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Probability — JEE Previous Year Questions

Every Probability question asked in JEE Main and JEE Advanced across the last 186 papers — 226 questions, each with its correct answer. Free to read, no account needed.

Questions

226

Papers it appeared in

176/186

Appearance rate

95%

All 226 Probability questions

Most recent papers first.

Q1·MathematicsMultiple correctJEE Advanced 2026
Suppose that Box I contains 6 red balls and 9 green balls, and Box II contains 8 red balls and 12 green balls. All the balls of Box I and Box II are mixed together and a ball is chosen at random from them. Let E1E_1E1​ be the event that the ball chosen belonged to Box I and let E2E_2E2​ be the event that the ball chosen belonged to Box II. Let F1F_1F1​ be the event that the ball chosen is red and let F2F_2F2​ be the event that the ball chosen is green. Then which of the following statements is (are) TRUE ?
  1. (A)The events E1E_1E1​ and F1F_1F1​ are independent
  2. (B)The events E2E_2E2​ and F2F_2F2​ are dependent
  3. (C)The conditional probability P(F1∣E1)P(F_1 \mid E_1)P(F1​∣E1​) is equal to the conditional probability P(F1∣E2)P(F_1 \mid E_2)P(F1​∣E2​)
  4. (D)The conditional probability P(F1∣E1)P(F_1 \mid E_1)P(F1​∣E1​) is greater than the conditional probability P(F2∣E2)P(F_2 \mid E_2)P(F2​∣E2​)

Correct answer: (A), (C)

Step-by-step solution →
Q2·MathematicsNumericalJEE Advanced 2026
A bookshelf contains 6 distinct books of Mathematics and 5 distinct books of Physics. From these 11 books, 6 books are chosen at random. Let XXX be the absolute value of the difference between the number of Mathematics books chosen and the number of Physics books chosen. If α\alphaα is the mean of the random variable XXX, then the value of 77α77\alpha77α is _____.

Correct answer: 100

Step-by-step solution →
Q3·MathematicsSingle correctJEE Main 2026
A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities of the candidate going by bus, scooter and car, respectively, are 25\frac{2}{5}52​, 15\frac{1}{5}51​ and 25\frac{2}{5}52​. The probabilities that the candidate reaches late at the examination centre are 15\frac{1}{5}51​, 13\frac{1}{3}31​ and 14\frac{1}{4}41​ if the candidate uses bus, scooter and car, respectively. Given that the candidate reached late at the examination centre, the probability that the candidate travelled by bus is:
  1. (A)1137\frac{11}{37}3711​
  2. (B)1237\frac{12}{37}3712​
  3. (C)1337\frac{13}{37}3713​
  4. (D)1437\frac{14}{37}3714​

Correct answer: (B)

Step-by-step solution →
Q4·MathematicsSingle correctJEE Main 2026
A bag contains (N + 1) coins − N fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is 916\frac{9}{16}169​, then N is equal to:
  1. (A)5
  2. (B)7
  3. (C)8
  4. (D)9

Correct answer: (B)

Step-by-step solution →
Q5·MathematicsSingle correctJEE Main 2026
A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is :
  1. (A)63925\frac{63}{925}92563​
  2. (B)17231\frac{17}{231}23117​
  3. (C)16231\frac{16}{231}23116​
  4. (D)64925\frac{64}{925}92564​

Correct answer: (C)

Step-by-step solution →
Q6·MathematicsSingle correctJEE Main 2026
A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:
  1. (A)710\frac{7}{10}107​
  2. (B)1017\frac{10}{17}1710​
  3. (C)1219\frac{12}{19}1912​
  4. (D)719\frac{7}{19}197​

Correct answer: (B)

Step-by-step solution →
Q7·MathematicsSingle correctJEE Main 2026
The probabilities that players A and B of a team are selected for the captaincy for a tournament are 0.6 and 0.4, respectively. If A is selected the captain, the probability that the team wins the tournament is 0.8 and if B is selected the captain, the probability that the team wins the tournament is 0.7. Then the probability, that the team wins the tournament, is :
  1. (A)0.74
  2. (B)0.76
  3. (C)0.72
  4. (D)0.78

Correct answer: (B)

Step-by-step solution →
Q8·MathematicsNumericalJEE Main 2026
From a month of 31 days, 3 different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to ab\frac{a}{b}ba​, where a,b∈Na, b \in \mathbb{N}a,b∈N and gcd⁡(a,b)=1\gcd(a, b) = 1gcd(a,b)=1, then a+ba + ba+b is equal to ______

Correct answer: 944

Step-by-step solution →
Q9·MathematicsNumericalJEE Main 2026
A coin is tossed 8 times. If the probability that exactly 4 heads appear in the first six tosses and exactly 3 heads appear in the last five tosses is ppp, then 96p96p96p is equal to _______.

Correct answer: 9

Step-by-step solution →
Q10·MathematicsNumericalJEE Main 2026
Let a,b,c∈{1,2,3,4}a, b, c \in \{1, 2, 3, 4\}a,b,c∈{1,2,3,4}. If the probability, that ax2+22 bx+c>0ax^{2} + 2\sqrt{2}\,bx + c > 0ax2+22​bx+c>0 for all x∈Rx \in \mathbb{R}x∈R, is mn\frac{m}{n}nm​, gcd⁡(m,n)=1\gcd(m, n) = 1gcd(m,n)=1, then m+nm + nm+n is equal to ________.

Correct answer: 81

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2026
A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is mn\frac{m}{n}nm​, gcd⁡(m,n)=1\gcd(m, n) = 1gcd(m,n)=1, then m+nm + nm+n is equal to:
  1. (A)535353
  2. (B)555555
  3. (C)107107107
  4. (D)105105105

Correct answer: (C)

Step-by-step solution →
Q12·MathematicsSingle correctJEE Main 2026
The probability distribution of a random variable X is given below : X4k307k327k347k367k387k407k6kP(X)2151152151511521515115\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X & 4k & \frac{30}{7}k & \frac{32}{7}k & \frac{34}{7}k & \frac{36}{7}k & \frac{38}{7}k & \frac{40}{7}k & 6k \\ \hline P(X) & \frac{2}{15} & \frac{1}{15} & \frac{2}{15} & \frac{1}{5} & \frac{1}{15} & \frac{2}{15} & \frac{1}{5} & \frac{1}{15} \\ \hline \end{array}XP(X)​4k152​​730​k151​​732​k152​​734​k51​​736​k151​​738​k152​​740​k51​​6k151​​​ If E(X)=26315E(X) = \dfrac{263}{15}E(X)=15263​, then P(X<20)P(X < 20)P(X<20) is equal to :
  1. (A)35\dfrac{3}{5}53​
  2. (B)815\dfrac{8}{15}158​
  3. (C)1115\dfrac{11}{15}1511​
  4. (D)1415\dfrac{14}{15}1514​

Correct answer: (C)

Step-by-step solution →
Q13·MathematicsNumericalJEE Main 2026
Let S be a set of 5 elements and P(S) denote the power set of S. Let E be an event of choosing an ordered pair (A, B) from the set P(S)×P(S)P(S) \times P(S)P(S)×P(S) such that A∩B=∅A \cap B = \varnothingA∩B=∅. If the probability of the event E is 3p2q\dfrac{3^{p}}{2^{q}}2q3p​, where p, q ∈N\in \mathbb{N}∈N, then p+qp + qp+q is equal to

Correct answer: 15

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2026
From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is :-
  1. (A)7107\frac{7}{10^{7}}1077​
  2. (B)81108\frac{81}{10^{8}}10881​
  3. (C)67108\frac{67}{10^{8}}10867​
  4. (D)73108\frac{73}{10^{8}}10873​

Correct answer: (D)

Step-by-step solution →
Q15·MathematicsSingle correctJEE Main 2026
Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A. Then a ball is randomly drawn from the bag A. If the probability, that the ball drawn is white, is pq\dfrac{p}{q}qp​, gcd(p, q) = 1, then p+qp + qp+q is equal to
  1. (A)22
  2. (B)23
  3. (C)24
  4. (D)21

Correct answer: (B)

Step-by-step solution →
Q16·MathematicsNumericalJEE Main 2026
From the first 100 natural numbers, two numbers first a and then b are selected randomly without replacement. If the probability that a−b≥10a - b \ge 10a−b≥10 is mn\frac{m}{n}nm​, gcd (m, n) = 1, then m+nm + nm+n is equal to _______.

Correct answer: 311

Step-by-step solution →
Q17·MathematicsSingle correctJEE Main 2026
Two distinct numbers a and b are selected at random from 1, 2, 3,......, 50. The probability, that their product ab is divisible by 3, is
  1. (A)5611225\frac{561}{1225}1225561​
  2. (B)6641225\frac{664}{1225}1225664​
  3. (C)2721225\frac{272}{1225}1225272​
  4. (D)825\frac{8}{25}258​

Correct answer: (B)

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2026
If a random variable x has the probability distribution x : 0, 1, 2, 3, 4, 5, 6, 7 p(x) : 0, 2k, k, 3k, 2k22k^22k2, 2k, k2+kk^2 + kk2+k, 7k27k^27k2 then P(3<x≤6)P(3 < x \le 6)P(3<x≤6) is equal to
  1. (A)0.34
  2. (B)0.22
  3. (C)0.64
  4. (D)0.33

Correct answer: (D)

Step-by-step solution →
Q19·MathematicsSingle correctJEE Main 2026
Let the mean and variance of 7 observations 2, 4, 10, x, 12, 14, y, x>yx > yx>y, be 8 and 16 respectively. Two numbers are chosen from {1,2,3,x–4,y,5}\{1, 2, 3, x–4, y, 5\}{1,2,3,x–4,y,5} one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is:
  1. (A)35\frac{3}{5}53​
  2. (B)45\frac{4}{5}54​
  3. (C)25\frac{2}{5}52​
  4. (D)13\frac{1}{3}31​

Correct answer: (B)

Step-by-step solution →
Q20·MathematicsSingle correctJEE Main 2026
A random variable X takes values 0, 1, 2, 3 with probabilities 2a+130,8a−130,4a+130\frac{2a+1}{30}, \frac{8a-1}{30}, \frac{4a+1}{30}302a+1​,308a−1​,304a+1​, b respectively, where a, b ∈R\mathbf{R}R. Let μ and σ respectively be the mean and standard deviation of X such that σ2+μ2=2\sigma^2 + \mu^2 = 2σ2+μ2=2. Then ab\frac{a}{b}ba​ is equal to :
  1. (A)30
  2. (B)3
  3. (C)60
  4. (D)12

Correct answer: (C)

Step-by-step solution →
Q21·MathematicsNumericalJEE Advanced 2025
A factory has a total of three manufacturing units, M1M_1M1​, M2M_2M2​, and M3M_3M3​, which produce bulbs independent of each other. The units M1M_1M1​, M2M_2M2​, and M3M_3M3​ produce bulbs in the proportions of 2 : 2 : 1. respectively. It is known that 20% of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by M1M_1M1​, 15% are defective. Suppose that, if a randomly chosen bulb produced in the factory is found to be defective, the probability that it was produced by M2M_2M2​ is 25\frac{2}{5}52​. If a bulb is chosen randomly from the bulbs produced by M3M_3M3​, then the probability that it is defective is ______

Correct answer: 0.30

Step-by-step solution →
Q22·MathematicsSingle correctJEE Advanced 2025
Three students S1S_1S1​, S2S_2S2​ and S3S_3S3​ are given a problem to solve. Consider the following events: U: At least one of S1S_1S1​, S2S_2S2​, and S3S_3S3​ can solve the problem, V: S1S_1S1​ can solve the problem, given that neither S2S_2S2​ nor S3S_3S3​ can solve the problem, W: S2S_2S2​ can solve the problem and S3S_3S3​ cannot solve the problem, T: S3S_3S3​ can solve the problem. for any event E, let P(E) denote the probability of E. If P(U)=12P(U) = \frac{1}{2}P(U)=21​, P(V)=110P(V) = \frac{1}{10}P(V)=101​, and P(W)=112P(W) = \frac{1}{12}P(W)=121​, then P(T) is equal to
  1. (A)1336\frac{13}{36}3613​
  2. (B)13\frac{1}{3}31​
  3. (C)1960\frac{19}{60}6019​
  4. (D)14\frac{1}{4}41​

Correct answer: (A)

Step-by-step solution →
Q23·MathematicsSingle correctJEE Main 2025
If AAA and BBB are two events such that P(A)=0.7P(A)=0.7P(A)=0.7, P(B)=0.4P(B)=0.4P(B)=0.4 and P(A∩Bˉ)=0.5P(A\cap\bar B)=0.5P(A∩Bˉ)=0.5, where Bˉ\bar BBˉ denotes the complement of BBB, then P(B ∣ (A∪Bˉ))P\big(B\,|\,(A\cup\bar B)\big)P(B∣(A∪Bˉ)) is equal to:
  1. (A)14\dfrac{1}{4}41​
  2. (B)12\dfrac{1}{2}21​
  3. (C)16\dfrac{1}{6}61​
  4. (D)13\dfrac{1}{3}31​

Correct answer: (A)

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2025
Let a random variable XXX take values 0,1,2,30,1,2,30,1,2,3 with P(X=0)=P(X=1)=pP(X=0)=P(X=1)=pP(X=0)=P(X=1)=p, P(X=2)=P(X=3)P(X=2)=P(X=3)P(X=2)=P(X=3) and E(X2)=2E(X)E(X^2)=2E(X)E(X2)=2E(X). Then the value of 8p−18p-18p−1 is:
  1. (A)0
  2. (B)2
  3. (C)1
  4. (D)3

Correct answer: (B)

Step-by-step solution →
Q25·MathematicsSingle correctJEE Main 2025
A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is mn\dfrac{m}{n}nm​, gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then n2−m2n^2-m^2n2−m2 is equal to:
  1. (A)80
  2. (B)60
  3. (C)72
  4. (D)64

Correct answer: (A)

Step-by-step solution →
Q26·MathematicsSingle correctJEE Main 2025
The probability, of forming a 12 persons committee from 4 engineers, 2 doctors and 10 professors containing at least 3 engineers and at least 1 doctor, is:
  1. (A)129182\dfrac{129}{182}182129​
  2. (B)103182\dfrac{103}{182}182103​
  3. (C)1726\dfrac{17}{26}2617​
  4. (D)1926\dfrac{19}{26}2619​

Correct answer: (A)

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2025
A box contains 10 pens of which 3 are defective. A sample of 2 pens is drawn at random and let XXX denote the number of defective pens. Then the variance of XXX is
  1. (A)1115\dfrac{11}{15}1511​
  2. (B)2875\dfrac{28}{75}7528​
  3. (C)215\dfrac{2}{15}152​
  4. (D)35\dfrac{3}{5}53​

Correct answer: (B)

Step-by-step solution →
Q28·MathematicsIntegerJEE Main 2025
A card from a pack of 52 cards is lost. From the remaining 51 cards, nnn cards are drawn and are found to be spades. If the probability of the lost card to be a spade is 1150\dfrac{11}{50}5011​, then nnn is equal to ______.

