $${{P(X = 15)} \over {P(X = 18)}} - {{P(X = 16)} \over {P(X = 17)}}$$ is equal to :</p> | [{"identifier": "A", "content": "1320"}, {"identifier": "B", "content": "1088"}, {"identifier": "C", "content": "$${{120} \\over {1331}}$$"}, {"identifier": "D", "content": "$${{1088} \\over {1089}}$$"}] | ["A"] | null | $3 P(X=0)=P(X=1)$
<br/><br/>
$$
\begin{aligned}
&3 \cdot{ }^{n} C_{0} P^{0}(1-P)^{n}={ }^{n} C_{1} P^{1}(1-P)^{n-1} \\\\
&\frac{3}{n}=\frac{P}{1-P} \Rightarrow \frac{1}{11}=\frac{P}{1-P} \\\\
&\Rightarrow 1-P=11 P \\\\
&\Rightarrow P=\frac{1}{12}
\end{aligned}
$$
<br/><br/>
$$
\frac{P(X=15)}{P(X=18)}-\frac{P(X=16)}{P(X=17)}
$$
<br/><br/>
$$
\begin{aligned}
&\Rightarrow \frac{{ }^{33} C_{15} P^{15}(1-P)^{18}}{{ }^{33} C_{18} P^{18}(1-P)^{15}}-\frac{{ }^{33} C_{16} P^{16}(1-P)^{17}}{{ }^{33} C_{17} P^{17}(1-P)^{16}} \\\\
&\Rightarrow\left(\frac{1-P}{P}\right)^{3}-\left(\frac{1-P}{P}\right) \\\\
&\Rightarrow \quad 11^{3}-11=1320
\end{aligned}
$$ | mcq | jee-main-2022-online-24th-june-morning-shift |
1l5w0d935 | maths | probability | binomial-distribution | <p>If a random variable X follows the Binomial distribution B(5, p) such that P(X = 0) = P(X = 1), then $${{P(X = 2)} \over {P(X = 3)}}$$ is equal to :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "5"}] | ["D"] | null | <p>Given,</p>
<p>$$n = 5$$</p>
<p>and $$P(X = 0) = P(X = 1)$$</p>
<p>We know, $$p(x = r) = {}^n{C_r}\,.\,{p^r}\,.\,{q^{n - r}}$$</p>
<p>where $$p + q = 1$$</p>
<p>$$\therefore$$ $$P(X = 0) = P(X = 1)$$</p>
<p>$$ \Rightarrow {}^5{C_0}\,.\,{p^0}\,.\,{q^5} = {}^5{C_1}\,.\,{p^1}\,.\,{q^4}$$</p>
<p>$$ \Rightarrow 1\,.\,1\,.\,{(1 - p)^5} = 5\,.\,p\,.\,{(1 - p)^4}$$</p>
<p>$$ \Rightarrow 1 - p = 5p$$</p>
<p>$$ \Rightarrow 6p = 1$$</p>
<p>$$ \Rightarrow p = {1 \over 6}$$</p>
<p>$$\therefore$$ $$q = 1 - p = 1 - {1 \over 6} = {5 \over 6}$$</p>
<p>Now, $${{P(X = 2)} \over {P(X = 3)}} = {{{}^5{C_2}\,.\,{p^2}\,.\,{q^3}} \over {{}^5{C_3}\,.\,{p^3}\,.\,{q^2}}}$$</p>
<p>$$ = {{10\,.\,q} \over {10\,.\,p}}$$</p>
<p>$$ = {5 \over 6} \times {6 \over 1}$$</p>
<p>$$ = 5$$</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6dwkw92 | maths | probability | binomial-distribution | <p>If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is :
</p> | [{"identifier": "A", "content": "$$\n\\frac{33}{2^{32}}\n$$"}, {"identifier": "B", "content": "$$\\frac{33}{2^{29}}$$"}, {"identifier": "C", "content": "$$\\frac{33}{2^{28}}$$"}, {"identifier": "D", "content": "$$\\frac{33}{2^{27}}$$"}] | ["C"] | null | If $n$ is number of trails, $p$ is probability of success and $q$ is probability of unsuccess then,
<br/><br/>
$$
\begin{aligned}
& \text { Mean }=n p \text { and variance }=n p q \text {. } \\\\
& \text { Here }\\\\
& n p+n p q=24 \quad \dots(i)\\\\
& n p . n p q=128 \quad \dots(ii)\\\\
&\text { and } q=1-p \quad \dots(iii)
\end{aligned}
$$
<br/><br/>
from eq. (i), (ii) and (iii) : $p=q=\frac{1}{2}$ and $n=32$.
<br/><br/>
$\therefore$ Required probability $=p(X=1)+p(X=2)$
<br/><br/>
$$
\begin{aligned}
& ={ }^{32} C_{1} \cdot\left(\frac{1}{2}\right)^{32}+{ }^{32} C_{2} \cdot\left(\frac{1}{2}\right)^{32} \\\\
& =\left(32+\frac{32 \times 31}{2}\right) \cdot \frac{1}{2^{32}} \\\\
& =\frac{33}{2^{28}}
\end{aligned}
$$ | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6gisnao | maths | probability | binomial-distribution | <p>The mean and variance of a binomial distribution are $$\alpha$$ and $$\frac{\alpha}{3}$$ respectively. If $$\mathrm{P}(X=1)=\frac{4}{243}$$, then $$\mathrm{P}(X=4$$ or 5$$)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{5}{9}$$"}, {"identifier": "B", "content": "$$\\frac{64}{81}$$"}, {"identifier": "C", "content": "$$\\frac{16}{27}$$"}, {"identifier": "D", "content": "$$\\frac{145}{243}$$"}] | ["C"] | null | <p>Given, mean $$ = np = \alpha $$.</p>
<p>and variance $$ = npq = {\alpha \over 3}$$</p>
<p>$$ \Rightarrow q = {1 \over 3}$$ and $$p = {2 \over 3}$$</p>
<p>$$P(X = 1) = n.{p^1}.{q^{n - 1}} = {4 \over {243}}$$</p>
<p>$$ \Rightarrow n.{2 \over 3}.{\left( {{1 \over 3}} \right)^{n - 1}} = {4 \over {243}}$$</p>
<p>$$ \Rightarrow n = 6$$</p>
<p>$$P(X = 4\,\mathrm{or}\,5) = {}^6{C_4}\,.\,{\left( {{2 \over 3}} \right)^4}\,.\,{\left( {{1 \over 3}} \right)^2} + {}^6{C_5}\,.\,{\left( {{2 \over 5}} \right)^5}\,.\,{1 \over 3}$$</p>
<p>$$ = {{16} \over {27}}$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6hz9u96 | maths | probability | binomial-distribution | <p>Let $$X$$ be a binomially distributed random variable with mean 4 and variance $$\frac{4}{3}$$. Then, $$54 \,P(X \leq 2)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{73}{27}$$"}, {"identifier": "B", "content": "$$\\frac{146}{27}$$"}, {"identifier": "C", "content": "$$\\frac{146}{81}$$"}, {"identifier": "D", "content": "$$\\frac{126}{81}$$"}] | ["B"] | null | <p>Mean $$ = 4 = \mu = np$$</p>
<p>Variance $$ = {\sigma ^2} = np(1 - P) = {4 \over 3}$$</p>
<p>$$4(1 - P) = {4 \over 3}$$</p>
<p>$$P = {2 \over 3}$$</p>
<p>$$n \times {2 \over 3} = 4$$</p>
<p>$$n = 6$$</p>
<p>$$P(X = k) = {}^n{C_k}\,{P^k}{(1 - P)^{n - k}}$$</p>
<p>$$P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)$$</p>
<p>$$ = {}^6{C_0}{P^0}{(1 - P)^6} + {}^6{C_1}{P^1}{(1 - P)^5} + {}^6{C_2}{P^2}{(1 - P)^4}$$</p>
