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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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571 Easy
571 of 1130
Application +4 | -1
Four identical dice are rolled once. The probability that atleast three different numbers appear on them, is
View answer and explanation
Correct option: D
2542\frac{25}{42}
Option D is correct because the probability is calculated by summing the favorable outcomes (at least three different numbers) and dividing by the total possible outcomes.
572 Easy
572 of 1130
Application +4 | -1
One hundred cards are numbered from 1 to 100. The probability that a randomly chosen card has a digit 5 is
View answer and explanation
Correct option: C
19100\frac{{19}}{{100}}
Option C (19/100) is correct. The probability that a randomly chosen card has a digit 5 is calculated by counting the number of cards with the digit 5 and dividing by the total number of cards.
573 Medium
573 of 1130
Application +4 | -1
If the integers mm and nn are chosen at random between 1 and 100, then the probability that a number of the form 7m+7n{7^m} + {7^n} is divisible by 5, equals
View answer and explanation
Correct option: A
14\frac{1}{4}
Option A (1/4) is correct because the probability that 7m+7n{7^m} + {7^n} is divisible by 5 depends on the last digits of 7m{7^m} and 7n{7^n}, which cycle through 7, 9, 3, 1.
574 Easy
574 of 1130
Application +4 | -1
The value of CC for which P(X=k)=C  k2P\left( {X = k} \right) = C\;{k^2} can serve as the probability function of a random variable XX that takes value 0, 1, 2, 3, 4 is
View answer and explanation
Correct option: A
130\frac{1}{{30}}
Option A (130\frac{1}{{30}}) is correct because it is the value of C that makes the sum of probabilities equal to 1, satisfying the condition for a probability function.
575 Medium
575 of 1130
Analyzing +4 | -1
Two dice are thrown independently. Let AA be the event that the number appeared on the 1st 1^{\text {st }} die is less than the number appeared on the 2nd 2^{\text {nd }} die, BB be the event that the number appeared on the 1st 1^{\text {st }} die is even and that one the second die is odd, and CC be the event that the number appeared on the 1st 1^{\text {st }} die is odd and that on the 2nd 2^{\text {nd }} is even. Then
View answer and explanation
Correct option: C
The number of favourable cases of the event (A∪B)∩C(A \cup B) \cap C is 6
Option C is correct. The number of favourable cases of the event (A∪B)∩C(A \cup B) \cap C is 6.
576 Easy
576 of 1130
Application +4 | -1
A die is tossed 5 times. Getting an odd number is considered a success. Then, the variance of distribution of success, is
View answer and explanation
Correct option: D
54\frac{5}{4}
Option D (54\frac{5}{4}) is correct because it's the variance calculated using the formula npq for a binomial distribution.
577 Medium
577 of 1130
Application +4 | -1
Three randomly chosen non negative integers x,yx, y and zz are found to satisfy eq. x+y+z=10x+y+z=10 and the probability that zz is even is p/qp / q (p,qp, q coprime) then p+q={p}+{q}= ____
View answer and explanation
Correct option: A
17
Option A (17) is correct because the probability that z is even is 6/11, and thus p+q = 6+11 = 17.
578 Easy
578 of 1130
Application +4 | -1
A group of 2n2 n boys and 2n2 n girls is randomly divided into two equal groups. The probability that each group contains the same number of boys and girls is
View answer and explanation
Correct option: D
(2nCn)24nCn\frac{\left({ }^{2 n} C_n\right)^2}{{ }^{4 n} C_n}
Option D ((2nCn)24nC2n\frac{\left({ }^{2 n} C_n\right)^2}{{ }^{4 n} C_{2 n}}) is correct because it represents the number of ways to form two equal groups with n boys and n girls each, divided by the total number of ways to form two equal groups.
579 Easy
579 of 1130
Application +4 | -1
Three dice are rolled. If the probability of getting different numbers on the three dice is pq\frac{p}{q}, where pp and qq are co-prime, then q−pq-p is equal to
View answer and explanation
Correct option: D
4
Option D (4) is correct because the probability of getting different numbers is 5/9, and q-p = 9-5 = 4.
580 Easy
580 of 1130
Application +4 | -1
Given P(A∪B)=0.6,P(A∩B)=0.2,P\left( {A \cup B} \right) = 0.6,P\left( {A \cap B} \right) = 0.2, the probability of exactly one of the event occurs is
View answer and explanation
Correct option: A
0.4
Option A (0.4) is correct because it calculates the probability of exactly one event occurring by subtracting the probability of the intersection from the probability of the union.