Correct answer: 2

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2025
If the probability that the random variable XXX takes the value xxx is given by P(X=x)=k(x+1)3−xP(X=x)=k(x+1)3^{-x}P(X=x)=k(x+1)3−x, x=0,1,2,3,…x=0,1,2,3,\ldotsx=0,1,2,3,…, where kkk is a constant, then P(X≥3)P(X\ge 3)P(X≥3) is equal to:
  1. (A)727\dfrac{7}{27}277​
  2. (B)49\dfrac{4}{9}94​
  3. (C)827\dfrac{8}{27}278​
  4. (D)19\dfrac{1}{9}91​

Correct answer: (D)

Step-by-step solution →
Q30·MathematicsIntegerJEE Main 2025
Three distinct numbers are selected randomly from the set {1,2,3,…,40}\{1,2,3,\ldots,40\}{1,2,3,…,40}. If the probability, that the selected numbers are in an increasing G.P. is mn\dfrac{m}{n}nm​, gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then m+nm+nm+n is equal to ______.

Correct answer: 4949

Step-by-step solution →
Q31·MathematicsSingle correctJEE Main 2025
Three identical bags, each containing 10 balls, have colours as follows — Bag I: 3 Red, 2 Blue, 5 Green; Bag II: 4 Red, 3 Blue, 3 Green; Bag III: 5 Red, 1 Blue, 4 Green. A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from Bag I is ppp, and if the ball is Green, the probability that it is from Bag III is qqq, then the value of (1p+1q)\left(\dfrac{1}{p}+\dfrac{1}{q}\right)(p1​+q1​) is:
  1. (A)6
  2. (B)9
  3. (C)7
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q32·MathematicsSingle correctJEE Main 2025
Let A=[aij]A=[a_{ij}]A=[aij​] be a 2×22\times 22×2 matrix such that aij∈{0,1}a_{ij}\in\{0, 1\}aij​∈{0,1} for all i and j. Let the random variable X denote the possible values of the determinant of the matrix A. Then the variance of X is:
  1. (A)14\dfrac{1}{4}41​
  2. (B)38\dfrac{3}{8}83​
  3. (C)58\dfrac{5}{8}85​
  4. (D)34\dfrac{3}{4}43​

Correct answer: (B)

Step-by-step solution →
Q33·MathematicsSingle correctJEE Main 2025
Bag 1 contains 4 white balls and 5 black balls, and Bag 2 contains n white balls and 3 black balls. One ball is drawn randomly from Bag 1 and transferred to Bag 2. A ball is then drawn randomly from Bag 2. If the probability, that the ball drawn from Bag 2, is white, is 2945\dfrac{29}{45}4529​, then n is equal to:
  1. (A)3
  2. (B)4
  3. (C)5
  4. (D)6

Correct answer: (D)

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2025
Bag B1B_1B1​ contains 6 white and 4 blue balls, Bag B2B_2B2​ contains 4 white and 6 blue balls, and Bag B3B_3B3​ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag B2B_2B2​, is:
  1. (A)13\frac{1}{3}31​
  2. (B)415\frac{4}{15}154​
  3. (C)23\frac{2}{3}32​
  4. (D)25\frac{2}{5}52​

Correct answer: (B)

Step-by-step solution →
Q35·MathematicsSingle correctJEE Main 2025
Two number k1k_1k1​ and k2k_2k2​ are randomly chosen from the set of natural numbers. Then, the probability that the value of ik1+ik2i^{k_1}+i^{k_2}ik1​+ik2​, (i=−1)(i=\sqrt{-1})(i=−1​) is non-zero, equals
  1. (A)12\frac{1}{2}21​
  2. (B)14\frac{1}{4}41​
  3. (C)34\frac{3}{4}43​
  4. (D)23\frac{2}{3}32​

Correct answer: (C)

Step-by-step solution →
Q36·MathematicsSingle correctJEE Main 2025
Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If x denote the number of defective oranges, then the variance of x is:
  1. (A)2875\frac{28}{75}7528​
  2. (B)1425\frac{14}{25}2514​
  3. (C)2675\frac{26}{75}7526​
  4. (D)1825\frac{18}{25}2518​

Correct answer: (A)

Step-by-step solution →
Q37·MathematicsSingle correctJEE Main 2025
A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is
  1. (A)917\dfrac{9}{17}179​
  2. (B)919\dfrac{9}{19}199​
  3. (C)817\dfrac{8}{17}178​
  4. (D)819\dfrac{8}{19}198​

Correct answer: (B)

Step-by-step solution →
Q38·MathematicsSingle correctJEE Main 2025
Let A=[aij]A=[a_{ij}]A=[aij​] be a square matrix of order 2 with entries 0 or 1. Let E be the event that A is an invertible matrix. Then the probability P(E)P(E)P(E) is :
  1. (A)58\dfrac{5}{8}85​
  2. (B)316\dfrac{3}{16}163​
  3. (C)18\dfrac{1}{8}81​
  4. (D)38\dfrac{3}{8}83​

Correct answer: (D)

Step-by-step solution →
Q39·MathematicsSingle correctJEE Main 2025
A board has 16 squares as shown in the figure : Out of these 16 squares, two squares are chosen at random. The probability that they have no side in common is :
  1. (A)45\dfrac{4}{5}54​
  2. (B)710\dfrac{7}{10}107​
  3. (C)35\dfrac{3}{5}53​
  4. (D)2330\dfrac{23}{30}3023​

Correct answer: (A)

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2025
One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is
  1. (A)12\dfrac{1}{2}21​
  2. (B)35\dfrac{3}{5}53​
  3. (C)23\dfrac{2}{3}32​
  4. (D)49\dfrac{4}{9}94​

Correct answer: (A)

Step-by-step solution →
Q41·MathematicsSingle correctJEE Main 2025
A coin is tossed three times. Let XXX be the number of times a tail follows a head. If μ,σ2\mu,\sigma^2μ,σ2 are the mean and variance of XXX, then 64(μ+σ2)64(\mu+\sigma^2)64(μ+σ2) equals:
  1. (A)51
  2. (B)48
  3. (C)32
  4. (D)64

Correct answer: (B)

Step-by-step solution →
Q42·MathematicsSingle correctJEE Main 2025
Two balls are drawn one by one without replacement from a bag of 4 white and 6 black balls. If the probability that the first ball is black given that the second ball is black is mn\tfrac{m}{n}nm​ (gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1), then m+nm+nm+n equals:
  1. (A)14
  2. (B)4
  3. (C)11
  4. (D)13

Correct answer: (A)

Step-by-step solution →
Q43·MathematicsSingle correctJEE Main 2025
If AAA and BBB are two events such that P(A∩B)=0.1P(A\cap B)=0.1P(A∩B)=0.1, and P(A∣B)P(A|B)P(A∣B) and P(B∣A)P(B|A)P(B∣A) are the roots of the equation 12x2−7x+1=012x^2-7x+1=012x2−7x+1=0, then the value of P(Aˉ∪Bˉ)P(Aˉ∩Bˉ)\dfrac{P\left(\bar{A}\cup\bar{B}\right)}{P\left(\bar{A}\cap\bar{B}\right)}P(Aˉ∩Bˉ)P(Aˉ∪Bˉ)​ is:
  1. (A)53\dfrac{5}{3}35​
  2. (B)43\dfrac{4}{3}34​
  3. (C)94\dfrac{9}{4}49​
  4. (D)74\dfrac{7}{4}47​

Correct answer: (C)

Step-by-step solution →
Q44·MathematicsIntegerJEE Advanced 2024
A bag contains N balls out of which 3 balls are white, 6 balls are green, and the remaining balls are blue. Assume that the balls are identical otherwise. Three balls are drawn randomly one after the other without replacement. For i=1,2,3i = 1, 2, 3i=1,2,3, let WiW_iWi​, GiG_iGi​, and BiB_iBi​ denote the events that the ball drawn in the ithi^{th}ith draw is a white ball, green ball, and blue ball, respectively. If the probability P(W1∩G2∩B3)=25NP\left(W_1 \cap G_2 \cap B_3\right) = \frac{2}{5N}P(W1​∩G2​∩B3​)=5N2​ and the conditional probability P(B3∣W1∩G2)=29P\left(B_3 \mid W_1 \cap G_2\right) = \frac{2}{9}P(B3​∣W1​∩G2​)=92​, then N equals ______

Correct answer: 11

Step-by-step solution →
Q45·MathematicsIntegerJEE Advanced 2024
Let XXX be a random variable, and let P(X=x)P(X = x)P(X=x) denote the probability that XXX takes the values xxx. Suppose that the points (x,P(X=x))(x, P(X = x))(x,P(X=x)), x=0,1,2,3,4x = 0, 1, 2, 3, 4x=0,1,2,3,4, lie on a fixed straight line in the xy-plane, and P(X=x)=0P(X = x) = 0P(X=x)=0 for all x∈R−{0,1,2,3,4}x \in R - \{0, 1, 2, 3, 4\}x∈R−{0,1,2,3,4}. If the mean of XXX is 52\frac{5}{2}25​, and the variance of XXX is α\alphaα, then the value of 24α24\alpha24α is ______ .

Correct answer: 42

Step-by-step solution →
Q46·MathematicsSingle correctJEE Advanced 2024
A student appears for a quiz consisting of only true-false type questions and answers all the questions. The student knows the answers of some questions and guesses the answers for the remaining questions. Whenever the student knows the answer of a question, he gives the correct answer. Assume that the probability of the student giving the correct answer for a question, given that he has guess it, is 12\frac{1}{2}21​. Also assume that the probability of the answer for a question being guessed, given that the student's answer is correct, is 16\frac{1}{6}61​. Then the probability that the student knows the answer of a randomly chosen question is
  1. (A)112\frac{1}{12}121​
  2. (B)17\frac{1}{7}71​
  3. (C)57\frac{5}{7}75​
  4. (D)512\frac{5}{12}125​

Correct answer: (C)

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2024
If an unbiased dice is rolled thrice, then the probability of getting a greater number in the ithi^{th}ith roll than the number obtained in the (i−1)th(i-1)^{th}(i−1)th roll, i=2,3i=2, 3i=2,3, is equal to:
  1. (A)354\dfrac{3}{54}543​
  2. (B)254\dfrac{2}{54}542​
  3. (C)554\dfrac{5}{54}545​
  4. (D)154\dfrac{1}{54}541​

Correct answer: (C)

Step-by-step solution →
Q48·MathematicsNumericalJEE Main 2024
Let aaa, bbb and ccc denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked 1,2,3,41, 2, 3, 41,2,3,4. If the probability that ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 has all real roots is mn\frac{m}{n}nm​, gcd⁡(m,n)=1\gcd(m, n) = 1gcd(m,n)=1, then m+nm + nm+n is equal to ________.

Correct answer: 19

Step-by-step solution →
Q49·MathematicsNumericalJEE Main 2024
Three balls are drawn at random from a bag containing 555 blue and 444 yellow balls. Let the random variables XXX and YYY respectively denote the number of blue and yellow balls. If Xˉ\bar XXˉ and Yˉ\bar YYˉ are the means of XXX and YYY respectively, then 7Xˉ+4Yˉ7\bar X+4\bar Y7Xˉ+4Yˉ is equal to ___

Correct answer: 17

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2024
Let the sum of two positive integers be 242424. If the probability, that their product is not less than 34\dfrac3443​ times their greatest possible product, is mn\dfrac{m}{n}nm​, where gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then n−mn-mn−m equals:
  1. (A)999
  2. (B)111111
  3. (C)888
  4. (D)101010

Correct answer: (D)

Step-by-step solution →
Q51·MathematicsSingle correctJEE Main 2024
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 Y, is
  1. (A)13\tfrac{1}{3}31​
  2. (B)12\tfrac{1}{2}21​
  3. (C)14\tfrac{1}{4}41​
  4. (D)512\tfrac{5}{12}125​

Correct answer: (A)

Step-by-step solution →
Q52·MathematicsSingle correctJEE Main 2024
If three letters can be posted to any one of the 5 different addresses, then the probability that the three letters are posted to exactly two addresses is:
  1. (A)1225\tfrac{12}{25}2512​
  2. (B)1825\tfrac{18}{25}2518​
  3. (C)425\tfrac{4}{25}254​
  4. (D)625\tfrac{6}{25}256​

Correct answer: (A)

Step-by-step solution →
Q53·MathematicsSingle correctJEE Main 2024
A company has two plants AAA and BBB to manufacture motorcycles. 60% motorcycles are manufactured at plant AAA and the remaining at plant BBB. 80% of the motorcycles manufactured at plant AAA are of standard quality, while 90% of those manufactured at plant BBB are of standard quality. If ppp is the probability that a motorcycle, found to be of standard quality, is manufactured at plant BBB, then 126p126p126p is
  1. (A)545454
  2. (B)646464
  3. (C)666666
  4. (D)565656

Correct answer: (A)

Step-by-step solution →
Q54·MathematicsNumericalJEE Main 2024
From a lot of 121212 items containing 333 defectives, a sample of 555 items is drawn at random. Let the random variable XXX denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of XXX is mn\frac{m}{n}nm​, where gcd⁡(m,n)=1\gcd(m,n)=1gcd(m,n)=1, then n−mn-mn−m is equal to ______.

Correct answer: 71

Step-by-step solution →
Q55·MathematicsNumericalJEE Main 2024
Let the mean and the standard deviation of the probability distribution with X={α,1,0,−3}X=\{\alpha,1,0,-3\}X={α,1,0,−3} and corresponding P(X)={13,K,16,14}P(X)=\left\{\dfrac{1}{3},K,\dfrac{1}{6},\dfrac{1}{4}\right\}P(X)={31​,K,61​,41​} (in the same order) be μ\muμ and σ\sigmaσ respectively. If σ−μ=2\sigma-\mu=2σ−μ=2, then σ+μ\sigma+\muσ+μ is equal to __________.

Correct answer: 5

Step-by-step solution →
Q56·MathematicsSingle correctJEE Main 2024
The coefficients a,b,ca, b, ca,b,c in the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 are chosen from the set {1,2,3,4,5,6,7,8}\{1, 2, 3, 4, 5, 6, 7, 8\}{1,2,3,4,5,6,7,8}. The probability of this equation having repeated roots is:
  1. (A)3256\dfrac{3}{256}2563​
  2. (B)1128\dfrac{1}{128}1281​
  3. (C)164\dfrac{1}{64}641​
  4. (D)3128\dfrac{3}{128}1283​

Correct answer: (C)

Step-by-step solution →
Q57·MathematicsNumericalJEE Main 2024
From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable X denote the number of defective items in the sample. If the variance of X is σ2\sigma^2σ2, then 96σ296\sigma^296σ2 is equal to ___.

