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JEE • MATHEMATICS • PROBABILITY

Probability practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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661 Easy
661 of 1130
Analyzing +4 | -1
Let AA be a set consisting of nn elements. The probability of selecting two subsets PP and QQ of set AA such that Q=PˉQ=\bar{P}, is
View answer and explanation
Correct option: B
12n−1\frac{1}{2^n-1}
Option B (12n−1\frac{{1}}{{2^n-1}}) is correct because it represents the probability of selecting two subsets P and Q of set A such that Q is the complement of P.
662 Easy
662 of 1130
Application +4 | -1
Let a random variable X have a binomial distribution with mean 8 and variance 4.
If P(X≤2)=k216,P\left( {X \le 2} \right) = \frac{k}{{{2^{16}}}}, then kk is equal to
View answer and explanation
Correct option: D
137
Option D (137) is correct because, with n=16 and p=1/2, the probability P(X <= 2) is calculated using the binomial distribution formula and simplifies to 137/216.
663 Easy
663 of 1130
Application +4 | -1
The numbers 1,2,3,…1,2,3, \ldots, nn are arranged in a random order. The probability that the digits 1,2,3,…,k(k<n)1,2,3, \ldots, k(k < n) appears as neighbours in that order is
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the probability that there are 4 persons between A and B is 1/7, which is not among the given options.
664 Easy
664 of 1130
Application +4 | -1
There are 100 students in a collage class of which 36 are boys studying statistics and 13 girls not studying statistics. If there are 55 girls in all, the Probability that a boy picked up at random is not studying, is
View answer and explanation
Correct option: C
15\frac{1}{5}
Option C (15\frac{1}{5}) is correct because it represents the probability that a randomly picked boy is not studying statistics, based on the given information.
665 Easy
665 of 1130
Application +4 | -1
Difference between mean and variance of a binomial variate is '1' and difference between their square is '11'. Then probability of getting exactly three success is:
View answer and explanation
Correct option: B
36C3(16)3×(56)33{ }^{36} C_3\left(\frac{1}{6}\right)^3 \times\left(\frac{5}{6}\right)^{33}
Option B (36C3(16)3×(56)33{ }^{36} C_3\left(\frac{1}{6}\right)^3 \times\left(\frac{5}{6}\right)^{33}) is correct because solving the equations for mean and variance gives n=36 and p=1/6, which are then used in the binomial probability formula.
666 Hard
666 of 1130
Application +4 | -1
An experiment has 10 equally likely outcomes. Let AA and BB be two non-empty events of the experiment. If AA consists of 4 outcomes, the number of outcomes that BB must have so that AA and BB are independent, is
View answer and explanation
Correct option: D
5 or 10
Option D (5 or 10) is correct because it identifies the number of outcomes for B that satisfy the independence condition P(A ∩ B) = P(A)P(B).
667 Medium
667 of 1130
Application +4 | -1
A person draws a card from a pack, replaces it, shuffles the pack, again draws a card, replaces it and draws again. This he does until he draws a heart. The probability that he will have to make at least four draws is
View answer and explanation
Correct option: C
2764\frac{27}{64}
Option C (27/64) is correct because it calculates the probability of not drawing a heart in the first three draws.
668 Easy
668 of 1130
Understanding +4 | -1
If AA and BB are two events, then PP (neither AA nor BB) equals
View answer and explanation
Correct option: A
1−P(A∪B)1 - P\left( {A \cup B} \right)
Option A, 1−P(A∪B)1 - P(A \cup B), is correct because the probability of neither A nor B occurring is the complement of the probability of A or B occurring.
669 Easy
669 of 1130
Application +4 | -1
If P(A∪B)=3/4P\left( {A \cup B} \right) = 3/4 and P(Aˉ)=2/3,P\left( {\bar A} \right) = 2/3, then P(Aˉ∩B)P\left( {\bar A \cap B} \right) is equal to
View answer and explanation
Correct option: C
5/125/12
Option C (5/12) is correct because it correctly uses the formula for the probability of the union of events to find the desired probability.
670 Easy
670 of 1130
Application +4 | -1
If out of 2n2 \mathrm{n} tickets, numbered as 1,2,3,…2n,31,2,3, \ldots 2 \mathrm{n}, 3 distinct are chosen at randomly, and probability that they are in A.P. is 110\frac{1}{10}, then &#39; nn &#39; is
View answer and explanation
Correct option: A
8
Option A (8) is correct because the probability of selecting three numbers in A.P. from 2n tickets is given as 1/10, which leads to n=8.
671 Easy
671 of 1130
Application +4 | -1
A biased coin with probability p,0<p<1p, 0 < p < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is 2/52 / 5, then pp is equal to
View answer and explanation
Correct option: C
1/31 / 3
Option C (13\frac{{1}}{{3}}) is correct because it solves the infinite geometric series representing the probability of the first head appearing on an even toss and equates it to 2/5.