<p>$$ = {}^6{C_0}{\left( {{1 \over 3}} \right)^6} + {}^6{C_1}\left( {{2 \over 3}} \right){\left( {{1 \over 3}} \right)^5} + {}^6{C_2}{\left( {{2 \over 3}} \right)^2}{\left( {{1 \over 3}} \right)^4}$$</p>
<p>$$ = {\left( {{1 \over 3}} \right)^6}[1 + 12 + 60] = {{73} \over {{3^6}}}$$</p>
<p>$$54P\,(X \le 2) = {{73} \over {{3^6}}} \times 54 = {{146} \over {27}}$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6kkqt6p | maths | probability | binomial-distribution | <p>Let X have a binomial distribution B(n, p) such that the sum and the product of the mean and variance of X are 24 and 128 respectively. If $$P(X>n-3)=\frac{k}{2^{n}}$$, then k is equal to :</p> | [{"identifier": "A", "content": "528"}, {"identifier": "B", "content": "529"}, {"identifier": "C", "content": "629"}, {"identifier": "D", "content": "630"}] | ["B"] | null | <p>Mean $$ = np = 16$$</p>
<p>Variance $$ = npq = 8$$</p>
<p>$$ \Rightarrow q = p = {1 \over 2}$$ and $$n = 32$$</p>
<p>$$P(x > n - 3) = p(x = n - 2) + p(x = n - 1) + p(x = n)$$</p>
<p>$$ = \left( {{}^{32}{C_2} + {}^{32}{C_1} + {}^{32}{C_0}} \right)\,.\,{1 \over {{2^n}}}$$</p>
<p>$$ = {{529} \over {{2^n}}}$$</p> | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6rffxn0 | maths | probability | binomial-distribution | <p>The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respectively. Then the number of trials in the binomial distribution is ____________.</p> | [] | null | 96 | Let two roots of a quadratic equation are mean = np and variance = npq.
<br/><br/>Given $n p+n p q=82.5$
<br/><br/>and $n p(n p q)=1350$
<br/><br/>$$ \therefore $$ Quadratic equation is
<br/><br/>$ x^{2}-82.5 x+1350=0$
<br/><br/>$\Rightarrow x^{2}-22.5 x-60 x+1350=0$
<br/><br/>$\Rightarrow x-(x-22.5)-60(x-22.5)=0$
<br/><br/>Mean $=60$ and Variance $=22.5$
<br/><br/>$$
n p=60, n p q=22.5
$$
<br/><br/>$$
\Rightarrow q=\frac{9}{24}=\frac{3}{8}, p=\frac{5}{8}
$$
<br/><br/>$$
\therefore \quad n \frac{5}{8}=60 \quad \Rightarrow n=96
$$ | integer | jee-main-2022-online-29th-july-evening-shift |
1ldomop5k | maths | probability | binomial-distribution | <p>In a binomial distribution $$B(n,p)$$, the sum and the product of the mean and the variance are 5 and 6 respectively, then $$6(n+p-q)$$ is equal to :</p> | [{"identifier": "A", "content": "52"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "51"}, {"identifier": "D", "content": "53"}] | ["A"] | null | $$
\begin{aligned}
& \text { Given } \\\\
& \mathrm{np}+\mathrm{npq}=5 \\\\
& \Rightarrow \mathrm{np}(1+\mathrm{q})=5 ........(i) \\\\
& \text { and (np) (npq) }=6 \\\\
& \Rightarrow \mathrm{n}^2 \mathrm{p}^2 \mathrm{q}=6 ........(ii) \\\\
& (\mathrm{i})^2 \div(\mathrm{ii}) \\\\
& \frac{(1+q)^2}{9}=\frac{25}{6} \\\\
& \Rightarrow 6 \mathrm{q}^2-13 \mathrm{q}+6=0 \\\\
& \Rightarrow \mathrm{q}=\frac{2}{3}, \frac{3}{2} \text { (rejected) } \\\\
& \mathrm{p}=1-\frac{2}{3}=\frac{1}{3} \\\\
& \frac{n}{3}\left(1+\frac{2}{3}\right)=5 \\\\
& \Rightarrow \mathrm{n}=9 \\\\
& 6(\mathrm{n}+\mathrm{p}-\mathrm{q})=52
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-morning-shift |
1lgoxr14i | maths | probability | binomial-distribution | <p>The random variable $$\mathrm{X}$$ follows binomial distribution $$\mathrm{B}(\mathrm{n}, \mathrm{p})$$, for which the difference of the mean and the variance is 1 . If $$2 \mathrm{P}(\mathrm{X}=2)=3 \mathrm{P}(\mathrm{X}=1)$$, then $$n^{2} \mathrm{P}(\mathrm{X}>1)$$ is equal to :</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "16"}] | ["C"] | null | $$
\begin{aligned}
& n p-n p q=1 \\\\
\Rightarrow & n p(1-q)=1 \\\\
\Rightarrow & n p^2=1
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& 2 P(X=2)=3 P(X=1) \\\\
& 2 \cdot{ }^n C_2 p^2 q{ }^{n-2}=3 \cdot{ }^n C_1 p \cdot q^{n-1} \\\\
\Rightarrow & 2 \cdot \frac{n \cdot(n-1)}{2} \cdot p=3 \cdot n \cdot q \\\\
\Rightarrow & (n-1) p=3(1-p) \\\\
\Rightarrow & \left(\frac{1}{p^2}-1\right) p=3(1-p) \\\\
\Rightarrow & \frac{(1-p)(1+p)}{p}=3(1-p) \\\\
\Rightarrow & 1+p=3 p \\\\
\Rightarrow & p=\frac{1}{2} \\\\
\therefore & n=4
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
n^2 P(x>1) & =n^2(1-P(x=1)-P(x=0)) \\\\
& =16\left(1-{ }^4 C_1 \cdot\left(\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^4\right)=11
\end{aligned}
$$ | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgvphj1r | maths | probability | binomial-distribution | <p> Let a die be rolled $$n$$ times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $$\frac{k}{2^{15}}$$, then $$\mathrm{k}$$ is equal to :</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "60"}, {"identifier": "C", "content": "30"}, {"identifier": "D", "content": "90"}] | ["B"] | null | Given that,
<br/><br/>$\mathrm{P}($ odd number seven times $)=\mathrm{P}($ Odd number nine times)
<br/><br/>$$
\begin{aligned}
& \Rightarrow{ }^n \mathrm{C}_7\left(\frac{1}{2}\right)^7\left(\frac{1}{2}\right)^{n-7}={ }^n \mathrm{C}_9\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)^{n-9} \\\\
& \Rightarrow{ }^n \mathrm{C}_7={ }^n \mathrm{C}_9 \\\\
& \Rightarrow n=7+9=16
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Hence, } \mathrm{P}(\text { Even number twice })=\frac{k}{2^{15}} \\\\
& \Rightarrow{ }^{16} \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{16-2}=\frac{k}{2^{15}} \\\\