Correct answer: 56

Step-by-step solution →
Q58·MathematicsSingle correctJEE Main 2024
The coefficients a,b,ca,b,ca,b,c in the quadratic equation ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0 are from the set {1,2,3,4,5,6}\{1,2,3,4,5,6\}{1,2,3,4,5,6}. If the probability of this equation having one real root bigger than the other is ppp, then 216p216p216p equals:
  1. (A)57
  2. (B)38
  3. (C)19
  4. (D)76

Correct answer: (B)

Step-by-step solution →
Q59·MathematicsNumericalJEE Main 2024
In a tournament, a team plays 10 matches with probabilities of winning and losing each match as 13\dfrac{1}{3}31​ and 23\dfrac{2}{3}32​ respectively. Let xxx be the number of matches that the team wins, and yyy be the number of matches that team loses. If the probability P(∣x−y∣≤2)P(|x-y|\le 2)P(∣x−y∣≤2) is ppp, then 39p3^9 p39p equals

Correct answer: 8288

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2024
Three urns AAA, BBB and CCC contain 777 red, 555 black; 555 red, 777 black and 666 red, 666 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 AAA is:
  1. (A)417\dfrac{4}{17}174​
  2. (B)518\dfrac{5}{18}185​
  3. (C)718\dfrac{7}{18}187​
  4. (D)516\dfrac{5}{16}165​

Correct answer: (B)

Step-by-step solution →
Q61·MathematicsSingle correctJEE Main 2024
If the mean of the following probability distribution of a random variable X: (XXX: 0, 2, 4, 6, 8 with P(X)P(X)P(X): 2a2a2a, 3a3a3a, a+ba+ba+b, 2b2b2b, 3b3b3b respectively) is 469\dfrac{46}{9}946​, then the variance of the distribution is
  1. (A)58181\tfrac{581}{81}81581​
  2. (B)56681\tfrac{566}{81}81566​
  3. (C)17327\tfrac{173}{27}27173​
  4. (D)15127\tfrac{151}{27}27151​

Correct answer: (B)

Step-by-step solution →
Q62·MathematicsSingle correctJEE Main 2024
Let Ajay will not appear in JEE exam with probability p=27p=\dfrac{2}{7}p=72​, while both Ajay and Vijay will appear in the exam with probability q=15q=\dfrac{1}{5}q=51​. Then the probability, that Ajay will appear in the exam and Vijay will not appear is:
  1. (A)935\dfrac{9}{35}359​
  2. (B)1835\dfrac{18}{35}3518​
  3. (C)2435\dfrac{24}{35}3524​
  4. (D)335\dfrac{3}{35}353​

Correct answer: (B)

Step-by-step solution →
Q63·MathematicsSingle correctJEE Main 2024
A bag contains 888 balls, whose colours are either white or black. 444 balls are drawn at random without replacement and it was found that 222 balls are white and 222 balls are black. The probability that the bag contains equal number of white and black balls is:
  1. (A)25\dfrac{2}{5}52​
  2. (B)27\dfrac{2}{7}72​
  3. (C)17\dfrac{1}{7}71​
  4. (D)15\dfrac{1}{5}51​

Correct answer: (B)

Step-by-step solution →
Q64·MathematicsSingle correctJEE Main 2024
Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability, that first drawn marble is red and second drawn marble is white, is
  1. (A)225\dfrac{2}{25}252​
  2. (B)425\dfrac{4}{25}254​
  3. (C)23\dfrac{2}{3}32​
  4. (D)475\dfrac{4}{75}754​

Correct answer: (D)

Step-by-step solution →
Q65·MathematicsSingle correctJEE Main 2024
Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable x to be the number of rotten apples in a draw of two apples, the variance of x is
  1. (A)37153\dfrac{37}{153}15337​
  2. (B)57153\dfrac{57}{153}15357​
  3. (C)47153\dfrac{47}{153}15347​
  4. (D)40153\dfrac{40}{153}15340​

Correct answer: (D)

Step-by-step solution →
Q66·MathematicsSingle correctJEE Main 2024
A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is
  1. (A)29\dfrac{2}{9}92​
  2. (B)19\dfrac{1}{9}91​
  3. (C)227\dfrac{2}{27}272​
  4. (D)127\dfrac{1}{27}271​

Correct answer: (A)

Step-by-step solution →
Q67·MathematicsSingle correctJEE Main 2024
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 bag A, if the ball drawn is white, is:
  1. (A)14\dfrac{1}{4}41​
  2. (B)19\dfrac{1}{9}91​
  3. (C)13\dfrac{1}{3}31​
  4. (D)310\dfrac{3}{10}103​

Correct answer: (C)

Step-by-step solution →
Q68·MathematicsSingle correctJEE Main 2024
Two integers xxx and yyy are chosen with replacement from the set {0,1,2,3,…,10}\{0,1,2,3,\ldots,10\}{0,1,2,3,…,10}. Then the probability that ∣x−y∣>5|x-y|>5∣x−y∣>5 is:
  1. (A)30121\dfrac{30}{121}12130​
  2. (B)62121\dfrac{62}{121}12162​
  3. (C)60121\dfrac{60}{121}12160​
  4. (D)31121\dfrac{31}{121}12131​

Correct answer: (A)

Step-by-step solution →
Q69·MathematicsSingle correctJEE Main 2024
A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is
  1. (A)56\dfrac{5}{6}65​
  2. (B)16\dfrac{1}{6}61​
  3. (C)511\dfrac{5}{11}115​
  4. (D)611\dfrac{6}{11}116​

Correct answer: (C)

Step-by-step solution →
Q70·MathematicsSingle correctJEE Main 2024
An integer is chosen at random from the integers 1,2,3,…,501,2,3,\ldots,501,2,3,…,50. The probability that the chosen integer is a multiple of atleast one of 4,64,64,6 and 777 is:
  1. (A)825\dfrac{8}{25}258​
  2. (B)2150\dfrac{21}{50}5021​
  3. (C)950\dfrac{9}{50}509​
  4. (D)1450\dfrac{14}{50}5014​

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correctJEE Main 2024
An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability, that the first draw gives all white balls and the second draw gives all black balls, is :
  1. (A)5256\dfrac{5}{256}2565​
  2. (B)5715\dfrac{5}{715}7155​
  3. (C)3715\dfrac{3}{715}7153​
  4. (D)3256\dfrac{3}{256}2563​

Correct answer: (C)

Step-by-step solution →
Q72·MathematicsNumericalJEE Main 2024
A fair die is tossed repeatedly until a six is obtained. Let XXX denote the number of tosses required and let a=P(X=3)a=P(X=3)a=P(X=3), b=P(X≥3)b=P(X\ge 3)b=P(X≥3) and c=P(X≥6∣X>3)c=P(X\ge 6\mid X>3)c=P(X≥6∣X>3). Then b+ca\dfrac{b+c}{a}ab+c​ is equal to __________.

Correct answer: 12

Step-by-step solution →
Q73·MathematicsSingle correctJEE Advanced 2023
Let X:{(x,y)∈Z×Z:x28+y220<1 and y2<5x}X : \left\{(x, y) \in Z \times Z : \frac{x^2}{8} + \frac{y^2}{20} < 1 \text{ and } y^2 < 5x \right\}X:{(x,y)∈Z×Z:8x2​+20y2​<1 and y2<5x} . Three distinct points P, Q and R are randomly chosen from X . Then the probability that P, Q and R form a triangle whose area is a positive integer, is
  1. (A)71220\frac{71}{220}22071​
  2. (B)73220\frac{73}{220}22073​
  3. (C)79220\frac{79}{220}22079​
  4. (D)83220\frac{83}{220}22083​

Correct answer: (B)

Step-by-step solution →
Q74·MathematicsSingle correctJEE Advanced 2023
Consider an experiment of tossing a coin repeatedly until the outcomes of two consecutive tosses are same. If the probability of a random toss resulting in head is 13\frac{1}{3}31​, then the probability that the experiment stops with head is
  1. (A)13\frac{1}{3}31​
  2. (B)521\frac{5}{21}215​
  3. (C)421\frac{4}{21}214​
  4. (D)27\frac{2}{7}72​

Correct answer: (B)

Step-by-step solution →
Q75·MathematicsNumericalJEE Advanced 2023
Consider the 6×66 \times 66×6 square in the figure. Let A1A_{1}A1​, A2A_{2}A2​, ....., A49A_{49}A49​ be the points of intersections (dots in the picture) in some order. We say that AiA_{i}Ai​ and AjA_{j}Aj​ are friends if they are adjacent along a row or along a column. Assume that each point AiA_{i}Ai​ has an equal chance of being chosen. Two distinct points are chosen randomly out of the points A1A_{1}A1​, A2A_{2}A2​, ....., A49A_{49}A49​. Let ppp be the probability that they are friends. Then the value of 7p7p7p is

Correct answer: 0.5

Step-by-step solution →
Q76·MathematicsNumericalJEE Advanced 2023
Consider the 6×66 \times 66×6 square in the figure. Let A1A_{1}A1​, A2A_{2}A2​, ....., A49A_{49}A49​ be the points of intersections (dots in the picture) in some order. We say that AiA_{i}Ai​ and AjA_{j}Aj​ are friends if they are adjacent along a row or along a column. Assume that each point AiA_{i}Ai​ has an equal chance of being chosen. Let pip_{i}pi​ be the probability that a randomly chosen point has iii many friends, i=0,1,2,3,4i = 0, 1, 2, 3, 4i=0,1,2,3,4. Let XXX be a random variable such that for i=0,1,2,3,4i = 0, 1, 2, 3, 4i=0,1,2,3,4, the probability P(X=i)=piP(X = i) = p_{i}P(X=i)=pi​. Then the value of 7E(X)7E(X)7E(X) is

Correct answer: 24

Step-by-step solution →
Q77·MathematicsSingle correctJEE Main 2023
A bag contains 6 white and 4 black balls. A die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is:
  1. (A)14\frac{1}{4}41​
  2. (B)950\frac{9}{50}509​
  3. (C)15\frac{1}{5}51​
  4. (D)1150\frac{11}{50}5011​

Correct answer: (C)

Step-by-step solution →
Q78·MathematicsSingle correctJEE Main 2023
A coin is biased so that a head is 333 times as likely to occur as a tail. This coin is tossed until a head or three tails occur. If XXX denotes the number of tosses of the coin, then the mean of XXX is:
  1. (A)2116\dfrac{21}{16}1621​
  2. (B)8164\dfrac{81}{64}6481​
  3. (C)1516\dfrac{15}{16}1615​
  4. (D)3716\dfrac{37}{16}1637​

Correct answer: (A)

Step-by-step solution →
Q79·MathematicsSingle correctJEE Main 2023
The random variable X follows binomial distribution B(n,p)B(n, p)B(n,p) for which the difference of the mean and the variance is 1. If 2P(X=2)=3P(X=1)2P(X = 2) = 3P(X = 1)2P(X=2)=3P(X=1), then n2P(X>1)n^2 P(X > 1)n2P(X>1) is equal to
  1. (A)12
  2. (B)15
  3. (C)11
  4. (D)16

Correct answer: (C)

Step-by-step solution →
Q80·MathematicsNumericalJEE Main 2023
A fair die with faces marked 111 to nnn (n>1n>1n>1) is tossed repeatedly until a number less than n appears. If the mean of the number of tosses required is n9\dfrac{n}{9}9n​, then n is equal to

Correct answer: 10

Step-by-step solution →
Q81·MathematicsSingle correctJEE Main 2023
Let R be a rectangle given by the lines x=0x = 0x=0, x=2x = 2x=2, y=0y = 0y=0 and y=5y = 5y=5. Let A(α,0)A(\alpha, 0)A(α,0) and B(0,β)B(0, \beta)B(0,β), α∈[0,2]\alpha \in [0, 2]α∈[0,2] and β∈[0,5]\beta \in [0, 5]β∈[0,5], be such that the line segment AB divides the area of the rectangle R in the ratio 4:14:14:1. Then, the mid-point of AB lies on a
  1. (A)parabola
  2. (B)hyberbola
  3. (C)straight line
  4. (D)circle

Correct answer: (B)

Step-by-step solution →
Q82·MathematicsSingle correctJEE Main 2023
Let S={M=[aij], aij∈{0,1,2}, 1≤i,j≤2}S = \{ M = [a_{ij}],\ a_{ij} \in \{0, 1, 2\},\ 1 \le i, j \le 2 \}S={M=[aij​], aij​∈{0,1,2}, 1≤i,j≤2} be a sample space and A={M∈S:M is invertible}A = \{ M \in S : M \text{ is invertible} \}A={M∈S:M is invertible} be an event. Then P(A)P(A)P(A) is equal to
  1. (A)5081\frac{50}{81}8150​
  2. (B)4781\frac{47}{81}8147​
  3. (C)4981\frac{49}{81}8149​
  4. (D)1627\frac{16}{27}2716​

Correct answer: (A)

Step-by-step solution →
Q83·MathematicsNumericalJEE Main 2023
The probability of getting a head for a biased coin is 14\dfrac{1}{4}41​. It is tossed repeatedly until a head appears. Let NNN be the number of tosses required. If the probability that the equation 64x2+5Nx+1=064x^2+5Nx+1=064x2+5Nx+1=0 has no real root is pq\dfrac{p}{q}qp​, where ppp and qqq are co-prime, then q−pq-pq−p is equal to

Correct answer: 27

Step-by-step solution →
Q84·MathematicsSingle correctJEE Main 2023
Let N denotes the sum of the numbers obtained when two dice are rolled. If the probability that 2N<N!2^N<N!2N<N! is mn\dfrac{m}{n}nm​, where m and n are coprime, then 4m−3n4m-3n4m−3n is equal to
  1. (A)888
  2. (B)161616
  3. (C)101010
  4. (D)121212

Correct answer: (A)

Step-by-step solution →
Q85·MathematicsSingle correctJEE Main 2023
Let a die be rolled nnn 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 k215\frac{k}{2^{15}}215k​, then kkk is equal to:
  1. (A)303030
  2. (B)909090
  3. (C)151515
  4. (D)606060

Correct answer: (D)

Step-by-step solution →
Q86·MathematicsSingle correctJEE Main 2023
In a bolt factory, machines A, B and C manufacture respectively 20%, 30% and 50% of the total bolts. Of their output 3, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. If the bolt drawn is found the defective, then the probability that it is manufactured by the machine C is
  1. (A)27\frac{2}{7}72​
  2. (B)928\frac{9}{28}289​
  3. (C)514\frac{5}{14}145​
  4. (D)37\frac{3}{7}73​

Correct answer: (C)

Step-by-step solution →
Q87·MathematicsSingle correctJEE Main 2023
If the probability that the random variable X takes values x is given by P(X=x)=k(x+1)3−xP(X=x)=k(x+1)3^{-x}P(X=x)=k(x+1)3−x, x=0,1,2,3,…x=0,1,2,3,\ldotsx=0,1,2,3,…, where k is a constant, then P(X≥2)P(X\geq2)P(X≥2) is equal to
  1. (A)127\dfrac{1}{27}271​
  2. (B)1118\dfrac{11}{18}1811​
  3. (C)718\dfrac{7}{18}187​
  4. (D)2027\dfrac{20}{27}2720​

Correct answer: (A)

Step-by-step solution →
Q88·MathematicsSingle correctJEE Main 2023
A pair of dice are thrown 555 times. For each throw, a total of 555 is considered a success. If the probability of at least 444 successes is k311\dfrac{k}{3^{11}}311k​, then kkk is equal to:
  1. (A)828282
  2. (B)123123123
  3. (C)164164164
  4. (D)757575