672 Easy
672 of 1130
Analyzing +4 | -1
If AA and BB are two events than the value of the determinant choosen at random from all the determinants of order 2 with entries 0 or 1 only is positive or negative respectively. Then,
View answer and explanation
Correct option: D
P(A)=P(B)P\left( A \right) = P\left( B \right)
Option D (P(A)=P(B)P(A) = P(B)) is correct because the number of positive and negative determinants that can be formed with entries 0 or 1 are equal.
673 Easy
673 of 1130
Application +4 | -1
Six boys and six girls sit in a row randomly. The probability that boys and girls sit alternatively is
View answer and explanation
Correct option: C
1462\frac{1}{462}
Option C (1462\frac{{1}}{{462}}) is correct because it represents the probability of boys and girls sitting alternatively, calculated by dividing the number of favorable arrangements by the total number of arrangements.
674 Easy
674 of 1130
Analyzing +4 | -1
Let AA be a set of all 4-digit natural numbers whose exactly one digit is 7. Then, the probability that a randomly chosen element of AA leaves remainder 2 when divided by 5 is
View answer and explanation
Correct option: C
97297\frac{{97}}{{297}}
Option D (97297\frac{{97}}{{297}}) is correct because it is the probability calculated by dividing the number of favorable cases (numbers with remainder 2 when divided by 5) by the total number of 4-digit numbers with exactly one digit as 7.
675 Easy
675 of 1130
Application +4 | -1
A carton contains 20 bulbs, 5 of which are defective. The probability that, if a sample of 3 bulbs is chosen at random from the carton, 2 will be defective, is
View answer and explanation
Correct option: C
Sep-64
Option C (Sep-64) is correct because the probability of selecting 2 defective bulbs out of 3 is calculated using combinations and basic probability principles.
676 Easy
676 of 1130
Application +4 | -1
A six faced die is biased such that 3×P3 \times P (a prime number) =6×P=6 \times P (a composite number) =2×P(1).=2 \times P(1).

Let XX 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 XX is :
View answer and explanation
Correct option: D
811\frac{8}{11}
Option D (811\frac{8}{11}) is correct because the mean is calculated as np, where n is the number of throws and p is the probability of getting a perfect square.
677 Easy
677 of 1130
Application +4 | -1
Five persons entered the lift cabin on the ground floor of an 8-floor house. Suppose that each of them independently and with equal probability can leave the cabin at any floor beginning with the first. The probability of all five persons leaving at different floors, is
View answer and explanation
Correct option: C
7C5×5!75\frac{{ }^7 C_5 \times 5 !}{7^5}
Option C (7C5×5!75\frac{{ }^7 C_5 \times 5 !}{7^5}) is correct because it represents the probability of all five persons leaving at different floors in an 8-floor house, starting from the first floor.
678 Easy
678 of 1130
Application +4 | -1
An experiment yields 3 mutually exclusive and exhaustive events A,B,C.A,B,C. If P(A)=2P(B)=3P(C),P\left( A \right) = 2P\left( B \right) = 3P\left( C \right), then P(A)P\left( A \right) is equal to
View answer and explanation
Correct option: D
611\frac{6}{{11}}
Option D (611\frac{6}{{11}}) is correct because it satisfies the given conditions for mutually exclusive and exhaustive events A, B, and C.
679 Easy
679 of 1130
Analyzing +4 | -1
Let A,BA, B and CC be three events such that P(A)=0.3,P(B)=0.4,P(C)=0.8,P(A∩B)=0.08P(A)=0.3, \quad P(B)=0.4, \quad P(C)=0.8, \quad P(A \cap B)=0.08, P(A∩C)=0.28,P(A∩B∩C)=0.09P(A \cap C)=0.28, P(A \cap B \cap C)=0.09. If P(A∪B∪C)≥0.75P(A \cup B \cup C) \geq 0.75, then show that P(B∩C)P(B \cap C) satisfies
View answer and explanation
Correct option: C
0.23≤P(B∩C)≤0.480.23 \leq P(B \cap C) \leq 0.48
Option C (0.23≤P(B∩C)≤0.480.23 \leq P(B \cap C) \leq 0.48) is correct because it represents the range of possible values for the probability of the intersection of events B and C, derived from the given probabilities and the properties of probability.
680 Easy
680 of 1130
Application +4 | -1
There are nn urns each containing (n+1)(n+1) balls such that the ith i^{\text {th }} urn contains ' ii ' white balls and (n+1−i)(n+1-i) red balls. Let UiU_i be the event of selecting ithurn,i=1,2,3,…,ni^{t h} u r n, i=1,2,3, \ldots, n and WW denotes the event of getting a white ball. If P(Ui)∝iP\left(U_i\right) \propto i, where i=1,2,3,…,ni=1,2,3, \ldots, n, then lim⁡n→∞P(W)\lim\limits_{n \to \infty } P(W) is equal to
View answer and explanation
Correct option: B
23\frac{2}{3}
Option B (2/3) is correct because it is the limit of the probability of drawing a white ball as n approaches infinity, calculated using the total probability theorem.