& \Rightarrow \frac{16 \times 15}{2} \times \frac{1}{2^{16}}=\frac{k}{2^{15}} \Rightarrow k=60
\end{aligned}
$$ | mcq | jee-main-2023-online-10th-april-evening-shift |
1lh203h68 | maths | probability | binomial-distribution | <p>A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is $$\frac{k}{3^{11}}$$, then $$k$$ is equal to :</p> | [{"identifier": "A", "content": "82"}, {"identifier": "B", "content": "164"}, {"identifier": "C", "content": "123"}, {"identifier": "D", "content": "75"}] | ["C"] | null | Given, a pair of dice is thrown once, then number of outcomes $=6 \times 6=36$
<br/><br/>A total of 5 are $\{(1,4),(2,3),(4,1),(3,2)\}$
<br/><br/>$$
\begin{aligned}
& \therefore p =p(\text { success })=\frac{4}{36}=\frac{1}{9} \\\\
& q =1-\frac{1}{9}=\frac{8}{9}
\end{aligned}
$$
<br/><br/>Let $X$ represent the number of success
<br/><br/>$$
\begin{aligned}
& \therefore P(\text { at least } 4 \text { success }) \\\\
&=P(X=4)+P(X=5) \\\\
&={ }^5 C_4 p^4 q+{ }^5 C_5 p^5 q^{5-5} \\\\
&=5\left(\frac{1}{9}\right)^4\left(\frac{8}{9}\right)+\left(\frac{1}{9}\right)^5 \\\\
&=\frac{40+1}{\left(3^2\right)^5}=\frac{41}{310} \\\\
&=\frac{41 \times 3}{3^{11}}=\frac{123}{3^{11}} \\\\
& \therefore k=123
\end{aligned}
$$ | mcq | jee-main-2023-online-6th-april-morning-shift |
lv2er9nl | maths | probability | binomial-distribution | <p>In a tournament, a team plays 10 matches with probabilities of winning and losing each match as $$\frac{1}{3}$$ and $$\frac{2}{3}$$ respectively. Let $$x$$ be the number of matches that the team wins, and $y$ be the number of matches that team loses. If the probability $$\mathrm{P}(|x-y| \leq 2)$$ is $$p$$, then $$3^9 p$$ equals _________.</p> | [] | null | 8288 | <p>$$\begin{aligned}
& x+y=10 \\
& A=x-y \\
& P(|A|<2) \text { is } P \\
& \Rightarrow|A|=2,1,0 \Rightarrow A=0,1,-1,2,-2 \\
& \Rightarrow x=\frac{10+A}{2} \Rightarrow A \in \text { even as } x \in \text { integer } \\
& \Rightarrow A=0,-2,2 \\
& \Rightarrow P(|A| \leq 2)=P(A=0)+P(A=-2)+P(A=2)
\end{aligned}$$</p>
<p>(1) $$A=0 \Rightarrow x=5=y$$</p>
<p>$$P(A=0)={ }^{10} C_5\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^5$$</p>
<p>(2) $$A=-2$$</p>
<p>$$\Rightarrow x=4$$ and $$y=6$$</p>
<p>$$P(A=-2)={ }^{10} C_4 \cdot\left(\frac{1}{3}\right)^4\left(\frac{2}{3}\right)^6$$ and</p>
<p>$$\begin{aligned}
& \text { Similarly, } P(A=2)={ }^{10} C_6\left(\frac{1}{3}\right)^6\left(\frac{2}{3}\right)^4 \\
& \Rightarrow P(|A| \leq 2) 3^9=3\left({ }^{10} C_5 \cdot 2^5+{ }^{10} C_4 \cdot 2^6+{ }^{10} C_6 \cdot 2^4\right) \\
& =8288
\end{aligned}$$</p> | integer | jee-main-2024-online-4th-april-evening-shift |
gj3cfkAKO4sn1XEq | maths | probability | classical-defininition-of-probability | Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is : | [{"identifier": "A", "content": "$${{2 \\over 5}}$$"}, {"identifier": "B", "content": "$${{4 \\over 5}}$$"}, {"identifier": "C", "content": "$${{3 \\over 5}}$$"}, {"identifier": "D", "content": "$${{1 \\over 5}}$$"}] | ["A"] | null | <p>X : Mr. A selected wining horse</p>
<p>$$\overline X $$ : Mr. A did not select wining horse</p>
<p>Mr. A selected two horses, now probability of not wining the first horse which Mr. A choses = $${4 \over 5}$$</p>
<p>And probability of not wining the second horse also which Mr. A choses = $${3 \over 4}$$ (Here Mr. A out of remaining 4 horses choses one horse among 3 horses which did not win)</p>
<p>$$\therefore$$ $$P(\overline X ) = {4 \over 5} \times {3 \over 4}$$</p>
<p>$$\therefore$$ $$P(X) = 1 - P(\overline X )$$</p>
<p>$$ = 1 - {4 \over 5} \times {3 \over 4}$$</p>
<p>$$ = 1 - {3 \over 5}$$</p>
<p>$$ = {2 \over 5}$$</p> | mcq | aieee-2003 |
9sHWR6rtY5FcXa4Y | maths | probability | classical-defininition-of-probability | The probability that $$A$$ speaks truth is $${4 \over 5},$$ while the probability for $$B$$ is $${3 \over 4}.$$ The probability that they contradict each other when asked to speak on a fact is : | [{"identifier": "A", "content": "$${4 \\over 5}$$ "}, {"identifier": "B", "content": "$${1 \\over 5}$$"}, {"identifier": "C", "content": "$${7 \\over 20}$$"}, {"identifier": "D", "content": "$${3 \\over 20}$$"}] | ["C"] | null | <p>The probability of speaking truth by A, P(A) = $${4 \over 5}$$. The probability of not speaking truth by A, P($$\overline A $$) = 1 $$-$$ $${4 \over 5} = {1 \over 5}$$.</p>
<p>The probability of speaking truth by B, P(B) = $${3 \over 4}$$. The probability of not speaking truth of B, P($$\overline B $$) $$ = {1 \over 4}$$.</p>
<p>The probability that they contradict each other</p>
<p>$$ = P(A) \times P(\overline B ) + P(\overline A ) \times P(B)$$</p>
<p>$$ = {4 \over 5} \times {1 \over 4} + {1 \over 5} \times {3 \over 4}$$</p>
<p>$$ = {1 \over 5} + {3 \over {20}} = {7 \over {20}}$$</p> | mcq | aieee-2004 |
4KumFwAAvt6eWfbO | maths | probability | classical-defininition-of-probability | Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is : | [{"identifier": "A", "content": "$${2 \\over 9}$$"}, {"identifier": "B", "content": "$${1 \\over 9}$$"}, {"identifier": "C", "content": "$${8 \\over 9}$$"}, {"identifier": "D", "content": "$${7 \\over 9}$$"}] | ["B"] | null | <p>Person 1st has three options to apply.</p>