Correct answer: (B)

Step-by-step solution →
Q89·MathematicsSingle correctJEE Main 2023
Three dice are rolled. If the probability of getting different numbers on the three dice is pq\frac{p}{q}qp​, where ppp and qqq are co-prime, then q−pq-pq−p is equal to
  1. (A)4
  2. (B)3
  3. (C)1
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q90·MathematicsSingle correctJEE Main 2023
Two dice are thrown independently. Let A be the event that the number appeared on the 1st1^{st}1st die is less than the number appeared on the 2nd2^{nd}2nd die, B be the event that the number appeared on the 1st1^{st}1st die is even and that on the second die is odd, and C be the event that the number appeared on the 1st1^{st}1st die is odd and that on the 2nd2^{nd}2nd die is even. Then:
  1. (A)the number of favourable cases of the events A, B and C are 15, 6 and 6 respectively
  2. (B)the number of favourable cases of the event (A∪B)∩C(A\cup B)\cap C(A∪B)∩C is 6
  3. (C)B and C are independent
  4. (D)A and B are mutually exclusive

Correct answer: (B)

Step-by-step solution →
Q91·MathematicsSingle correctJEE Main 2023
In a binomial distribution B(n,p)B(n,p)B(n,p), the sum and the product of the mean and the variance are 555 and 666 respectively, then 6(n+p−q)6(n+p-q)6(n+p−q) is equal to
  1. (A)525252
  2. (B)505050
  3. (C)515151
  4. (D)535353

Correct answer: (A)

Step-by-step solution →
Q92·MathematicsNumericalJEE Main 2023
Let AAA be the event that the absolute difference between two randomly chosen real numbers in the sample space [0,60][0,60][0,60] is less than or equal to aaa. If P(A)=1136P(A)=\dfrac{11}{36}P(A)=3611​, then aaa is equal to

Correct answer: 10

Step-by-step solution →
Q93·MathematicsSingle correctJEE Main 2023
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
  1. (A)37\frac{3}{7}73​
  2. (B)57\frac{5}{7}75​
  3. (C)56\frac{5}{6}65​
  4. (D)27\frac{2}{7}72​

Correct answer: (B)

Step-by-step solution →
Q94·MathematicsSingle correctJEE Main 2023
If an unbiased die, marked with −2,−1,0,1,2,3-2,-1,0,1,2,3−2,−1,0,1,2,3 on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is:
  1. (A)8812592\dfrac{881}{2592}2592881​
  2. (B)27288\dfrac{27}{288}28827​
  3. (C)4402592\dfrac{440}{2592}2592440​
  4. (D)5212592\dfrac{521}{2592}2592521​

Correct answer: (D)

Step-by-step solution →
Q95·MathematicsNumericalJEE Main 2023
A bag contains six balls of different colours. Two balls are drawn in succession with replacement. The probability that both the balls are of the same colour is ppp. Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colour is qqq. If p:q=m:np:q=m:np:q=m:n, where mmm and nnn are coprime, then m+nm + nm+n is equal to _______.

Correct answer: 14

Step-by-step solution →
Q96·MathematicsSingle correctJEE Main 2023
Fifteen football players of a club-team are given 15 T-shirts with their names written on the backside. If the players pick up the T-shirts randomly, then the probability that at least 3 players pick the correct T-shirt is
  1. (A)524\dfrac{5}{24}245​
  2. (B)16\dfrac{1}{6}61​
  3. (C)536\dfrac{5}{36}365​
  4. (D)215\dfrac{2}{15}152​

Correct answer: (B)

Step-by-step solution →
Q97·MathematicsSingle correctJEE Main 2023
Three rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable XXX denote the number of rotten apples. If μ\muμ and σ2\sigma^2σ2 represent mean and variance of XXX, respectively, then 10(μ2+σ2)10(\mu^2 + \sigma^2)10(μ2+σ2) is equal to
  1. (A)250
  2. (B)25
  3. (C)30
  4. (D)20

Correct answer: (D)

Step-by-step solution →
Q98·MathematicsSingle correctJEE Main 2023
Let S={w1,w2,… }S=\{w_1,w_2,\dots\}S={w1​,w2​,…} be the sample space associated to a random experiment. Let P(wn)=P(wn−1)2, n≥2P(w_n)=\dfrac{P(w_{n-1})}{2},\,n\ge 2P(wn​)=2P(wn−1​)​,n≥2. Let A={2k+3l: k,l∈N}A=\{2k+3l:\,k,l\in\mathbb{N}\}A={2k+3l:k,l∈N} and B={wn: n∈A}B=\{w_n:\,n\in A\}B={wn​:n∈A}. Then P(B)P(B)P(B) is equal to:
  1. (A)364\dfrac{3}{64}643​
  2. (B)116\dfrac{1}{16}161​
  3. (C)132\dfrac{1}{32}321​
  4. (D)332\dfrac{3}{32}323​

Correct answer: (A)

Step-by-step solution →
Q99·MathematicsNumericalJEE Main 2023
25%25\%25% of the population are smokers. A smoker has 272727 times more chances to develop lung cancer than a non smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is k10\dfrac{k}{10}10k​. Then the value of kkk is.

Correct answer: 9

Step-by-step solution →
Q100·MathematicsSingle correctJEE Main 2023
Let NNN be the sum of the numbers appeared when two fair dice are rolled and let the probability that N−2N - 2N−2, 3N\sqrt{3N}3N​, N+2N + 2N+2 are in geometric progression be k48\dfrac{k}{48}48k​. Then the value of kkk is
  1. (A)8
  2. (B)16
  3. (C)2
  4. (D)4

Correct answer: (D)

Step-by-step solution →
Q101·MathematicsSingle correctJEE Main 2023
Let MMM be the maximum value of the product of two positive integers when their sum is 666666. Let the sample space S={x∈Z:x(66−x)≥59M}S=\Big\{x\in\mathbb{Z}:x(66-x)\ge\dfrac{5}{9}M\Big\}S={x∈Z:x(66−x)≥95​M} and the event A={x∈S:x is a multiple of 3}A=\{x\in S:x\text{ is a multiple of }3\}A={x∈S:x is a multiple of 3}. Then P(A)P(A)P(A) is equal to:
  1. (A)1522\dfrac{15}{22}2215​
  2. (B)15\dfrac{1}{5}51​
  3. (C)1544\dfrac{15}{44}4415​
  4. (D)13\dfrac{1}{3}31​

Correct answer: (D)

Step-by-step solution →
Q102·MathematicsSingle correctJEE Main 2023
Let NNN denote the number that turns up when a fair die is rolled. If the probability that the system of equations x+y+z=1x+y+z=1x+y+z=1, 2x+Ny+2z=22x+Ny+2z=22x+Ny+2z=2, 3x+3y+Nz=33x+3y+Nz=33x+3y+Nz=3 has unique solution is k6\dfrac{k}{6}6k​, then the sum of value of kkk and all possible values of NNN is
  1. (A)212121
  2. (B)181818
  3. (C)202020
  4. (D)191919

Correct answer: (C)

Step-by-step solution →
Q103·MathematicsNumericalJEE Main 2023
Three urns A, B and C contain 444 red, 666 black; 555 red, 555 black; and λ\lambdaλ red, 444 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 0.40.40.4, then the square of the length of the side of the largest equilateral triangle, inscribed in the parabola y2=λxy^2=\lambda xy2=λx with one vertex at the vertex of the parabola, is

Correct answer: 432

Step-by-step solution →
Q104·MathematicsSingle correctJEE Main 2023
Let Ω\OmegaΩ be the sample space and A⊆ΩA\subseteq\OmegaA⊆Ω be an event. Given below are two statements: (S1): If P(A)=0P(A)=0P(A)=0, then A=∅A=\varnothingA=∅. (S2): If P(A)=1P(A)=1P(A)=1, then A=ΩA=\OmegaA=Ω. Then
  1. (A)both (S1) and (S2) are true
  2. (B)only (S1) is true
  3. (C)only (S2) is true
  4. (D)both (S1) and (S2) are false

Correct answer: (D)

Step-by-step solution →
Q105·MathematicsSingle correctJEE Advanced 2022
Suppose that Box-I contains 8 red, 3 blue and 5 green balls, Box-II contains 24 red, 9 blue and 15 green balls, Box-III contains 1 blue, 12 green and 3 yellow balls, Box-IV contains 10 green, 16 orange and 6 white balls, A ball is chosen randomly for Box-I; call that ball b. If b is red then a ball is chosen randomly from Box-II, if b is blue then a ball is chosen randomly from Box-III, and if b is green then a ball is chosen randomly from Box-IV. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen ball is green' has happened, is equal to
  1. (A)15256\frac{15}{256}25615​
  2. (B)316\frac{3}{16}163​
  3. (C)552\frac{5}{52}525​
  4. (D)18\frac{1}{8}81​ .

Correct answer: (C)

Step-by-step solution →
Q106·MathematicsSingle correctJEE Advanced 2022
Two players, P1P_{1}P1​ and P2P_{2}P2​, play a game against each other. In every round of the game, each player rolls a fair die once, where the six faces of the die have six distinct numbers. Let xxx and yyy denote the readings on the die rolled by P1P_{1}P1​ and P2P_{2}P2​, respectively. If x > y, then P1P_{1}P1​ scores 5 points and P2P_{2}P2​ scores 0 points. If x=yx = yx=y, then each player scores 2 points. If x<yx < yx<y, then P1P_{1}P1​ scores 0 point and P2P_{2}P2​ scores 5 points. Let XiX_{i}Xi​ and YiY_{i}Yi​ be the total scores of P1P_{1}P1​ and P2P_{2}P2​, respectively, after playing the ithi^{\text{th}}ith round. The correct option is:
List-IList-II
I.Probability of (X2≥Y2)\left(X_{2} \geq Y_{2}\right)(X2​≥Y2​) isP.38\frac{3}{8}83​
II.Probability of (X2>Y2)\left(X_{2} > Y_{2}\right)(X2​>Y2​) isQ.1116\frac{11}{16}1611​
III.Probability of (X3=Y3)\left(X_{3} = Y_{3}\right)(X3​=Y3​) isR.516\frac{5}{16}165​
IV.Probability of (X3>Y3)\left(X_{3} > Y_{3}\right)(X3​>Y3​) isS.355864\frac{355}{864}864355​
T.77432\frac{77}{432}43277​
  1. (A)(I) → (Q; (II) → (R); (III) → (T); (IV) → (S)
  2. (B)(I) → (Q; (II) → (R); (III) → (T); (IV) → (T)
  3. (C)(I) → (P); (II) → (R); (III) → (Q); (IV) → (S)
  4. (D)(I) → (P); (II) → (R); (III) → (Q); (IV) → (T)

Correct answer: (A)

Step-by-step solution →
Q107·MathematicsSingle correctJEE Main 2022
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 draw from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is:
  1. (A)49\frac{4}{9}94​
  2. (B)518\frac{5}{18}185​
  3. (C)16\frac{1}{6}61​
  4. (D)310\frac{3}{10}103​

Correct answer: (B)

Step-by-step solution →
Q108·MathematicsSingle correctJEE Main 2022
Let S={1, 2, 3, …, 2022}S = \{1,\,2,\,3,\,\ldots,\,2022\}S={1,2,3,…,2022}. Then the probability, that a randomly chosen number n from the set S such that HCF (n, 2022)=1(n,\,2022) = 1(n,2022)=1, is :
  1. (A)1281011\frac{128}{1011}1011128​
  2. (B)1661011\frac{166}{1011}1011166​
  3. (C)127337\frac{127}{337}337127​
  4. (D)112337\frac{112}{337}337112​

Correct answer: (D)

Step-by-step solution →
Q109·MathematicsNumericalJEE Main 2022
The sum and product of the mean and variance of a binomial distribution are 82.5 and 1350 respectively. They the number of trials in the binomial distribution is:

Correct answer: 96

Step-by-step solution →
Q110·MathematicsNumericalJEE Main 2022
A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X be the number of white balls, among the drawn balls. If σ2\sigma^{2}σ2 is the variance of X, then 100 σ2100\,\sigma^{2}100σ2 is equal to

Correct answer: 56

Step-by-step solution →
Q111·MathematicsSingle correctJEE Main 2022
Let A and B be two events such that P(B∣A)=25P(B|A)=\frac{2}{5}P(B∣A)=52​, P(A∣B)=17P(A|B)=\frac{1}{7}P(A∣B)=71​ and P(A∩B)=19P(A \cap B)=\frac{1}{9}P(A∩B)=91​. Consider (S1) P(A′∪B)=56P(A' \cup B)=\frac{5}{6}P(A′∪B)=65​, (S2) P(A′∩B′)=118P(A' \cap B')=\frac{1}{18}P(A′∩B′)=181​. Then
  1. (A)Both (S1) and (S2) are true
  2. (B)Both (S1) and (S2) are false
  3. (C)Only (S1) is true
  4. (D)Only (S2) is true

Correct answer: (A)

Step-by-step solution →
Q112·MathematicsSingle correctJEE Main 2022
Let S be the sample space of all five digit numbers. If ppp is the probability that a randomly selected number from S, is a multiple of 7 but not divisible by 5, then 9p9p9p is equal to
  1. (A)1.0146
  2. (B)1.2085
  3. (C)1.0285
  4. (D)1.1521

Correct answer: (C)

Step-by-step solution →
Q113·MathematicsSingle correctJEE Main 2022
A six faced die is biased such that 3×P(a prime number)=6×P(a composite number)=2×P(1)3 \times P(\text{a prime number}) = 6 \times P(\text{a composite number}) = 2 \times P(1)3×P(a prime number)=6×P(a composite number)=2×P(1). Let X be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of X is :
  1. (A)311\frac{3}{11}113​
  2. (B)511\frac{5}{11}115​
  3. (C)711\frac{7}{11}117​
  4. (D)811\frac{8}{11}118​

Correct answer: (D)

Step-by-step solution →
Q114·MathematicsSingle correctJEE Main 2022
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)=k2nP(X > n - 3) = \frac{k}{2^{n}}P(X>n−3)=2nk​, then k is equal to
  1. (A)528
  2. (B)529
  3. (C)629
  4. (D)630

Correct answer: (B)

Step-by-step solution →
Q115·MathematicsSingle correctJEE Main 2022
Let E1_{1}1​, E2_{2}2​, E3_{3}3​ be three mutually exclusive events such that P(E1)=2+3p6P(E_{1}) = \frac{2+3p}{6}P(E1​)=62+3p​, P(E2)=2−p8P(E_{2}) = \frac{2-p}{8}P(E2​)=82−p​ and P(E3)=1−p2P(E_{3}) = \frac{1-p}{2}P(E3​)=21−p​. If the maximum and minimum values of p are p1_{1}1​ and p2_{2}2​, then (p1_{1}1​ + p2_{2}2​) is equal to :
  1. (A)23\frac{2}{3}32​
  2. (B)53\frac{5}{3}35​
  3. (C)54\frac{5}{4}45​
  4. (D)1

Correct answer: (B)