681 Easy
681 of 1130
Application +4 | -1
A biased die is tossed and the respective probabilities for various faces to turn up are
Face                      :  1            2                3              4              5              6Probability:  0.1      0.24      0.19    0.18    0.15     0.14\begin{array}{llllllllllllllllllll} {Face\;\;\;\;\;\;\;\;\;\;\;:\;1\;\;\;\;\;\;2\;\;\;\;\;\;\;\;3\;\;\;\;\;\;\;4\;\;\;\;\;\;\;5\;\;\;\;\;\;\;6}\\ {Probability\mathrm{ }:\;0.1\;\;\,\,0.24\;\;\,\,0.19\;\,\,0.18\;\,\,0.15\;\,\,\,0.14} \end{array}
If an even face has turned up, then the probability that it is face 2 or face 4, is
View answer and explanation
Correct option: C
0.75
Option C (0.75) is correct because it correctly calculates the conditional probability of an even face being 2 or 4.
682 Easy
682 of 1130
Application +4 | -1
Two numbers are selected randomly from the set S={1,  2,  3,  4,  5,  6}S = \left\{ {1,\,\,2,\,\,3,\,\,4,\,\,5,\,\,6} \right\} } without replacement one by one. The probability that minimum of the two number is less than 4, is
View answer and explanation
Correct option: D
45\frac{4}{{5}}
Option D (45\frac{4}{{5}}) is correct because it calculates the probability that the minimum of two randomly selected numbers from the set is less than 4.
683 Medium
683 of 1130
Application +4 | -1
Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane, is
View answer and explanation
Correct option: C
0.32
Option B (0.14) is incorrect. The correct answer is 0.32 because it accounts for the probability of the second plane hitting the target only if the first plane misses.
684 Easy
684 of 1130
Application +4 | -1
If the probability of AA to fail in an examination is 0.2 and that for BB is 0.3, then probability that either AA or BB is fail, is
View answer and explanation
Correct option: B
0.44
Option B (0.44) is correct because it calculates the probability of either A or B failing by using the formula P(A or B) = P(A) + P(B) - P(A and B), considering A and B are independent events.
685 Easy
685 of 1130
Application +4 | -1
If two numbers pp and qq are choosen randomly from the set {1,2,3,4}\{1,2,3,4\} with replacement, then the probability that p2≥4qp^2 \geq 4 q is equal to
View answer and explanation
Correct option: D
716\frac{7}{16}
Option D (716\frac{{7}}{{16}}) is correct because by listing all possible outcomes and counting the favorable ones, the probability is found to be 7/16.
686 Easy
686 of 1130
Application +4 | -1
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:
View answer and explanation
Correct option: A
1225\frac{12}{25}
Option A (1225\frac{{12}}{{25}}) is correct because it represents the probability calculated by dividing the number of favorable outcomes (letters posted to exactly two addresses) by the total possible outcomes.
687 Easy
687 of 1130
Application +4 | -1
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 nn , so that the probability of guessing at least ′n′'n' correct answer is less than 12\frac{1}{2} , is
View answer and explanation
Correct option: A
5
Option 5 is the smallest value of n such that the probability of guessing at least n correct answers is less than 1/2.
688 Easy
688 of 1130
Understanding +4 | -1
There are nn letters and n\mathrm{n} addressed envelopes, the probability that all the letters are not kept in the right envelope, is
View answer and explanation
Correct option: B
1−1n!1-\frac{1}{n !}
Option B (1−1n!1-\frac{1}{n !}) is correct because the probability that all letters are in the correct envelope is 1n!\frac{1}{n!}, so the probability that they are NOT all in the correct envelope is 1−1n!1 - \frac{1}{n!}.
689 Medium
689 of 1130
Application +4 | -1
A letter is taken out at random from ‘ASSISTANT’ and another is taken out from ‘STATISTICS’. The probability that they are same letters, is
View answer and explanation
Correct option: C
1990\frac{{19}}{{90}}
Option C (1990\frac{{19}}{{90}}) is correct because it sums the probabilities of drawing the same letter (A, I, S, or T) from both words, considering the frequency of each letter in each word.
690 Easy
690 of 1130
Understanding +4 | -1
If A\mathrm{A} and B\mathrm{B} are two events such that A⊂B\mathrm{A} \subset \mathrm{B} and P(B)≠0P(B) \neq 0, then which of the following is correct:
View answer and explanation
Correct option: C
P(A∣B)≥P(A)\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \geq \mathrm{P}(\mathrm{A})
Option C (P(A∣B)≥P(A)\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \geq \mathrm{P}(\mathrm{A})) is correct because if A is a subset of B, then the probability of A given B is P(A)/P(B), and since P(B) <= 1, P(A|B) >= P(A).
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