<p>Similarly, person 2nd has three options t apply</p>
<p>and person 3rd has three options to apply.</p>
<p>Total cases = 3<sup>3</sup></p>
<p>Now, favourable cases = 3 (An either all has applied for house 1 or 2 or 3)</p>
<p>So, probability $$ = {3 \over {{3^3}}} = {1 \over 9}$$.</p> | mcq | aieee-2005 |
5g0YZMGZ3okC9vsb | maths | probability | classical-defininition-of-probability | A die is thrown. Let $$A$$ be the event that the number obtained is greater than $$3.$$ Let $$B$$ be the event that the number obtained is less than $$5.$$ Then $$P\left( {A \cup B} \right)$$ is : | [{"identifier": "A", "content": "$${3 \\over 5}$$ "}, {"identifier": "B", "content": "$$0$$"}, {"identifier": "C", "content": "$$1$$"}, {"identifier": "D", "content": "$${2 \\over 5}$$"}] | ["C"] | null | No of outcome for a die = { 1, 2, 3, 4, 5, 6 }
<br>According to the question,
<br><br>A = { 4, 5, 6 }
<br><br>$$\therefore$$ P(A) = $${3 \over 6}$$
<br><br>B = { 1, 2, 3, 4 }
<br><br>$$\therefore$$ P(A) = $${4 \over 6}$$
<br><br>A $$ \cap $$ B = { 4 }
<br><br>So P(A $$ \cap $$ B) = $${1 \over 6}$$
<br><br>We know, $$P\left( {A \cup B} \right)$$ = $$P\left( A \right)$$ + $$P\left( B \right)$$ - $$P\left( {A \cap B} \right)$$
<br><br>$$\therefore$$ $$P\left( {A \cup B} \right)$$ = $${3 \over 6}$$ + $${4 \over 6}$$ - $${1 \over 6}$$ = $${{7 - 1} \over 6}$$ = $${6 \over 6}$$ = 1
<br><br>$$\therefore$$ Option (C) is correct. | mcq | aieee-2008 |
K13pUy0l2sZ1pymP | maths | probability | classical-defininition-of-probability | If two different numbers are taken from the set {0, 1, 2, 3, ........, 10}; then the probability that their sum as
well as absolute difference are both multiple of 4, is : | [{"identifier": "A", "content": "$${{12} \\over {55}}$$"}, {"identifier": "B", "content": "$${{14} \\over {45}}$$"}, {"identifier": "C", "content": "$${{7} \\over {55}}$$"}, {"identifier": "D", "content": "$${{6} \\over {55}}$$"}] | ["D"] | null | Let A = {0, 1, 2, 3, 4, ......., 10}
<br><br>Total number of ways of selecting 2 different numbers from A is
<br><br>n (S) = <sup>11</sup>C<sub>2</sub> = 55, where 'S' denotes sample space
<br><br>Let E be the given event
<br><br>$$ \therefore $$ E = {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}
<br><br>$$ \Rightarrow $$ n (E) = 6
<br><br>$$ \therefore $$ P(E) = $${{n\left( E \right)} \over {n\left( S \right)}}$$ = $${6 \over {55}}$$ | mcq | jee-main-2017-offline |
yaHW5wRBHJjSQGuLekwWw | maths | probability | classical-defininition-of-probability | Three persons P, Q and R independently try to hit a target. I the probabilities of their hitting the target are $${3 \over 4},{1 \over 2}$$ and $${5 \over 8}$$ respectively, then the probability that the target is hit by P or Q but not by R is : | [{"identifier": "A", "content": "$${{21} \\over {64}}$$ "}, {"identifier": "B", "content": "$${{9} \\over {64}}$$"}, {"identifier": "C", "content": "$${{15} \\over {64}}$$"}, {"identifier": "D", "content": "$${{39} \\over {64}}$$"}] | ["A"] | null | <p>We have the following probabilities:</p>
<p>$$\bullet$$ The probability that the target is hit by the person P is $${3 \over 4}$$.</p>
<p>$$\bullet$$ The probability that the target is not hit by the person P is $$1 - {3 \over 4} = {1 \over 4}$$.</p>
<p>$$\bullet$$ The probability that the target is hit by the person Q is $${1 \over 2}$$.</p>
<p>$$\bullet$$ The probability that the target is not hit by the person Q is $$1 - {1 \over 2} = {1 \over 2}$$.</p>
<p>$$\bullet$$ The probability that the target is hit by the person R is $${5 \over 8}$$.</p>
<p>$$\bullet$$ The probability that the target is not hit by the person R is $$1 - {5 \over 8} = {3 \over 8}$$.</p>
<p>Here, we have used the fact that if the probability of occurrence of an event is p, then the probability of non-occurrence of an event is $$q = 1 - p$$.</p>
<p>Therefore, the probability that the target is hit by P or Q and not by R is</p>
<p>(Probability that the target is hit by P and not by Q and R) + (Probability that the target is hit by Q and not by P and R) + (Probability that the target is hit by both P and Q and not by R)</p>
<p>$$ = \left( {{3 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right) + \left( {{1 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right) + \left( {{3 \over 4}} \right)\left( {{1 \over 2}} \right)\left( {{3 \over 8}} \right)$$</p>
<p>$$ = {9 \over {64}} + {3 \over {64}} + {9 \over {64}} = {{9 + 3 + 9} \over {64}} = {{21} \over {64}}$$</p> | mcq | jee-main-2017-online-8th-april-morning-slot |
QRIX9P2KGdebyvwez5VFA | maths | probability | classical-defininition-of-probability | An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is : | [{"identifier": "A", "content": "$${{255} \\over {256}}$$"}, {"identifier": "B", "content": "$${{127} \\over {128}}$$"}, {"identifier": "C", "content": "$${{63} \\over {64}}$$"}, {"identifier": "D", "content": "$${{1} \\over {2}}$$"}] | ["B"] | null | An unbiased coin is tossed 8 times which is same as 8 coins tossed 1 times.
<br><br>$$\therefore\,\,\,$$ Possible no. of out come = 2<sup>8</sup>
<br><br>$$\therefore\,\,\,$$ Sample space = 2<sup>8</sup>
<br><br>Here in this condition, all head or all tail out come is not acceptable.