Step-by-step solution →
Q116·MathematicsSingle correctJEE Main 2022
Let X be a binomially distributed random variable with mean 4 and variance 43\frac{4}{3}34​. Then 54 P(X≤2)54\,P(X\le 2)54P(X≤2) is equal to
  1. (A)7327\frac{73}{27}2773​
  2. (B)14627\frac{146}{27}27146​
  3. (C)14681\frac{146}{81}81146​
  4. (D)12681\frac{126}{81}81126​

Correct answer: (B)

Step-by-step solution →
Q117·MathematicsSingle correctJEE Main 2022
The mean and variance of a binomial distribution are α and α3\frac{\alpha}{3}3α​ respectively. If P(X=1)=4243P\left(X = 1\right) = \frac{4}{243}P(X=1)=2434​, then P(X = 4 or 5) is equal to :
  1. (A)59\frac{5}{9}95​
  2. (B)6481\frac{64}{81}8164​
  3. (C)1627\frac{16}{27}2716​
  4. (D)145243\frac{145}{243}243145​

Correct answer: (C)

Step-by-step solution →
Q118·MathematicsSingle correctJEE Main 2022
If the numbers appeared on the two throws of a fair six faced die are α\alphaα and β\betaβ, then the probability that x2+αx+β>0x^{2} + \alpha x + \beta > 0x2+αx+β>0, for all x∈Rx \in Rx∈R, is :
  1. (A)1736\frac{17}{36}3617​
  2. (B)49\frac{4}{9}94​
  3. (C)12\frac{1}{2}21​
  4. (D)1936\frac{19}{36}3619​

Correct answer: (A)

Step-by-step solution →
Q119·MathematicsSingle correctJEE Main 2022
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 :
  1. (A)33232\frac{33}{2^{32}}23233​
  2. (B)33229\frac{33}{2^{29}}22933​
  3. (C)33228\frac{33}{2^{28}}22833​
  4. (D)33227\frac{33}{2^{27}}22733​

Correct answer: (C)

Step-by-step solution →
Q120·MathematicsSingle correctJEE Main 2022
The probability that a randomly chosen 2 × 2 matrix with all the entries from the set of first 10 primes, is singular, is equal to :
  1. (A)133104\frac{133}{10^{4}}104133​
  2. (B)18103\frac{18}{10^{3}}10318​
  3. (C)19103\frac{19}{10^{3}}10319​
  4. (D)271104\frac{271}{10^{4}}104271​

Correct answer: (C)

Step-by-step solution →
Q121·MathematicsSingle correctJEE Main 2022
The probability that a relation R from {x,y}\{x,y\}{x,y} to {x,y}\{x,y\}{x,y} is both symmetric and transitive, is equal to:
  1. (A)516\frac{5}{16}165​
  2. (B)916\frac{9}{16}169​
  3. (C)1116\frac{11}{16}1611​
  4. (D)1316\frac{13}{16}1613​

Correct answer: (A)

Step-by-step solution →
Q122·MathematicsSingle correctJEE Main 2022
The probability that a randomly chosen one-one function from the set {a,b,c,d}\{a, b, c, d\}{a,b,c,d} to the set {1,2,3,4,5}\{1, 2, 3, 4, 5\}{1,2,3,4,5} satisfies f(a)+2f(b)−f(c)=f(d)f(a) + 2f(b) - f(c) = f(d)f(a)+2f(b)−f(c)=f(d) is :
  1. (A)124\frac{1}{24}241​
  2. (B)140\frac{1}{40}401​
  3. (C)130\frac{1}{30}301​
  4. (D)120\frac{1}{20}201​

Correct answer: (D)

Step-by-step solution →
Q123·MathematicsSingle correctJEE Main 2022
The probability, that in a randomly selected 3-digit number at least two digits are odd, is
  1. (A)1936\frac{19}{36}3619​
  2. (B)1536\frac{15}{36}3615​
  3. (C)1336\frac{13}{36}3613​
  4. (D)2336\frac{23}{36}3623​

Correct answer: (A)

Step-by-step solution →
Q124·MathematicsSingle correctJEE Main 2022
If a point A(x,y)A(x, y)A(x,y) lies in the region bounded by the y-axis, straight lines 2y+x=62y+x=62y+x=6 and 5x−6y=305x-6y=305x−6y=30, then the probability that y<1y<1y<1 is :
  1. (A)16\dfrac{1}{6}61​
  2. (B)56\dfrac{5}{6}65​
  3. (C)23\dfrac{2}{3}32​
  4. (D)67\dfrac{6}{7}76​

Correct answer: (B)

Step-by-step solution →
Q125·MathematicsSingle correctJEE Main 2022
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 :
  1. (A)10516\frac{105}{16}16105​
  2. (B)716\frac{7}{16}167​
  3. (C)7736\frac{77}{36}3677​
  4. (D)4916\frac{49}{16}1649​

Correct answer: (C)

Step-by-step solution →
Q126·MathematicsNumericalJEE Main 2022
Let S={E,E2....E8}S=\{E, E_{2}....E_{8}\}S={E,E2​....E8​} be a sample space of random experiment such that P(En)=n36P(E_{n})=\dfrac{n}{36}P(En​)=36n​ for every n = 1, 2....8. Then the number of elements in the set {A⊂S:P(A)≥45}\left\{A\subset S:P(A)\ge\dfrac{4}{5}\right\}{A⊂S:P(A)≥54​} is ______

Correct answer: 19

Step-by-step solution →
Q127·MathematicsSingle correctJEE Main 2022
Five numbers x1x_1x1​, x2x_2x2​, x3x_3x3​, x4x_4x4​, x5x_5x5​ are randomly selected from the numbers 1, 2, 3,......, 18 and are arranged in the increasing order (x1<x2<x3<x4<x5x_1 < x_2 < x_3 < x_4 < x_5x1​<x2​<x3​<x4​<x5​). The probability that x2=7x_2 = 7x2​=7 and x4=11x_4 = 11x4​=11 is :
  1. (A)1136\frac{1}{136}1361​
  2. (B)172\frac{1}{72}721​
  3. (C)168\frac{1}{68}681​
  4. (D)134\frac{1}{34}341​

Correct answer: (C)

Step-by-step solution →
Q128·MathematicsSingle correctJEE Main 2022
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:
  1. (A)27565\frac{275}{6^{5}}65275​
  2. (B)3654\frac{36}{5^{4}}5436​
  3. (C)18155\frac{181}{5^{5}}55181​
  4. (D)4664\frac{46}{6^{4}}6446​

Correct answer: (D)

Step-by-step solution →
Q129·MathematicsNumericalJEE Main 2022
If the probability that a randomly chosen 6-digit number formed by using digits 1 and 8 only is a multiple of 21 is p, then 96 p is equal to ______ .

Correct answer: 33

Step-by-step solution →
Q130·MathematicsSingle correctJEE Main 2022
Let E1E_1E1​ and E2E_2E2​ be two events such that the conditional probabilities P(E1∣E2)=12P(E_1 | E_2) = \frac{1}{2}P(E1​∣E2​)=21​, P(E2∣E1)=34P(E_2 | E_1) = \frac{3}{4}P(E2​∣E1​)=43​ and P(E1∩E2)=18P(E_1 \cap E_2) = \frac{1}{8}P(E1​∩E2​)=81​. Then:
  1. (A)P(E1∩E2)=P(E1)⋅P(E2)P(E_1 \cap E_2) = P(E_1)\cdot P(E_2)P(E1​∩E2​)=P(E1​)⋅P(E2​)
  2. (B)P(E1′∩E2′)=P(E1′)⋅P(E2)P(E'_1 \cap E'_2) = P(E'_1)\cdot P(E_2)P(E1′​∩E2′​)=P(E1′​)⋅P(E2​)
  3. (C)P(E1∩E2′)=P(E1)⋅P(E2)P(E_1 \cap E'_2) = P(E_1)\cdot P(E_2)P(E1​∩E2′​)=P(E1​)⋅P(E2​)
  4. (D)P(E1′∩E2)=P(E1)⋅P(E2)P(E'_1 \cap E_2) = P(E_1)\cdot P(E_2)P(E1′​∩E2​)=P(E1​)⋅P(E2​)

Correct answer: (C)

Step-by-step solution →
Q131·MathematicsSingle correctJEE Main 2022
A biased die is marked with numbers 2,4, 8, 16, 32, 32 on its faces and the probability of getting a face with mark n is 1n\frac{1}{n}n1​. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is
  1. (A)7211\frac{7}{2^{11}}2117​
  2. (B)7212\frac{7}{2^{12}}2127​
  3. (C)3210\frac{3}{2^{10}}2103​
  4. (D)13212\frac{13}{2^{12}}21213​

Correct answer: (D)

Step-by-step solution →
Q132·MathematicsSingle correctJEE Main 2022
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 611\dfrac{6}{11}116​, then n is equal to ______ .
  1. (A)13
  2. (B)6
  3. (C)4
  4. (D)3

Correct answer: (C)

Step-by-step solution →
Q133·MathematicsNumericalJEE Main 2022
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 34\dfrac{3}{4}43​ and the remaining 6 questions correctly with probability 14\dfrac{1}{4}41​. If the probability that the student guesses the answers of exactly 8 questions correctly out of 10 is 27k410\dfrac{27k}{4^{10}}41027k​, then k is equal to ______.

Correct answer: 479

Step-by-step solution →
Q134·MathematicsSingle correctJEE Main 2022
If a random variable X follows the Binomial distribution B (33, p) such that 3P(X=0)=P(X=1)3P(X=0)=P(X=1)3P(X=0)=P(X=1), then the value of P(X=15)P(X=18)−P(X=16)P(X=17)\dfrac{P(X=15)}{P(X=18)}-\dfrac{P(X=16)}{P(X=17)}P(X=18)P(X=15)​−P(X=17)P(X=16)​ is equal to
  1. (A)1320
  2. (B)1088
  3. (C)1201331\dfrac{120}{1331}1331120​
  4. (D)10881089\dfrac{1088}{1089}10891088​

Correct answer: (A)

Step-by-step solution →
Q135·MathematicsNumericalJEE Advanced 2021
Three numbers are chosen at random, one after another with replacement, from the set S={1,2,3,...,100}S = \{1, 2, 3, ..., 100\}S={1,2,3,...,100}. Let p1p_1p1​ be the probability that the maximum of chosen numbers is at least 81 and p2p_2p2​ be the probability that the minimum of chosen numbers is at most 40. The value of 1254p2\frac{125}{4}p_24125​p2​ is ________.

Correct answer: 24.50

Step-by-step solution →
Q136·MathematicsIntegerJEE Advanced 2021
A number is chosen at random from the set {1, 2, 3, ... , 2000}. Let p be the probability that the chosen number is a multiple of 3 or a multiple of 7. Then the value of 500p is ___.

Correct answer: 214

Step-by-step solution →
Q137·MathematicsNumericalJEE Advanced 2021
Three numbers are chosen at random, one after another with replacement, from the set S={1,2,3,...,100}S = \{1, 2, 3, ..., 100\}S={1,2,3,...,100}. Let p1p_1p1​ be the probability that the maximum of chosen numbers is at least 81 and p2p_2p2​ be the probability that the minimum of chosen numbers is at most 40. The value of 6254p1\frac{625}{4}p_14625​p1​ is ________.

Correct answer: 76.25

Step-by-step solution →
Q138·MathematicsMultiple correctJEE Advanced 2021
Let E,F and G be three events having probabilities P(E)=18,P(F)=16P(E) = \frac{1}{8}, P(F) = \frac{1}{6}P(E)=81​,P(F)=61​ and P(G)=14P(G) = \frac{1}{4}P(G)=41​, and let P(E∩F∩G)=110P(E \cap F \cap G) = \frac{1}{10}P(E∩F∩G)=101​. For any event H, if HCH^CHC denotes its complement, then which of the following statements is(are) TRUE ?
  1. (A)P(E∩F∩GC)≤140P\left(E \cap F \cap G^C\right) \le \frac{1}{40}P(E∩F∩GC)≤401​
  2. (B)P(EC∩F∩G)≤115P\left(E^C \cap F \cap G\right) \le \frac{1}{15}P(EC∩F∩G)≤151​
  3. (C)P(E∪F∪G)≤1324P\left(E \cup F \cup G\right) \le \frac{13}{24}P(E∪F∪G)≤2413​
  4. (D)P(EC∩FC∩GC)≤512P\left(E^C \cap F^C \cap G^C\right) \le \frac{5}{12}P(EC∩FC∩GC)≤125​

Correct answer: (A), (B), (C)

Step-by-step solution →
Q139·MathematicsSingle correctJEE Advanced 2021
Consider three sets E1={1,2,3}E_1 = \{1, 2, 3\}E1​={1,2,3}, F1={1,3,4}F_1 = \{1, 3, 4\}F1​={1,3,4} and G1={2,3,4,5}G_1 = \{2, 3, 4, 5\}G1​={2,3,4,5}. Two elements are chosen at random, without replacement, from the set E1E_1E1​, and let S1S_1S1​ denote the set of these chosen elements. Let E2=E1−S1E_2 = E_1 - S_1E2​=E1​−S1​ and F2=F1∪S1F_2 = F_1 \cup S_1F2​=F1​∪S1​. Now two elements are chosen at random, without replacement, from the set F2F_2F2​ and let S2S_2S2​ denote the set of these chosen elements. Let G2=G1∪S2G_2 = G_1 \cup S_2G2​=G1​∪S2​. Finally, two elements are chosen at random, without replacement, from the set G2G_2G2​ and let S3S_3S3​ denote the set of these chosen elements. Let E3=E2∪S3E_3 = E_2 \cup S_3E3​=E2​∪S3​. Given that E1=E3E_1 = E_3E1​=E3​, let p be the conditional probability of the event S1={1,2}S_1 = \{1, 2\}S1​={1,2}. Then the value of p is
  1. (A)15\frac{1}{5}51​
  2. (B)35\frac{3}{5}53​
  3. (C)12\frac{1}{2}21​
  4. (D)25\frac{2}{5}52​

Correct answer: (A)

Step-by-step solution →
Q140·MathematicsSingle correctJEE Main 2021
Two squares are chosen at random on a chessboard (see figure). The probability that they have a side in common is :
  1. (A)27\frac{2}{7}72​
  2. (B)118\frac{1}{18}181​
  3. (C)17\frac{1}{7}71​
  4. (D)19\frac{1}{9}91​

Correct answer: (B)

Step-by-step solution →
Q141·MathematicsNumericalJEE Main 2021
Let X be a random variable with distribution. If the mean of X is 2.3 and variance of X is σ2\sigma^{2}σ2, then 100 σ2100\,\sigma^{2}100σ2 is equal to ________ .
x−2−1346
P(X = x)15\frac{1}{5}51​a13\frac{1}{3}31​15\frac{1}{5}51​b

Correct answer: 781

Step-by-step solution →
Q142·MathematicsNumericalJEE Main 2021
An electric instrument consists of two units. Each unit must function independently for the instrument to operate. The probability that the first unit functions is 0.9 and that of the second unit is 0.8. The instrument is switched on and it fails to operate. If the probability that only the first unit failed and second unit is functioning is p, then 98 p is equal to ______.

Correct answer: 28

Step-by-step solution →
Q143·MathematicsSingle correctJEE Main 2021
Let S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}S={1,2,3,4,5,6}. Then the probability that a randomly chosen onto function ggg from S to S satisfies g(3)=2g(1)g(3) = 2g(1)g(3)=2g(1) is :
  1. (A)110\frac{1}{10}101​
  2. (B)115\frac{1}{15}151​
  3. (C)15\frac{1}{5}51​
  4. (D)130\frac{1}{30}301​

Correct answer: (A)

Step-by-step solution →
Q144·MathematicsSingle correctJEE Main 2021
When a certain biased die is rolled, a particular face occurs with probability 16−x\frac{1}{6} - x61​−x and its opposite face occurs with probability 16+x\frac{1}{6} + x61​+x . All other faces occur with probability 16\frac{1}{6}61​ . Note that opposite faces sum to 7 in any die. If 0<x<160 < x < \frac{1}{6}0<x<61​ , and the probability of obtaining total sum =7= 7=7, when such a die is rolled twice, is 1396\frac{13}{96}9613​ , then the value of x is:
  1. (A)116\frac{1}{16}161​
  2. (B)18\frac{1}{8}81​
  3. (C)19\frac{1}{9}91​
  4. (D)112\frac{1}{12}121​

Correct answer: (B)

Step-by-step solution →
Q145·MathematicsNumericalJEE Main 2021
The probability distribution of random variable X is given by: X | 1 | 2 | 3 | 4 | 5 P(X) | K | 2K | 2K | 3K | K Let p = P(1 < X < 4 | X < 3). If 5p = λK, then λ equal to _________ .

Correct answer: 30

Step-by-step solution →
Q146·MathematicsSingle correctJEE Main 2021
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 :
  1. (A)18\frac{1}{8}81​
  2. (B)58\frac{5}{8}85​
  3. (C)516\frac{5}{16}165​
  4. (D)1

Correct answer: (C)

Step-by-step solution →
Q147·MathematicsSingle correctJEE Main 2021
A fair die is tossed until six is obtained on it. Let X be the number of required tosses, then the conditional probability P(X≥5∣X>2)P\left(X \geq 5 \mid X > 2\right)P(X≥5∣X>2) is :
  1. (A)125216\frac{125}{216}216125​
  2. (B)1136\frac{11}{36}3611​
  3. (C)56\frac{5}{6}65​
  4. (D)2536\frac{25}{36}3625​

Correct answer: (D)

Step-by-step solution →
Q148·MathematicsSingle correctJEE Main 2021
Let A and B be independent events such that P(A)=pP(A) = pP(A)=p, P(B)=2pP(B) = 2pP(B)=2p. The largest value of p, for which P (exactly one of A, B occurs) =59= \frac{5}{9}=95​, is :
  1. (A)13\frac{1}{3}31​
  2. (B)29\frac{2}{9}92​
  3. (C)49\frac{4}{9}94​
  4. (D)512\frac{5}{12}125​

Correct answer: (D)

Step-by-step solution →
Q149·MathematicsSingle correctJEE Main 2021
The probability that a randomly selected 2-digit number belongs to the set {n∈N:(2n−2) is a multiple of 3}\left\{ n \in N : \left(2^n - 2\right) \text{ is a multiple of } 3 \right\}{n∈N:(2n−2) is a multiple of 3} is equal to :
  1. (A)23\frac{2}{3}32​
  2. (B)12\frac{1}{2}21​
  3. (C)16\frac{1}{6}61​
  4. (D)13\frac{1}{3}31​

Correct answer: (B)

Step-by-step solution →
Q150·MathematicsSingle correctJEE Main 2021
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 12\frac{1}{2}21​, is :
  1. (A)5
  2. (B)6
  3. (C)3
  4. (D)4

Correct answer: (A)

Step-by-step solution →
Q151·MathematicsSingle correctJEE Main 2021
Let X be a random variable such that the probability function of a distribution is given by P(X=0)=12,P(X=j)=13j(j=1,2,3,.....,∞)P\left(X = 0\right) = \frac{1}{2}, P\left(X = j\right) = \frac{1}{3^j}\left(j = 1,2,3,.....,\infty\right)P(X=0)=21​,P(X=j)=3j1​(j=1,2,3,.....,∞). Then the mean of the distribution and P(X is positive and even) respectively are :
  1. (A)34\frac{3}{4}43​ and 19\frac{1}{9}91​
  2. (B)34\frac{3}{4}43​ and 116\frac{1}{16}161​
  3. (C)34\frac{3}{4}43​ and 18\frac{1}{8}81​
  4. (D)38\frac{3}{8}83​ and 18\frac{1}{8}81​

Correct answer: (C)

Step-by-step solution →
Q152·MathematicsSingle correctJEE Main 2021
Let 9 distinct balls be distributed among 4 boxes, B1,B2,B3B_1, B_2, B_3B1​,B2​,B3​ and B4B_4B4​. If the probability that B3B_3B3​ contains exactly 3 balls is k(34)9k\left(\frac{3}{4}\right)^{9}k(43​)9 then k lies in the set :
  1. (A){x∈R:∣x−2∣≤1}\left\{x \in R : \left|x - 2\right| \leq 1\right\}{x∈R:∣x−2∣≤1}
  2. (B){x∈R:∣x−5∣≤1}\left\{x \in R : \left|x - 5\right| \leq 1\right\}{x∈R:∣x−5∣≤1}
  3. (C){x∈R:∣x−3∣<1}\left\{x \in R : \left|x - 3\right| < 1\right\}{x∈R:∣x−3∣<1}
  4. (D){x∈R:∣x−1∣<1}\left\{x \in R : \left|x - 1\right| < 1\right\}{x∈R:∣x−1∣<1}

Correct answer: (C)

Step-by-step solution →
Q153·MathematicsNumericalJEE Main 2021
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...........

Correct answer: 4

Step-by-step solution →
Q154·MathematicsSingle correctJEE Main 2021
Four dice are thrown simultaneously and the numbers shown on these dice are recorded in 2 x 2 matrices. The probability that such formed matrices have all different entries and are non-singular, is :
  1. (A)2281\dfrac{22}{81}8122​
  2. (B)45162\dfrac{45}{162}16245​
  3. (C)43162\dfrac{43}{162}16243​
  4. (D)2381\dfrac{23}{81}8123​

Correct answer: (C)

Step-by-step solution →
Q155·MathematicsSingle correctJEE Main 2021
Words with or without meaning are to be formed using all the letters of the word EXAMINATION. The probability that the letter M appears at the fourth position in any such word is :
  1. (A)19\frac{1}{9}91​
  2. (B)111\frac{1}{11}111​
  3. (C)211\frac{2}{11}112​
  4. (D)166\frac{1}{66}661​

Correct answer: (B)

Step-by-step solution →
Q156·MathematicsSingle correctJEE Main 2021
Let A, B and C be three events such that the probability that exactly one of A and B occurs is (1−k)(1-k)(1−k), the probability that exactly one of B and C occurs is (1−2k)(1-2k)(1−2k), the probability that exactly one of C and A occurs is (1−k)(1-k)(1−k) and the probability of all A, B and C occur simultaneously is k2k^{2}k2, where 0<k<10<k<10<k<1. Then the probability that at least one of A, B and C occur is :
  1. (A)greater than 12\frac{1}{2}21​
  2. (B)greater than 14\frac{1}{4}41​ but less than 12\frac{1}{2}21​
  3. (C)exactly equal to 12\frac{1}{2}21​
  4. (D)greater than 18\frac{1}{8}81​ but less than 14\frac{1}{4}41​

Correct answer: (A)

Step-by-step solution →
Q157·MathematicsSingle correctJEE Main 2021
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 :
  1. (A)32625\frac{32}{625}62532​
  2. (B)80243\frac{80}{243}24380​
  3. (C)40243\frac{40}{243}24340​
  4. (D)128625\frac{128}{625}625128​

Correct answer: (A)

Step-by-step solution →
Q158·MathematicsSingle correctJEE Main 2021
Two dices 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 :
  1. (A)49\frac{4}{9}94​
  2. (B)1736\frac{17}{36}3617​
  3. (C)512\frac{5}{12}125​
  4. (D)12\frac{1}{2}21​

Correct answer: (B)

Step-by-step solution →
Q159·MathematicsSingle correctJEE Main 2021
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 12\frac{1}{2}21​ and probability of occurrence of 0 at the odd place be 13\frac{1}{3}31​. Then the probability that '10' is followed by '01' is equal to :
  1. (A)118\frac{1}{18}181​
  2. (B)13\frac{1}{3}31​
  3. (C)16\frac{1}{6}61​
  4. (D)19\frac{1}{9}91​

Correct answer: (D)

Step-by-step solution →
Q160·MathematicsNumericalJEE Main 2021
Let there be three independent events E1E_1E1​, E2E_2E2​ and E3E_3E3​. The probability that only E1E_1E1​ occurs is α\alphaα, only E2E_2E2​ occurs is β\betaβ and only E3E_3E3​ occurs is γ\gammaγ. Let 'p' denote the probability of none of events occurs that satisfies the equations (α−2β)p=αβ(\alpha - 2\beta)p = \alpha\beta(α−2β)p=αβ and (β−3γ)p=2βγ(\beta - 3\gamma)p = 2\beta\gamma(β−3γ)p=2βγ. All the given probabilities are assumed to lie in the interval (0, 1). Then, Probability of occurrence of E1Probability of occurrence of E3\frac{\text{Probability of occurrence of } E_1}{\text{Probability of occurrence of } E_3}Probability of occurrence of E3​Probability of occurrence of E1​​ is equal to _____.

Correct answer: 6

Step-by-step solution →
Q161·MathematicsSingle correctJEE Main 2021
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 :
  1. (A)956\frac{9}{56}569​
  2. (B)49\frac{4}{9}94​
  3. (C)37\frac{3}{7}73​
  4. (D)1127\frac{11}{27}2711​

Correct answer: (B)

Step-by-step solution →
Q162·MathematicsSingle correctJEE Main 2021
A pack of cards has one card missing. Two cards are drawn randomly and are found to be spades. The probability that the missing card is not a spade, is :
  1. (A)34\frac{3}{4}43​
  2. (B)52867\frac{52}{867}86752​
  3. (C)3950\frac{39}{50}5039​
  4. (D)22425\frac{22}{425}42522​

Correct answer: (C)

Step-by-step solution →
Q163·MathematicsSingle correctJEE Main 2021
A seven digit number is formed using digit 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is :
  1. (A)67\frac{6}{7}76​
  2. (B)47\frac{4}{7}74​
  3. (C)37\frac{3}{7}73​
  4. (D)17\frac{1}{7}71​

Correct answer: (C)

Step-by-step solution →
Q164·MathematicsSingle correctJEE Main 2021
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 :
  1. (A)15212\frac{15}{2^{12}}21215​
  2. (B)15213\frac{15}{2^{13}}21315​
  3. (C)15214\frac{15}{2^{14}}21415​
  4. (D)1528\frac{15}{2^{8}}2815​

Correct answer: (B)

Step-by-step solution →
Q165·MathematicsSingle correctJEE Main 2021
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:
  1. (A)457
  2. (B)458
  3. (C)1445
  4. (D)2845

Correct answer: (D)

Step-by-step solution →
Q166·MathematicsSingle correctJEE Main 2021
The coefficients a, b and c of the quadratic equation, ax2+bx+c=0ax^{2} + bx + c = 0ax2+bx+c=0 are obtained by throwing a dice three times. The probability that this equation has equal roots is :
  1. (A)154\frac{1}{54}541​
  2. (B)172\frac{1}{72}721​
  3. (C)136\frac{1}{36}361​
  4. (D)5216\frac{5}{216}2165​

Correct answer: (D)

Step-by-step solution →
Q167·MathematicsSingle correctJEE Main 2021
Let A be a set of all 4-digit natural numbers whose exactly one digit is 7. Then the probability that a randomly chosen element of A leaves remainder 2 when divided by 5 is:
  1. (A)15
  2. (B)29
  3. (C)29797
  4. (D)122297

Correct answer: (C)

Step-by-step solution →
Q168·MathematicsSingle correctJEE Main 2021
When a missile is fired from a ship, the probability that it is intercepted is 13\frac{1}{3}31​ and the probability that the missile hits the target, given that it is not intercepted, is 34\frac{3}{4}43​. If three missiles are fired independently from the ship, then the probability that all three hit the target, is:
  1. (A)18\frac{1}{8}81​
  2. (B)127\frac{1}{27}271​
  3. (C)34\frac{3}{4}43​
  4. (D)38\frac{3}{8}83​

Correct answer: (A)

Step-by-step solution →
Q169·MathematicsSingle correctJEE Main 2021
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 :
  1. (A)316\frac{3}{16}163​
  2. (B)12\frac{1}{2}21​
  3. (C)516\frac{5}{16}165​
  4. (D)132\frac{1}{32}321​

Correct answer: (B)

Step-by-step solution →
Q170·MathematicsSingle correctJEE Main 2021
The probability that two randomly selected subsets of the set {1,2,3,4,5}\{1,2,3,4,5\}{1,2,3,4,5} have exactly two elements in their intersection, is:
  1. (A)6527\frac{65}{2^{7}}2765​
  2. (B)13529\frac{135}{2^{9}}29135​
  3. (C)6528\frac{65}{2^{8}}2865​
  4. (D)3527\frac{35}{2^{7}}2735​

Correct answer: (B)

Step-by-step solution →
Q171·MathematicsNumericalJEE Main 2021
Let Bi(i=1,2,3)B_i(i = 1, 2, 3)Bi​(i=1,2,3) be three independent events in a sample space. The probability that only B1B_1B1​ occur is α\alphaα, only B2B_2B2​ occurs is β\betaβ and only B3B_3B3​ occurs is γ\gammaγ. Let p be the probability that none of the events BiB_iBi​ occurs and these 4 probabilities satisfy the equations (α−2β)p=αβ(\alpha - 2\beta)p = \alpha\beta(α−2β)p=αβ and (β−3γ)p=2βγ(\beta - 3\gamma)p = 2\beta\gamma(β−3γ)p=2βγ (All the probabilities are assumed to lie in the interval (0, 1)). Then P(B1)P(B3)\frac{P(B_1)}{P(B_3)}P(B3​)P(B1​)​ is equal to _____

Correct answer: 6

Step-by-step solution →
Q172·MathematicsSingle correctJEE Advanced 2020
Let C1C_{1}C1​ and C2C_{2}C2​ be two biased coins such that the probabilities of getting head in a single toss are 23\frac{2}{3}32​ and 13\frac{1}{3}31​, respectively. Suppose α\alphaα is the number of heads that appear when C1C_{1}C1​ is tossed twice, independently, and suppose β\betaβ is the number of heads that appear when C2C_{2}C2​ is tossed twice, independently, Then probability that the roots of the quadratic polynomial x2−αx+βx^{2} - \alpha x + \betax2−αx+β are real and equal, is
  1. (A)4081\frac{40}{81}8140​
  2. (B)2081\frac{20}{81}8120​
  3. (C)12\frac{1}{2}21​
  4. (D)14\frac{1}{4}41​