<br><br>No. of times all head can occur
<br><br>(H H H H H H H H) = 1
<br><br>$$\therefore\,\,\,$$ Probability (all head) = $${1 \over {{2^8}}}$$ = $${1 \over {256}}$$
<br><br>No. of times all tail can occur
<br><br>(T T T T T T T T) = 1
<br><br>$$\therefore\,\,\,$$ Probability (all tail) = $${1 \over {{2^8}}}$$ = $${{1 \over {256}}}$$
<br><br>$$\therefore\,\,\,$$ Required probability
<br><br>= 1 $$-$$ (P (All head) + P (All tail))
<br><br>= 1 $$-$$ ( $${{1 \over {256}}}$$ + $${{1 \over {256}}}$$)
<br><br>= 1 $$-$$ $${{1 \over {128}}}$$
<br><br>= $${{127} \over {128}}$$ | mcq | jee-main-2017-online-8th-april-morning-slot |
e5HmDrCrJDgX09wpKD5Po | maths | probability | classical-defininition-of-probability | In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to : | [{"identifier": "A", "content": "$${{200} \\over {{6^5}}}$$"}, {"identifier": "B", "content": "$${{225} \\over {{6^5}}}$$"}, {"identifier": "C", "content": "$${{150} \\over {{6^5}}}$$"}, {"identifier": "D", "content": "$${{175} \\over {{6^5}}}$$"}] | ["D"] | null | $$\underline {} \,\,\,\underline {} \,\,\,\underline {} \,\,\underline 4 \,\,\underline 4 $$
<br><br>$${1 \over {{6^2}}}\left( {{{{5^3}} \over {{6^3}}} + {{2{C_1}{{.5}^2}} \over {{6^3}}}} \right) = {{175} \over {{6^5}}}$$ | mcq | jee-main-2019-online-12th-january-morning-slot |
yAQH5naWMpMAGXWMQjDko | maths | probability | classical-defininition-of-probability | In a game, a man wins Rs. 100 if he gets 5 or 6 on a throw of a fair die and loses Rs. 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is :
| [{"identifier": "A", "content": "$${{400} \\over 3}$$ loss"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$${{400} \\over 9}$$ loss"}, {"identifier": "D", "content": "$${{400} \\over 3}$$ gain"}] | ["B"] | null | Expected Gain/Loss =
<br><br>= w $$ \times $$ 100 + Lw($$-$$ 50 + 100)
<br><br> + L<sup>2</sup>w ($$-$$ 50 $$-$$ 50 + 100) + L<sup>3</sup>($$-$$ 150)
<br><br>= $${1 \over 3} \times 100 + {2 \over 3}.{1 \over 3}\left( {50} \right) + {\left( {{2 \over 3}} \right)^2}\left( {{1 \over 3}} \right)\left( 0 \right)$$
<br><br> $$ + {\left( {{2 \over 3}} \right)^3}\left( { - 150} \right) = 0$$
<br><br>here w denotes probability that outcome 5 or 6 (w = $${2 \over 6} = {1 \over 3}$$)
<br><br>here L denotes probability that outcome
<br><br>1,2,3,4 (L = $${4 \over 6}$$ = $${2 \over 3}$$) | mcq | jee-main-2019-online-12th-january-evening-slot |
m2nwnLqVyIMJTL11MZrgl | maths | probability | classical-defininition-of-probability | The minimum number of times one has to toss a
fair coin so that the probability of observing at least
one head is at least 90% is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "5"}] | ["C"] | null | Probablity of getting head P(H) = $${1 \over 2}$$
<br><br>Probablity of getting tail P(T) = 1 - $${1 \over 2}$$ = $${1 \over 2}$$
<br><br>Probability of observing at least one head out of n tosses
<br><br>= 1 - Probability of observing no head occurs out of n tosses
<br><br>= 1 - $${\left( {{1 \over 2}} \right)^n}$$
<br><br>According to the question,
<br><br>1 - $${\left( {{1 \over 2}} \right)^n}$$ $$ \ge $$ $${{90} \over {100}}$$
<br><br>$$ \Rightarrow $$ $${\left( {{1 \over 2}} \right)^n} \le {1 \over {10}}$$
<br><br>$$ \Rightarrow $$ $${2^n} \ge 10$$
<br><br>As 2<sup>3</sup> = 8
<br><br>2<sup>4</sup> = 16
<br><br>$$ \therefore $$ Minimum value of n = 4
| mcq | jee-main-2019-online-8th-april-evening-slot |
9n3LPHYPclxu3TKzKw3rsa0w2w9jx22oarv | maths | probability | classical-defininition-of-probability | Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is
more than 99% is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "7"}] | ["D"] | null | $$1 - {\left( {{1 \over 2}} \right)^n} > {{99} \over {100}}$$<br><br>
$${\left( {{1 \over 2}} \right)^n} < {1 \over {100}}$$<br><br>
$$ \Rightarrow $$ n = 7 | mcq | jee-main-2019-online-10th-april-evening-slot |
TlVWYzrt0ahn22L9h7jgy2xukf901fr4 | maths | probability | classical-defininition-of-probability | The probability of a man hitting a target is $${1 \over {10}}$$. The least number of shots required, so that the
probability of his hitting the target at least once is greater than $${1 \over {4}}$$, is ____________. | [] | null | 3 | We have, $$1 - $$(probability of all shots results in failure out of n trials) > $${1 \over 4}$$<br><br>$$ \Rightarrow 1 - {\left( {{9 \over {10}}} \right)^n} > {1 \over 4}$$<br><br>$$ \Rightarrow {3 \over 4} > {\left( {{9 \over {10}}} \right)^n} \Rightarrow n \ge 3$$ | integer | jee-main-2020-online-4th-september-morning-slot |
lNy86hckKMSF39JnDJ7k9k2k5ip6e44 | maths | probability | classical-defininition-of-probability | In a box, there are 20 cards, out of which 10
are lebelled as A and the remaining 10 are
labelled as B. Cards are drawn at random, one
after the other and with replacement, till a
second A-card is obtained. The probability that
the second A-card appears before the third
B-card is : | [{"identifier": "A", "content": "$${{13} \\over {16}}$$"}, {"identifier": "B", "content": "$${{11} \\over {16}}$$"}, {"identifier": "C", "content": "$${{15} \\over {16}}$$"}, {"identifier": "D", "content": "$${{9} \\over {16}}$$"}] | ["B"] | null | Possibilities that the second A card appears before the third B card are
<br>=AA + ABA + BAA + ABBA + BBAA + BABA
<br><br>= $${\left( {{1 \over 2}} \right)^2}$$ + $${\left( {{1 \over 2}} \right)^3}$$ + $${\left( {{1 \over 2}} \right)^3}$$ + $${\left( {{1 \over 2}} \right)^4}$$ + $${\left( {{1 \over 2}} \right)^4}$$ + $${\left( {{1 \over 2}} \right)^4}$$
<br><br>= $${1 \over 4}$$ + $${1 \over 8}$$ + $${1 \over 8}$$ + $${1 \over {16}}$$ + $${1 \over {16}}$$ + $${1 \over {16}}$$
<br><br>= $${{11} \over {16}}$$ | mcq | jee-main-2020-online-9th-january-morning-slot |