Correct answer: (B)

Step-by-step solution →
Q173·MathematicsNumericalJEE Advanced 2020
Two fair dice, each with faces numbered 1,2,3,4,5 and 6, are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If ppp is the probability that this perfect square is an odd number, then the value of 14p14p14p is ________

Correct answer: 8.00

Step-by-step solution →
Q174·MathematicsIntegerJEE Advanced 2020
The probability that a missile hits a target successfully is 0.75. In order to destroy the target completely, at least three successful hits are required. Then the minimum number of missiles that have to be fired so that the probability of completely destroying the target is NOT less than 0.95, is ________

Correct answer: 6

Step-by-step solution →
Q175·MathematicsSingle correctJEE Main 2020
The probabilities of three events A, B and C are given by P(A)=0.6P(A) = 0.6P(A)=0.6, P(B)=0.4P(B) = 0.4P(B)=0.4 and P(C)=0.5P(C) = 0.5P(C)=0.5. If P(A∪B)=0.8P(A \cup B) = 0.8P(A∪B)=0.8, P(A∩C)=0.3P(A \cap C) = 0.3P(A∩C)=0.3, P(A∩B∩C)=0.2P(A \cap B \cap C) = 0.2P(A∩B∩C)=0.2, P(B∩C)=βP(B \cap C) = \betaP(B∩C)=β and P(A∪B∪C)=αP(A \cup B \cup C) = \alphaP(A∪B∪C)=α, where 0.85≤α≤0.950.85 \leq \alpha \leq 0.950.85≤α≤0.95, then β\betaβ lies in the interval:
  1. (A)[0.35,0.36][0.35, 0.36][0.35,0.36]
  2. (B)[0.25,0.35][0.25, 0.35][0.25,0.35]
  3. (C)[0.20,0.25][0.20, 0.25][0.20,0.25]
  4. (D)[0.36,0.40][0.36, 0.40][0.36,0.40]

Correct answer: (B)

Step-by-step solution →
Q176·MathematicsSingle correctJEE Main 2020
Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is:
  1. (A)15101\dfrac{15}{101}10115​
  2. (B)5101\dfrac{5}{101}1015​
  3. (C)533\dfrac{5}{33}335​
  4. (D)1099\dfrac{10}{99}9910​

Correct answer: (C)

Step-by-step solution →
Q177·MathematicsNumericalJEE Main 2020
In a bombing attack, there is 50% chance that a bomb will hit the target. At least 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 __________.

Correct answer: 11.00

Step-by-step solution →
Q178·MathematicsNumericalJEE Main 2020
Four fair dice are thrown independently 27 times. Then the expected number of times, at least two dice show up a three or a five, a ____.

Correct answer: 11.00

Step-by-step solution →
Q179·MathematicsSingle correctJEE Main 2020
In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws a total of 6 before B throws a total of 7 and B wins the game if he throws a total of 7 before A throws a total of six. The game stops as soon as either of the players wins. The probability of A winning the game is:
  1. (A)56\frac{5}{6}65​
  2. (B)3161\frac{31}{61}6131​
  3. (C)3061\frac{30}{61}6130​
  4. (D)531\frac{5}{31}315​

Correct answer: (C)

Step-by-step solution →
Q180·MathematicsNumericalJEE Main 2020
The probability of a man hitting a target is 110\frac{1}{10}101​. The least number of shots required, so that the probability of his hitting the target at least once is greater than 14\frac{1}{4}41​, is __________.

Correct answer: 3

Step-by-step solution →
Q181·MathematicsSingle correctJEE Main 2020
A die is thrown two times and the sum of the scorers appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is
  1. (A)19\frac{1}{9}91​
  2. (B)13\frac{1}{3}31​
  3. (C)18\frac{1}{8}81​
  4. (D)14\frac{1}{4}41​

Correct answer: (A)

Step-by-step solution →
Q182·MathematicsSingle correctJEE Main 2020
The probability that a randomly chosen 5 - digit number is made from exactly two digits is:
  1. (A)134104\frac{134}{10^4}104134​
  2. (B)121104\frac{121}{10^4}104121​
  3. (C)150104\frac{150}{10^4}104150​
  4. (D)135104\frac{135}{10^4}104135​

Correct answer: (D)

Step-by-step solution →
Q183·MathematicsSingle correctJEE Main 2020
In a box, there are 20 cards, out of which 10 are labeled 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:
  1. (A)916\dfrac{9}{16}169​
  2. (B)1516\dfrac{15}{16}1615​
  3. (C)1316\dfrac{13}{16}1613​
  4. (D)1116\dfrac{11}{16}1611​

Correct answer: (D)

Step-by-step solution →
Q184·MathematicsSingle correctJEE Main 2020
A random variable X has the following probability distribution: Then P(X>2)P(X > 2)P(X>2) is equal to:
X12345
P(X)K2K^{2}K22KK2K5K25K^{2}5K2
  1. (A)136\dfrac{1}{36}361​
  2. (B)712\dfrac{7}{12}127​
  3. (C)2336\dfrac{23}{36}3623​
  4. (D)16\dfrac{1}{6}61​

Correct answer: (C)

Step-by-step solution →
Q185·MathematicsSingle correctJEE Main 2020
Let A and B be two independent events such that P(A)=13P(A)=\dfrac{1}{3}P(A)=31​ and P(B)=16P(B)=\dfrac{1}{6}P(B)=61​. Then, which of the following is TRUE?
  1. (A)P(AB)=23P\left(\dfrac{A}{B}\right)=\dfrac{2}{3}P(BA​)=32​
  2. (B)P(A′B′)=13P\left(\dfrac{A'}{B'}\right)=\dfrac{1}{3}P(B′A′​)=31​
  3. (C)P(AB′)=13P\left(\dfrac{A}{B'}\right)=\dfrac{1}{3}P(B′A​)=31​
  4. (D)P(A(A∪B))=14P\left(\dfrac{A}{(A\cup B)}\right)=\dfrac{1}{4}P((A∪B)A​)=41​

Correct answer: (C)

Step-by-step solution →
Q186·MathematicsSingle correctJEE Main 2020
Let A and B be two events such that the probability that exactly one of them occurs is 25\frac{2}{5}52​ and the probability that A or B occurs is 12\frac{1}{2}21​, then the probability of both of them occur together is:
  1. (A)0.01
  2. (B)0.10
  3. (C)0.20
  4. (D)0.02

Correct answer: (B)

Step-by-step solution →
Q187·MathematicsSingle correctJEE Main 2020
In a workshop, there are five machines and the probability of any one of them to be out of service on a day is 14\dfrac{1}{4}41​. If the probability that at most two machines will be out of service on the same day is (34)3k\left(\dfrac{3}{4}\right)^{3}k(43​)3k, then k is equal to:
  1. (A)444
  2. (B)178\dfrac{17}{8}817​
  3. (C)172\dfrac{17}{2}217​
  4. (D)174\dfrac{17}{4}417​

Correct answer: (B)

Step-by-step solution →
Q188·MathematicsSingle correctJEE Main 2020
An unbiased coin is tossed 5 time. Suppose that a variable X is assigned the value k when k consecutive heads are obtained for k = 3, 4, 5, otherwise X takes the value −1. Then the expected value of X, is
  1. (A)−316-\frac{3}{16}−163​
  2. (B)−18-\frac{1}{8}−81​
  3. (C)18\frac{1}{8}81​
  4. (D)316\frac{3}{16}163​

Correct answer: (C)

Step-by-step solution →
Q189·MathematicsNumericalJEE Advanced 2019
Let ∣x∣|x|∣x∣ denote the number of elements in a set X. Let S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}S={1,2,3,4,5,6} be a sample space, where each element is equally likely to occur. If A and B are independent events associated with S, then the number of ordered pairs (A, B) such that 1≤∣B∣<∣A∣1 \le |B| < |A|1≤∣B∣<∣A∣, equals ______

Correct answer: 422.00

Step-by-step solution →
Q190·MathematicsMultiple correctJEE Advanced 2019
There are three bags B1B_1B1​, B2B_2B2​ and B3B_3B3​. The bag B1B_1B1​ contains 5 red and 5 green balls, B2B_2B2​ contains 3 red and 5 green balls and B3B_3B3​ contains 5 red and 3 green balls. Bags B1B_1B1​, B2B_2B2​ and B3B_3B3​ have probabilities 310\frac{3}{10}103​, 310\frac{3}{10}103​ and 410\frac{4}{10}104​ respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?
  1. (A)Probability that the chosen ball is green equals 3980\frac{39}{80}8039​
  2. (B)Probability that the chosen ball is green, given that the selected bag is B3B_3B3​, equals 38\frac{3}{8}83​
  3. (C)Probability that the selected bag is B3B_3B3​ and the chosen ball is green equals 310\frac{3}{10}103​
  4. (D)Probability that the selected bag is B3B_3B3​, given that the chosen ball is green, equals 513\frac{5}{13}135​

Correct answer: (A), (B)

Step-by-step solution →
Q191·MathematicsNumericalJEE Advanced 2019
Let S be the sample space of all 3×33 \times 33×3 matrices with entries from the set {0,1}\{0, 1\}{0,1}. Let the events E1E_1E1​ and E2E_2E2​ be given by E1={A∈S:det⁡A=0}E_1 = \{A \in S : \det A = 0\}E1​={A∈S:detA=0} and E2={A∈S:E_2 = \{A \in S :E2​={A∈S: sum of entries of A is 7}7\}7} If a matrix is chosen at random from S, then the conditional probability P(E1/E2)P(E_1/E_2)P(E1​/E2​) equals ____

Correct answer: 0.50

Step-by-step solution →
Q192·MathematicsSingle correctJEE Main 2019
Let a random variable X have a binomial distribution with mean 8 and variance 4. If P(X≤2)=k216P(X \le 2) = \frac{k}{2^{16}}P(X≤2)=216k​, then k is equal to :
  1. (A)17
  2. (B)137
  3. (C)1
  4. (D)121

Correct answer: (B)

Step-by-step solution →
Q193·MathematicsSingle correctJEE Main 2019
If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is :
  1. (A)310\frac{3}{10}103​
  2. (B)15\frac{1}{5}51​
  3. (C)110\frac{1}{10}101​
  4. (D)320\frac{3}{20}203​

Correct answer: (C)

Step-by-step solution →
Q194·MathematicsSingle correctJEE Main 2019
A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins Rs.12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is :
  1. (A)14\dfrac{1}{4}41​ loss
  2. (B)2 gain
  3. (C)12\dfrac{1}{2}21​ gain
  4. (D)12\dfrac{1}{2}21​ loss

Correct answer: (D)

Step-by-step solution →
Q195·MathematicsSingle correctJEE Main 2019
For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is 45\dfrac{4}{5}54​, then the probability that he is unable to solve less than two problems is :
  1. (A)16425(15)48\dfrac{164}{25}\left(\dfrac{1}{5}\right)^{48}25164​(51​)48
  2. (B)2015(15)49\dfrac{201}{5}\left(\dfrac{1}{5}\right)^{49}5201​(51​)49
  3. (C)545(45)49\dfrac{54}{5}\left(\dfrac{4}{5}\right)^{49}554​(54​)49
  4. (D)31625(45)48\dfrac{316}{25}\left(\dfrac{4}{5}\right)^{48}25316​(54​)48

Correct answer: (C)

Step-by-step solution →
Q196·MathematicsSingle correctJEE Main 2019
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
  1. (A)8
  2. (B)6
  3. (C)7
  4. (D)5

Correct answer: (C)

Step-by-step solution →
Q197·MathematicsSingle correctJEE Main 2019
Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is:
  1. (A)110\dfrac{1}{10}101​
  2. (B)117\dfrac{1}{17}171​
  3. (C)112\dfrac{1}{12}121​
  4. (D)111\dfrac{1}{11}111​

Correct answer: (D)

Step-by-step solution →
Q198·MathematicsSingle correctJEE Main 2019
Four persons can hit a target correctly with probabilities 12,13,14\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}21​,31​,41​ and 18\dfrac{1}{8}81​ respectively. If all hit at the target independently, then the probability that the target would be hit, is:
  1. (A)2532\dfrac{25}{32}3225​
  2. (B)25192\dfrac{25}{192}19225​
  3. (C)732\dfrac{7}{32}327​
  4. (D)1192\dfrac{1}{192}1921​

Correct answer: (A)

Step-by-step solution →
Q199·MathematicsSingle correctJEE Main 2019
Two newspaper A and B are published in a city. It is known that 25% of the city populations reads A and 20% reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisements and 40% of those who read B but not A also look into advertisements, while 50% of those who read both A and B look into advertisements. Then the percentage of the population who look into advertisement is:
  1. (A)12.8
  2. (B)13.5
  3. (C)13.9
  4. (D)13

Correct answer: (C)

Step-by-step solution →
Q200·MathematicsSingle correctJEE Main 2019
Let A and b be two non-null events such that A⊂BA\subset BA⊂B. Then, which of the following statements is always correct?
  1. (A)P(A∣B)=1P(A|B)=1P(A∣B)=1
  2. (B)P(A∣B)≤P(A)P(A|B)\le P(A)P(A∣B)≤P(A)
  3. (C)P(A∣B)=P(B)−P(A)P(A|B)=P(B)-P(A)P(A∣B)=P(B)−P(A)
  4. (D)P(A∣B)≥P(A)P(A|B)\ge P(A)P(A∣B)≥P(A)

Correct answer: (D)

Step-by-step solution →
Q201·MathematicsSingle correctJEE Main 2019
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:
  1. (A)2
  2. (B)3
  3. (C)4
  4. (D)5

Correct answer: (C)

Step-by-step solution →
Q202·MathematicsSingle correctJEE Main 2019
In a class of 60 students, 40 opted for NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the student selected has opted neither for NCC nor for NSS is :
  1. (A)16\frac{1}{6}61​
  2. (B)13\frac{1}{3}31​
  3. (C)23\frac{2}{3}32​
  4. (D)56\frac{5}{6}65​

Correct answer: (A)

Step-by-step solution →
Q203·MathematicsSingle correctJEE Main 2019
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 :
  1. (A)4009\dfrac{400}{9}9400​ loss
  2. (B)000
  3. (C)4003\dfrac{400}{3}3400​ gain
  4. (D)4003\dfrac{400}{3}3400​ loss

Correct answer: (B)

Step-by-step solution →
Q204·MathematicsSingle correctJEE Main 2019
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:
  1. (A)20065\dfrac{200}{6^{5}}65200​
  2. (B)15065\dfrac{150}{6^{5}}65150​
  3. (C)22565\dfrac{225}{6^{5}}65225​
  4. (D)17565\dfrac{175}{6^{5}}65175​

Correct answer: (D)

Step-by-step solution →
Q205·MathematicsSingle correctJEE Main 2019
Two integers are selected at random from the set {1, 2, …, 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is:
  1. (A)710\frac{7}{10}107​
  2. (B)12\frac{1}{2}21​
  3. (C)25\frac{2}{5}52​
  4. (D)35\frac{3}{5}53​