Hlrmx8ls5tFwjNNkM61klrk41dk | maths | probability | classical-defininition-of-probability | Let B<sub>i</sub> (i = 1, 2, 3) be three independent events in a sample space. The probability that only B<sub>1</sub> occur is $$\alpha $$, only B<sub>2</sub> occurs is $$\beta $$ and only B<sub>3</sub> occurs is $$\gamma $$. Let p be the probability that none of the events B<sub>i</sub> occurs and these 4 probabilities satisfy the equations $$\left( {\alpha - 2\beta } \right)p = \alpha \beta $$ and $$\left( {\beta - 3\gamma } \right)p = 2\beta \gamma $$ (All the probabilities are assumed to lie in the interval (0, 1)).<br/> Then $${{P\left( {{B_1}} \right)} \over {P\left( {{B_3}} \right)}}$$ is equal to ________. | [] | null | 6 | Let x, y, z be probability of B<sub>1</sub>, B<sub>2</sub>, B<sub>3</sub> respectively
<br><br>$$\alpha $$ = P(B<sub>1</sub> $$ \cap $$ $$\overline {{B_2}} \cap \overline {{B_3}} $$) = $$P\left( {{B_1}} \right)P\left( {\overline {{B_2}} } \right)P\left( {\overline {{B_3}} } \right)$$
<br><br>$$ \Rightarrow $$ x(1 $$-$$ y)(1 $$-$$ z) = $$\alpha$$<br><br>Similarly, y(1 $$-$$ x)(1 $$-$$ z) = $$\beta$$<br><br> z(1 $$-$$ x)(1 $$-$$ y) = $$\gamma$$<br><br>and (1 $$-$$ x)(1 $$-$$ y)(1 $$-$$ z) = p<br><br>($$\alpha$$ $$-$$ 2$$\beta$$)p = $$\alpha$$$$\beta$$<br><br>(x(1 $$-$$ y)(1 $$-$$ z) $$-$$2y(1 $$-$$ x)(1 $$-$$ z)) (1 $$-$$ x)(1 $$-$$ y)(1 $$-$$ z) = xy(1 $$-$$ x)(1 $$-$$ y)(1 $$-$$ z)<br><br>x $$-$$ xy $$-$$ 2y + 2xy = xy<br><br>x = 2y ...... (1)<br><br>Similarly ($$\beta$$ $$-$$ 3$$\gamma $$)p = 2$$\beta$$$$\gamma $$<br><br>$$ \Rightarrow $$ y = 3z .... (2)<br><br>From (1) & (2)<br><br>x = 6z<br><br>Now <br><br>$${x \over z} = 6$$ | integer | jee-main-2021-online-24th-february-morning-slot |
yV5PXKqTuDnpvTD1Mr1kls5afdp | maths | probability | classical-defininition-of-probability | The coefficients a, b and c of the quadratic equation, ax<sup>2</sup> + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is : | [{"identifier": "A", "content": "$${1 \\over {72}}$$"}, {"identifier": "B", "content": "$${5 \\over {216}}$$"}, {"identifier": "C", "content": "$${1 \\over {36}}$$"}, {"identifier": "D", "content": "$${1 \\over {54}}$$"}] | ["B"] | null | ax<sup>2</sup> + bx + c = 0
<br><br>a, b, c $$ \in $$ {1,2,3,4,5,6}
<br><br>n(s) = 6 Γ 6 Γ 6 = 216
<br><br>For equal roots, D = 0 $$ \Rightarrow $$ b<sup>2</sup> = 4ac
<br><br>$$ \Rightarrow $$ ac = $${{{b^2}} \over 4}$$
<br><br>Favourable case :
<br><br>If b = 2, ac = 1 $$ \Rightarrow $$ a = 1, c = 1
<br><br>If b = 4, ac = 4 :
<br>a = 1, c = 4
<br>a = 4, c = 1
<br>a = 2, c = 2
<br><br>If b = 6, ac = 9 $$ \Rightarrow $$ a = 3, c = 3
<br><br>$$ \therefore $$ Favorable cases = 5
<br><br>$$ \therefore $$ Required probability = $${5 \over {216}}$$ | mcq | jee-main-2021-online-25th-february-morning-slot |
UBXmKU6xOGiAcbU8No1kmiyynw3 | maths | probability | classical-defininition-of-probability | Let A denote the event that a 6-digit integer formed by 0, 1, 2, 3, 4, 5, 6 without repetitions, be divisible by 3. Then probability of event A is equal to : | [{"identifier": "A", "content": "$${4 \\over {9}}$$"}, {"identifier": "B", "content": "$${9 \\over {56}}$$"}, {"identifier": "C", "content": "$${11 \\over {27}}$$"}, {"identifier": "D", "content": "$${3 \\over {7}}$$"}] | ["A"] | null | Total cases :<br><br>$$\underline 6 $$ . $$\underline 6 $$ . $$\underline 5 $$ . $$\underline 4 $$ . $$\underline 3 $$ . $$\underline 2 $$<br><br>n(s) = 6 . 6!<br><br>Favourable cases :<br><br>Number divisible by 3 $$ \equiv $$ Sum of digits must be divisible by 3<br><br>Case - I<br><br>1, 2, 3, 4, 5, 6<br><br>Number of ways = 6!<br><br>Case - II<br><br>0, 1, 2, 4, 5, 6<br><br>Number of ways = 5 . 5!<br><br>Case - III<br><br>0, 1, 2, 3, 4, 5<br><br>Number of ways = 5 . 5!<br><br>n(favourable) = 6! + 2 . 5 . 5!<br><br>$$P = {{6! + 2.\,5.\,5!} \over {6\,.\,6!}} = {4 \over 9}$$ | mcq | jee-main-2021-online-16th-march-evening-shift |
sgzSQk4kJbYhlNtQ1e1kmjazetr | maths | probability | classical-defininition-of-probability | Two dies are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is : | [{"identifier": "A", "content": "$${4 \\over 9}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${5 \\over {12}}$$"}, {"identifier": "D", "content": "$${17 \\over {36}}$$"}] | ["D"] | null | n(S) = 36<br><br>possible ordered pair :
<br><br> (1, 1), (1, 2), (1, 3), (1, 5), (1, 7), <br>(2, 1), (2, 2), (2, 3), (2, 5), <br>(3, 1), (3, 2), (3, 3), (3, 5), <br>(5, 1), (5, 2), (5, 3), <br>(7, 1)<br><br>Number of favourable outcomes = 17<br><br>Probability = $${{17} \over {36}}$$ | mcq | jee-main-2021-online-17th-march-morning-shift |
VDtLlmJEgfbNCOYH831kmjbmvhl | maths | probability | classical-defininition-of-probability | Let there be three independent events E<sub>1</sub>, E<sub>2</sub> and E<sub>3</sub>. The probability that only E<sub>1</sub> occurs is $$\alpha$$, only E<sub>2</sub> occurs is $$\beta$$ and only E<sub>3</sub> occurs is $$\gamma$$. Let 'p' denote the probability of none of events occurs that satisfies the equations <br/>($$\alpha$$ $$-$$ 2$$\beta$$)p = $$\alpha$$$$\beta$$ and ($$\beta$$ $$-$$ 3$$\gamma$$)p = 2$$\beta$$$$\gamma$$. All the given probabilities are assumed to lie in the interval (0, 1).<br/><br/>Then, $$\frac{Probability\ of\ occurrence\ of\ E_{1}}{Probability\ of\ occurrence\ of\ E_{3}} $$ is equal to _____________. | [] | null | 6 | Let P(E<sub>1</sub>) = x, P(E<sub>2</sub>) = y and P(E<sub>3</sub>) = z
<br><br>$$\alpha $$ = P$$\left( {{E_1} \cap {{\overline E }_2} \cap {{\overline E }_3}} \right)$$ = $$P\left( {{E_1}} \right).P\left( {{{\overline E }_2}} \right).P\left( {{{\overline E }_3}} \right)$$
<br><br>$$ \Rightarrow $$ $$\alpha $$ = x(1 $$-$$ y) (1 $$-$$ z) ......(i)
<br><br>Similarly
<br><br>Ξ² = (1 β x).y(1 β z) ...(ii)
<br><br>$$\gamma $$ = (1 β x)(1 β y).z ...(iii)
<br><br>p = (1 β x)(1 β y)(1 β z) ...(iv)
<br><br>From (i) and (iv)
<br><br>$${x \over {1 - x}} = {\alpha \over p}$$
<br><br>$$ \Rightarrow $$ x = $${\alpha \over {\alpha + p}}$$