Correct answer: (C)

Step-by-step solution →
Q206·MathematicsSingle correctJEE Main 2019
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 (mean of Xstandard deviation of X)\left(\dfrac{\text{mean of }X}{\text{standard deviation of }X}\right)(standard deviation of Xmean of X​) is equal to:
  1. (A)4
  2. (B)434\sqrt{3}43​
  3. (C)323\sqrt{2}32​
  4. (D)433\dfrac{4\sqrt{3}}{3}343​​

Correct answer: (B)

Step-by-step solution →
Q207·MathematicsSingle correctJEE Main 2019
let S = {1, 2, … 20}. A subset B of S is said to be “nice”, if the sum of the elements of B is 203. Then the probability that a randomly chosen subset of S is ‘nice’ is:
  1. (A)7220\frac{7}{2^{20}}2207​
  2. (B)5220\frac{5}{2^{20}}2205​
  3. (C)4220\frac{4}{2^{20}}2204​
  4. (D)6220\frac{6}{2^{20}}2206​

Correct answer: (B)

Step-by-step solution →
Q208·MathematicsSingle correctJEE Main 2019
An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is:
  1. (A)2649\dfrac{26}{49}4926​
  2. (B)3249\dfrac{32}{49}4932​
  3. (C)2749\dfrac{27}{49}4927​
  4. (D)2149\dfrac{21}{49}4921​

Correct answer: (B)

Step-by-step solution →
Q209·MathematicsSingle correctJEE Main 2019
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)P(X=1)+P(X=2)P(X=1)+P(X=2) equals:
  1. (A)49169\dfrac{49}{169}16949​
  2. (B)52169\dfrac{52}{169}16952​
  3. (C)24169\dfrac{24}{169}16924​
  4. (D)25169\dfrac{25}{169}16925​

Correct answer: (D)

Step-by-step solution →
Q210·MathematicsSingle correctJEE Advanced 2018
PARAGRAPH "A" There are five students S1S_{1}S1​, S2S_{2}S2​, S3S_{3}S3​, S4S_{4}S4​ and S5S_{5}S5​ in a music class and for them there are five seats R1R_{1}R1​, R2R_{2}R2​, R3R_{3}R3​, R4R_{4}R4​ and R5R_{5}R5​ arranged in a row, where initially the seat RiR_{i}Ri​ is allotted to the student SiS_{i}Si​, i=1,2,3,4,5i = 1, 2, 3, 4, 5i=1,2,3,4,5. But, on the examination day, the five students are randomly allotted the five seats. (There are two questions based on PARAGRAPH "A", the question given below is one of them) For i=1,2,3,4i = 1, 2, 3, 4i=1,2,3,4, let TiT_{i}Ti​ denote the event that the students SiS_{i}Si​ and Si+1S_{i+1}Si+1​ do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event T1∩T2∩T3∩T4T_{1} \cap T_{2} \cap T_{3} \cap T_{4}T1​∩T2​∩T3​∩T4​ is
  1. (A)115\frac{1}{15}151​
  2. (B)110\frac{1}{10}101​
  3. (C)760\frac{7}{60}607​
  4. (D)15\frac{1}{5}51​

Correct answer: (C)

Step-by-step solution →
Q211·MathematicsSingle correctJEE Advanced 2018
PARAGRAPH "A" There are five students S1S_{1}S1​, S2S_{2}S2​, S3S_{3}S3​, S4S_{4}S4​ and S5S_{5}S5​ in a music class and for them there are five seats R1R_{1}R1​, R2R_{2}R2​, R3R_{3}R3​, R4R_{4}R4​ and R5R_{5}R5​ arranged in a row, where initially the seat RiR_{i}Ri​ is allotted to the student SiS_{i}Si​, i=1,2,3,4,5i = 1, 2, 3, 4, 5i=1,2,3,4,5. But, on the examination day, the five students are randomly allotted the five seats. (There are two questions based on PARAGRAPH "A", the question given below is one of them) The probability that, on the examination day, the student S1S_{1}S1​ gets the previously allotted seat R1R_{1}R1​, and NONE of the remaining students gets the seat previously allotted to him/her is
  1. (A)340\frac{3}{40}403​
  2. (B)18\frac{1}{8}81​
  3. (C)740\frac{7}{40}407​
  4. (D)15\frac{1}{5}51​

Correct answer: (A)

Step-by-step solution →
Q212·MathematicsMultiple correctJEE Advanced 2017
Let X and Y be two events such that P(X)=13P(X) = \frac{1}{3}P(X)=31​, P(X∣Y)=12P(X|Y) = \frac{1}{2}P(X∣Y)=21​ and P(Y∣X)=25P(Y|X) = \frac{2}{5}P(Y∣X)=52​. Then
  1. (A)P(X′∣Y)=12P(X'|Y) = \frac{1}{2}P(X′∣Y)=21​
  2. (B)P(X∩Y)=15P(X \cap Y) = \frac{1}{5}P(X∩Y)=51​
  3. (C)P(X∪Y)=25P(X \cup Y) = \frac{2}{5}P(X∪Y)=52​
  4. (D)P(Y)=415P(Y) = \frac{4}{15}P(Y)=154​

Correct answer: (A), (D)

Step-by-step solution →
Q213·MathematicsSingle correctJEE Advanced 2017
Three randomly chosen nonnegative integers xxx, yyy and zzz are found to satisfy the equation x+y+z=10x + y + z = 10x+y+z=10. Then the probability that zzz is even, is
  1. (A)3655\dfrac{36}{55}5536​
  2. (B)611\dfrac{6}{11}116​
  3. (C)12\dfrac{1}{2}21​
  4. (D)511\dfrac{5}{11}115​

Correct answer: (B)

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Q214·MathematicsSingle correctJEE Advanced 2016
Football teams T1T_{1}T1​ and T2T_{2}T2​ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1T_{1}T1​ winning, drawing and losing a game against T2T_{2}T2​ are 12\frac{1}{2}21​, 16\frac{1}{6}61​ and 13\frac{1}{3}31​, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1T_{1}T1​ and T2T_{2}T2​, respectively, after two games. P(X>Y)P(X > Y)P(X>Y) is
  1. (A)14\frac{1}{4}41​
  2. (B)512\frac{5}{12}125​
  3. (C)12\frac{1}{2}21​
  4. (D)712\frac{7}{12}127​

Correct answer: (B)

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Q215·MathematicsSingle correctJEE Advanced 2016
Football teams T1T_{1}T1​ and T2T_{2}T2​ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T1T_{1}T1​ winning, drawing and losing a game against T2T_{2}T2​ are 12\frac{1}{2}21​, 16\frac{1}{6}61​ and 13\frac{1}{3}31​, respectively. Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T1T_{1}T1​ and T2T_{2}T2​, respectively, after two games. P(X=Y)P(X = Y)P(X=Y) is
  1. (A)1136\frac{11}{36}3611​
  2. (B)13\frac{1}{3}31​
  3. (C)1336\frac{13}{36}3613​
  4. (D)12\frac{1}{2}21​

Correct answer: (C)

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Q216·MathematicsSingle correctJEE Advanced 2016
A computer producing factory has only two plants T1T_1T1​ and T2T_2T2​. Plant T1T_1T1​ produces 20% and plant T2T_2T2​ produces 80% of the total computers produced. 7% of computers produced in the factory turn out to be defective. It is known that P(computer turns out to be defective given that it is produced in plant T1T_1T1​) = 10 P(computer turns out to be defective given that it is produced in plant T2T_2T2​), where P(E) denotes the probability of an event E. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant T2T_2T2​ is
  1. (A)3673\frac{36}{73}7336​
  2. (B)4779\frac{47}{79}7947​
  3. (C)7893\frac{78}{93}9378​
  4. (D)7583\frac{75}{83}8375​

Correct answer: (C)

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Q217·MathematicsIntegerJEE Advanced 2015
The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96 is

Correct answer: 8

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Q218·MathematicsMultiple correctJEE Advanced 2015
Let n1n_{1}n1​ and n2n_{2}n2​ be the number of red and black balls, respectively, in box I. Let n3n_{3}n3​ and n4n_{4}n4​ be the number of red and black balls, respectively, in box II. A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 13\dfrac{1}{3}31​, then the correct option(s) with the possible values of n1n_{1}n1​ and n2n_{2}n2​ is(are)
  1. (A)n1=4n_{1} = 4n1​=4, n2=6n_{2} = 6n2​=6
  2. (B)n1=2n_{1} = 2n1​=2, n2=3n_{2} = 3n2​=3
  3. (C)n1=10n_{1} = 10n1​=10, n2=20n_{2} = 20n2​=20
  4. (D)n1=3n_{1} = 3n1​=3, n2=6n_{2} = 6n2​=6

Correct answer: (C), (D)

Step-by-step solution →
Q219·MathematicsMultiple correctJEE Advanced 2015
Let n1n_{1}n1​ and n2n_{2}n2​ be the number of red and black balls, respectively, in box I. Let n3n_{3}n3​ and n4n_{4}n4​ be the number of red and black balls, respectively, in box II. One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is 13\dfrac{1}{3}31​, then the correct option(s) with the possible values of n1n_{1}n1​, n2n_{2}n2​, n3n_{3}n3​ and n4n_{4}n4​ is(are)
  1. (A)n1=3n_{1} = 3n1​=3, n2=3n_{2} = 3n2​=3, n3=5n_{3} = 5n3​=5, n4=15n_{4} = 15n4​=15
  2. (B)n1=3n_{1} = 3n1​=3, n2=6n_{2} = 6n2​=6, n3=10n_{3} = 10n3​=10, n4=50n_{4} = 50n4​=50
  3. (C)n1=8n_{1} = 8n1​=8, n2=6n_{2} = 6n2​=6, n3=5n_{3} = 5n3​=5, n4=20n_{4} = 20n4​=20
  4. (D)n1=6n_{1} = 6n1​=6, n2=12n_{2} = 12n2​=12, n3=5n_{3} = 5n3​=5, n4=20n_{4} = 20n4​=20

Correct answer: (A), (B)

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Q220·MathematicsSingle correctJEE Advanced 2014
Box 1 contains three cards bearing numbers 1, 2, 3 ; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5 ; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xix_{i}xi​ be the number on the card drawn from the ithi^{th}ith box, i=1,2,3i = 1, 2, 3i=1,2,3. The probability that x1+x2+x3x_{1} + x_{2} + x_{3}x1​+x2​+x3​ is odd, is
  1. (A)29105\frac{29}{105}10529​
  2. (B)53105\frac{53}{105}10553​
  3. (C)57105\frac{57}{105}10557​
  4. (D)12\frac{1}{2}21​

Correct answer: (B)

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Q221·MathematicsSingle correctJEE Advanced 2014
Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is
  1. (A)12\frac{1}{2}21​
  2. (B)13\frac{1}{3}31​
  3. (C)23\frac{2}{3}32​
  4. (D)34\frac{3}{4}43​

Correct answer: (A)

Step-by-step solution →
Q222·MathematicsSingle correctJEE Advanced 2014
Box 1 contains three cards bearing numbers 1, 2, 3 ; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5 ; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let xix_{i}xi​ be the number on the card drawn from the ithi^{th}ith box, i=1,2,3i = 1, 2, 3i=1,2,3. The probability that x1,x2,x3x_{1}, x_{2}, x_{3}x1​,x2​,x3​ are in an arithmetic progression, is
  1. (A)9105\frac{9}{105}1059​
  2. (B)10105\frac{10}{105}10510​
  3. (C)11105\frac{11}{105}10511​
  4. (D)7105\frac{7}{105}1057​

Correct answer: (C)

Step-by-step solution →
Q223·MathematicsSingle correctJEE Advanced 2013
A box B1B_{1}B1​ contains 1 white ball, 3 red balls and 2 black balls. Another box B2B_{2}B2​ contains 2 white balls, 3 red balls and 4 black balls. A third box B3B_{3}B3​ contains 3 white balls, 4 red balls and 5 black balls. If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B2B_{2}B2​ is
  1. (A)116181\frac{116}{181}181116​
  2. (B)126181\frac{126}{181}181126​
  3. (C)65181\frac{65}{181}18165​
  4. (D)55181\frac{55}{181}18155​

Correct answer: (D)

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Q224·MathematicsIntegerJEE Advanced 2013
Of the three independent events E1E_1E1​, E2E_2E2​, and E3E_3E3​, the probability that only E1E_1E1​ occurs is α\alphaα, only E2E_2E2​ occurs is β\betaβ and only E3E_3E3​ occurs is γ\gammaγ. Let the probability ppp that none of events E1E_1E1​, E2E_2E2​ or E3E_3E3​ occurs satisfy the equations (α−2β)p=αβ(\alpha-2\beta)p=\alpha\beta(α−2β)p=αβ and (β−3γ)p=2βγ(\beta-3\gamma)p=2\beta\gamma(β−3γ)p=2βγ. All the given probabilities are assumed to lie in the interval (0,1)(0,1)(0,1). Then Probability of occurrence of E1Probability of occurrence of E3=\dfrac{\text{Probability of occurrence of }E_1}{\text{Probability of occurrence of }E_3}=Probability of occurrence of E3​Probability of occurrence of E1​​= ___________

Correct answer: 6

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Q225·MathematicsSingle correctJEE Advanced 2013
A box B1B_{1}B1​ contains 1 white ball, 3 red balls and 2 black balls. Another box B2B_{2}B2​ contains 2 white balls, 3 red balls and 4 black balls. A third box B3B_{3}B3​ contains 3 white balls, 4 red balls and 5 black balls. If 1 ball is drawn from each of the boxes B1B_{1}B1​, B2B_{2}B2​ and B3B_{3}B3​, the probability that all 3 drawn balls are of the same colour is
  1. (A)82648\frac{82}{648}64882​
  2. (B)90648\frac{90}{648}64890​
  3. (C)558648\frac{558}{648}648558​
  4. (D)566648\frac{566}{648}648566​

Correct answer: (A)

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Q226·MathematicsSingle correctJEE Advanced 2013
Four persons independently solve a certain problem correctly with probabilities 12,34,14,18\frac{1}{2},\frac{3}{4},\frac{1}{4},\frac{1}{8}21​,43​,41​,81​. Then the probability that the problem is solved correctly by at least one of them is
  1. (A)235256\frac{235}{256}256235​
  2. (B)21256\frac{21}{256}25621​
  3. (C)3256\frac{3}{256}2563​
  4. (D)253256\frac{253}{256}256253​

Correct answer: (A)

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Probability — frequently asked

How many questions from Probability appear in JEE?

Probability has appeared in 176 of the last 186 JEE Main and JEE Advanced papers — about 95% of them — contributing 226 questions in total across those papers.

Is Probability an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 95% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Probability questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Mathematics chapters

  • Three Dimensional Geometry 344
  • Matrices and Determinants 342
  • Sets, Relations and Functions 313
  • Sequence and Series 287
  • Definite Integration 267
  • Vector Algebra 245
  • Differential Equations 240
  • Permutations and Combinations 220

All 26 Mathematics chapters →

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