<br><br>From (iii) and (iv)
<br><br>$${z \over {1 - z}} = {\gamma \over p}$$
<br><br>$$ \Rightarrow $$ z = $${\gamma \over {\gamma + p}}$$
<br><br>$$ \therefore $$ $${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}} = {x \over z} = {{{\alpha \over {\alpha + p}}} \over {{\gamma \over {\gamma + p}}}}$$ $$ = {{{{\gamma + p} \over \gamma }} \over {{{\alpha + p} \over \alpha }}} = {{1 + {p \over \gamma }} \over {1 + {p \over \alpha }}}$$ ..(v)
<br><br>Also given,
<br><br>($$\alpha$$ $$-$$ 2$$\beta$$)p = $$\alpha$$$$\beta$$ $$ \Rightarrow $$ $$\alpha $$p = ($$\alpha $$ + 2p)$$\beta $$ ....(vi)
<br><br>$$\beta$$ $$-$$ 3$$\gamma$$)p = 2$$\beta$$$$\gamma$$ $$ \Rightarrow $$ 3$$\gamma $$p = (p - 2$$\gamma $$)$$\beta $$ .....(vii)
<br><br>From (vi) and (vii),
<br><br>$${\alpha \over {3\gamma }} = {{\alpha + 2p} \over {p - 2\gamma }}$$
<br><br>$$ \Rightarrow $$ p$$\alpha $$ - 6p$$\gamma $$ = 5$$\gamma $$$$\alpha $$
<br><br>$$ \Rightarrow $$ $${p \over \gamma } - {{6p} \over \alpha } = 5$$
<br><br>$$ \Rightarrow $$ $${p \over \gamma } + 1 = 6\left( {{p \over \alpha } + 1} \right)$$ ....(viii)
<br><br>Now from (v) and (viii),
<br><br>$${{P\left( {{E_1}} \right)} \over {P\left( {{E_3}} \right)}}$$ = 6 | integer | jee-main-2021-online-17th-march-morning-shift |
2X5uWEUa0odmBJe9ZR1kmkl6i35 | maths | probability | classical-defininition-of-probability | Let a computer program generate only the digits 0 and 1 to form a string of binary numbers with probability of occurrence of 0 at even places be $${1 \over 2}$$ and probability of occurrence of 0 at the odd place be $${1 \over 3}$$. Then the probability that '10' is followed by '01' is equal to : | [{"identifier": "A", "content": "$${1 \\over 18}$$"}, {"identifier": "B", "content": "$${1 \\over 3}$$"}, {"identifier": "C", "content": "$${1 \\over 9}$$"}, {"identifier": "D", "content": "$${1 \\over 6}$$"}] | ["C"] | null | P(0 at even place) = $${1 \over 2}$$
<br><br>P(0 at odd place) = $${1 \over 3}$$
<br><br> P(1 at even place) = $${1 \over 2}$$
<br><br>P(1 at odd place) = $${2 \over 3}$$
<br><br>P(10 is followed by 01)
<br><br>= $$\left( {{2 \over 3} \times {1 \over 2} \times {1 \over 3} \times {1 \over 2}} \right) + \left( {{1 \over 2} \times {1 \over 3} \times {1 \over 2} \times {2 \over 3}} \right)$$
<br><br>= $${1 \over {18}} + {1 \over {18}}$$
<br><br>= $${1 \over 9}$$ | mcq | jee-main-2021-online-17th-march-evening-shift |
1krpzo9s2 | maths | probability | classical-defininition-of-probability | The probability of selecting integers a$$\in$$[$$-$$ 5, 30] such that x<sup>2</sup> + 2(a + 4)x $$-$$ 5a + 64 > 0, for all x$$\in$$R, is : | [{"identifier": "A", "content": "$${7 \\over {36}}$$"}, {"identifier": "B", "content": "$${2 \\over {9}}$$"}, {"identifier": "C", "content": "$${1 \\over {6}}$$"}, {"identifier": "D", "content": "$${1 \\over {4}}$$"}] | ["B"] | null | D < 0<br><br>$$\Rightarrow$$ 4(a + 4)<sup>2</sup> $$-$$ 4($$-$$5a + 64) < 0<br><br>$$\Rightarrow$$ a<sup>2</sup> + 16 + 8a + 5a $$-$$ 64 < 0<br><br>$$\Rightarrow$$ a<sup>2</sup> + 13a $$-$$ 48 < 0<br><br>$$\Rightarrow$$ (a + 16) (a $$-$$ 3) < 0<br><br>$$\Rightarrow$$ a $$\in$$ ($$-$$16, 3)<br><br>$$\therefore$$ Possible a : {$$-$$5, $$-$$4, ............., 3}<br><br>$$\therefore$$ Required probability = $${8 \over {36}}$$ = $${2 \over {9}}$$ | mcq | jee-main-2021-online-20th-july-morning-shift |
1krzrj7ab | maths | probability | classical-defininition-of-probability | A fair coin is tossed n-times such that the probability of getting at least one head is at least 0.9. Then the minimum value of n is ______________. | [] | null | 4 | P(Head) = $${1 \over 2}$$<br><br>1 $$-$$ P(All tail) $$\ge$$ 0.9<br><br>$$1 - {\left( {{1 \over 2}} \right)^n}$$ $$\ge$$ 0.9<br><br>$$ \Rightarrow {\left( {{1 \over 2}} \right)^n} \le {1 \over {10}}$$<br><br>$$\Rightarrow$$ n<sub>min</sub> = 4 | integer | jee-main-2021-online-25th-july-evening-shift |
1ks0ai6ce | maths | probability | classical-defininition-of-probability | The probability that a randomly selected 2-digit number belongs to the set {n $$\in$$ N : (2<sup>n</sup> $$-$$ 2) is a multiple of 3} is equal to : | [{"identifier": "A", "content": "$${1 \\over 6}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["C"] | null | Total number of cases = $${}^{90}{C_1} = 90$$<br><br>Now, $${2^n} - 2 = {(3 - 1)^n} - 2$$<br><br>$${}^n{C_0}{3^n} - {}^n{C_1}{.3^{n - 1}} + .... + {( - 1)^{n - 1}}.{}^n{C_{n - 1}}3 + {( - 1)^n}.{}^n{C_n} - 2$$<br><br>= $$3\left( {{3^{n - 1}} - n{3^{n - 2}} + ... + {{( - 1)}^{n - 1}}.n} \right) + {( - 1)^n} - 2$$<br><br>$$({2^n} - 2)$$ is multiply of 3 only when n is odd<br><br>Required Probability $$ = {{45} \over {90}} = {1 \over 2}$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktd0dr87 | maths | probability | classical-defininition-of-probability | Two fair dice are thrown. The numbers on them are taken as $$\lambda$$ and $$\mu$$, and a system of linear equations<br/><br/>x + y + z = 5<br/><br/>x + 2y + 3z = $$\mu$$ <br/><br/>x + 3y + $$\lambda$$z = 1<br/><br/>is constructed. If p is the probability that the system has a unique solution and q is the probability that the system has no solution, then : | [{"identifier": "A", "content": "$$p = {1 \\over 6}$$ and $$q = {1 \\over 36}$$"}, {"identifier": "B", "content": "$$p = {5 \\over 6}$$ and $$q = {5 \\over 36}$$"}, {"identifier": "C", "content": "$$p = {5 \\over 6}$$ and $$q = {1 \\over 36}$$"}, {"identifier": "D", "content": "$$p = {1 \\over 6}$$ and $$q = {5 \\over 36}$$"}] | ["B"] | null | $$D \ne 0 \Rightarrow \left| {\matrix{
1 & 1 & 1 \cr
1 & 2 & 3 \cr
1 & 3 & \lambda \cr
} } \right| \ne 0 \Rightarrow \lambda \ne 5$$<br><br>For no solution D = 0 $$\Rightarrow$$ $$\lambda$$ = 5<br><br>$${D_1} = \left| {\matrix{
1 & 1 & 5 \cr
1 & 2 & \mu \cr
1 & 3 & 1 \cr
} } \right| \ne 0 \Rightarrow \mu \ne 3$$<br><br>$$p = {5 \over 6}$$<br><br>$$q = {1 \over 6} \times {5 \over 6} = {5 \over {36}}$$<br><br>Option (b). | mcq | jee-main-2021-online-26th-august-evening-shift |
1l56744uu | maths | probability | classical-defininition-of-probability | <p>The probability, that in a randomly selected 3-digit number at least two digits are odd, is :</p> | [{"identifier": "A", "content": "$${{19} \\over {36}}$$"}, {"identifier": "B", "content": "$${{15} \\over {36}}$$"}, {"identifier": "C", "content": "$${{13} \\over {36}}$$"}, {"identifier": "D", "content": "$${{23} \\over {36}}$$"}] | ["A"] | null | At least two digits are odd = exactly two digits are odd + exactly there 3 digits
are odd <br><br>For exactly three digits are odd <br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc8eefiq/e8de7882-e991-47ff-999d-2ea9ce2e3c3f/ccd33f20-8716-11ed-92bb-e57e64e1a06d/file-1lc8eefir.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc8eefiq/e8de7882-e991-47ff-999d-2ea9ce2e3c3f/ccd33f20-8716-11ed-92bb-e57e64e1a06d/file-1lc8eefir.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th June Morning Shift Mathematics - Probability Question 83 English Explanation"><br>
For exactly two digits odd :<br><br>
If 0 is used then $: 2 \times 5 \times 5=50$<br><br>
If 0 is not used then : ${ }^3 \mathrm{C}_1 \times 4 \times 5 \times 5=300$<br><br>
Required Probability $=\frac{475}{900}=\frac{19}{36}$ | mcq | jee-main-2022-online-28th-june-morning-shift |
1l56rh8nv | maths | probability | classical-defininition-of-probability | <p>If a point A(x, y) lies in the region bounded by the y-axis, straight lines 2y + x = 6 and 5x $$-$$ 6y = 30, then the probability that y < 1 is :</p> | [{"identifier": "A", "content": "$${1 \\over 6}$$"}, {"identifier": "B", "content": "$${5 \\over 6}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${6 \\over 7}$$"}] | ["B"] | null | <p>The required probability</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5p7g5yy/9381a58a-1f21-4b26-abca-dc94a3e72b11/6c8ce9a0-05bf-11ed-8617-d71e6444d1a0/file-1l5p7g5yz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5p7g5yy/9381a58a-1f21-4b26-abca-dc94a3e72b11/6c8ce9a0-05bf-11ed-8617-d71e6444d1a0/file-1l5p7g5yz.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th June Evening Shift Mathematics - Probability Question 82 English Explanation"></p>
<p>$$ = {\text{Area of Region PQCAP} \over \text{Area of Region ABCA}}$$</p>
<p>$$ = {{{1 \over 2} \times 8 \times 6 - {1 \over 2} \times 2 \times 4} \over {{1 \over 2} \times 8 \times 6}}$$</p>
<p>$$ = {5 \over 6}$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l56uaobr | maths | probability | classical-defininition-of-probability | <p>Let S = {E<sub>1</sub>, E<sub>2</sub>, ........., E<sub>8</sub>} be a sample space of a random experiment such that $$P({E_n}) = {n \over {36}}$$ for every n = 1, 2, ........, 8. Then the number of elements in the set $$\left\{ {A \subseteq S:P(A) \ge {4 \over 5}} \right\}$$ is ___________.</p> | [] | null | 19 | <p>Here $$P({E_n}) = {n \over {36}}$$ for n = 1, 2, 3, ......, 8</p>
<p>Here $$P(A) = {{Any\,possible\,sum\,of\,(1,2,3,\,...,\,8)( = a\,say)} \over {36}}$$</p>
<p>$$\because$$ $${a \over {36}} \ge {4 \over 5}$$</p>
<p>$$\therefore$$ $$a \ge 29$$</p>
<p>If one of the number from {1, 2, ......, 8} is left then total $$a \ge 29$$ by 3 ways.</p>
<p>Similarly by leaving terms more 2 or 3 we get 16 more combinations.</p>
<p>$$\therefore$$ Total number of different set A possible is 16 + 3 = 19</p> | integer | jee-main-2022-online-27th-june-evening-shift |
1l6dwn5ao | maths | probability | classical-defininition-of-probability | <p>If the numbers appeared on the two throws of a fair six faced die are $$\alpha$$ and $$\beta$$, then the probability that $$x^{2}+\alpha x+\beta>0$$, for all $$x \in \mathbf{R}$$, is :
</p> | [{"identifier": "A", "content": "$$\\frac{17}{36}$$"}, {"identifier": "B", "content": "$$\n\\frac{4}{9}\n$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{2}$$"}, {"identifier": "D", "content": "$$\\frac{19}{36}$$"}] | ["A"] | null | For $x^{2}+\alpha x+\beta>0 \forall x \in R$ to hold, we should have $\alpha^{2}-4 \beta<0$
<br/><br/>
If $\alpha=1, \beta$ can be 1, 2, 3, 4, 5, 6 i.e., 6 choices
<br/><br/>
If $\alpha=2, \beta$ can be 2, 3, 4, 5, 6 i.e., 5 choices
<br/><br/>
If $\alpha=3, \beta$ can be $3,4,5,6$ i.e., 4 choices
<br/><br/>
If $\alpha=4, \beta$ can be 5 or 6 i.e., 2 choices
<br/><br/>
If $\alpha=6$, No possible value for $\beta$ i.e., 0 choices<br/><br/>
Hence total favourable outcomes
<br/><br/>
$$
\begin{aligned}
&=6+5+4+2+0+0 \\\\
&=17
\end{aligned}
$$
<br/><br/>
Total possible choices for $\alpha$ and $\beta=6 \times 6=36$
<br/><br/>
Required probability $=\frac{17}{36}$ | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6jc7ci7 | maths | probability | classical-defininition-of-probability | <p>Let $$S$$ be the sample space of all five digit numbers. It $$p$$ is the probability that a randomly selected number from $$S$$, is a multiple of 7 but not divisible by 5 , then $$9 p$$ is equal to :</p> | [{"identifier": "A", "content": "1.0146"}, {"identifier": "B", "content": "1.2085"}, {"identifier": "C", "content": "1.0285"}, {"identifier": "D", "content": "1.1521"}] | ["C"] | null | <p>Among the 5 digit numbers,</p>
<p>First number divisible by 7 is 10003 and last is 99995.</p>
<p>$$\Rightarrow$$ Number of numbers divisible by 7.</p>
<p>$$ = {{99995 - 10003} \over 7} + 1$$</p>
<p>$$ = 12857$$</p>
<p>First number divisible by 35 is 10010 and last is 99995.</p>
<p>$$\Rightarrow$$ Number of numbers divisible by 35</p>
<p>$$ = {{99995 - 10010} \over {35}} + 1$$</p>
<p>$$ = 2572$$</p>
<p>Hence number of number divisible by 7 but not by 5</p>
<p>$$ = 12857 - 2572$$</p>
<p>$$ = 10285$$</p>
<p>$$9P. = {{10285} \over {90000}} \times 9$$</p>
<p>$$ = 1.0285$$</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
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