581 Easy
581 of 1130
Analyzing +4 | -1
A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its faces and the probability of getting a face with mark n is 1/n. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is:
View answer and explanation
Correct option: D
13212\frac{{13}}{{{2^{12}}}}
Option D (13212\frac{{13}}{{{{2^{12}}}}}) is correct because it represents the probability of obtaining a sum of 48 when the biased die is thrown thrice, considering the two possible combinations and their probabilities.
582 Medium
582 of 1130
Application +4 | -1
The probability that a certain kind of component will survive a given shock test is 34\frac{3}{4}. The probability that exactly 2 of the next 4 components tested survive is
View answer and explanation
Correct option: D
27/128
Option D (27/128) is correct because it is the probability calculated using the binomial probability formula for exactly 2 successes out of 4 trials.
583 Easy
583 of 1130
Analyzing +4 | -1
Let S={1,2,3,…,2022}S=\{1,2,3, \ldots, 2022\}. Then the probability, that a randomly chosen number nn from the set SS such that HCF⁡(n,2022)=1\operatorname{HCF}(n, 2022)=1, is
View answer and explanation
Correct option: D
112337\frac{112}{337}
Option D (112337\frac{112}{337}) is correct because it represents the probability of choosing a number from S that is relatively prime to 2022, calculated using the principle of inclusion-exclusion.
584 Easy
584 of 1130
Application +4 | -1
A student takes his examination in four subjects α,β\alpha, \beta, γ,δ\gamma, \delta. He estimates his chance of passing in α\alpha as 45\frac{4}{5}, in β\beta as 34\frac{3}{4}, in γ\gamma as 56\frac{5}{6} and in δ\delta as 23\frac{2}{3}. To qualify he must pass in α\alpha and at least two other subjects. The Probability that he qualifies is
View answer and explanation
Correct option: B
6190\frac{61}{90}
Option B (6190\frac{61}{90}) is correct because it sums the probabilities of all qualifying scenarios where the student passes in alpha and at least two other subjects.
585 Hard
585 of 1130
Application +4 | -1
Three six-faced fair dice are thrown together. The probability that the sum of the numbers appearing on the dice is k(3≤k≤8)k(3 \leq k \leq 8), is
View answer and explanation
Correct option: A
(k−1)(k−2)432\frac{(k-1)(k-2)}{432}
Option A is correct because the probability is calculated by finding the coefficient of x^k in the generating function and dividing by the total number of outcomes.
586 Easy
586 of 1130
Application +4 | -1
Four positive integers are taken at random and are multiplied together. Then the probability that the product ends in an odd digit other than 5, is
View answer and explanation
Correct option: B
16/625
Option C (17\frac{1}{7}) is correct because it calculates the probability of drawing balls of alternating colors by considering both possible sequences (WBWB and BWBW) and summing their probabilities.
587 Easy
587 of 1130
Application +4 | -1
If mm rupee coins and nn ten paise coins are placed in a line, then the probability that the extreme coins are ten paise coins, is
View answer and explanation
Correct option: B
n(n−1)(m+n)(m+n−1)\frac{{n\left( {n - 1} \right)}}{{\left( {m + n} \right)\left( {m + n - 1} \right)}}
Option B (n(n−1)(m+n)(m+n−1)\frac{{n(n - 1)}}{{(m + n)(m + n - 1)}}) is correct because it represents the probability of selecting a ten paise coin for both extreme positions out of the total possible arrangements.
588 Easy
588 of 1130
Application +4 | -1
Two cards are drawn successively with replacement from a well shuffled deck of 52 cards, then the mean of the number of aces is
View answer and explanation
Correct option: C
213\frac{2}{{13}}
Option C (213\frac{2}{{13}}) is correct because it calculates the mean number of aces by considering the probability of drawing an ace in each draw with replacement.
589 Easy
589 of 1130
Application +4 | -1
Let E1,E2{E_1},{E_2} be two mutually exclusive events of an experiment with PP(not E2{E_2}) =0.6=P(E1∪E2)= 0.6 = P\left( {{E_1} \cup {E_2}} \right).Then, P(E1)P\left( {{E_1}} \right) is equal to
View answer and explanation
Correct option: D
0.20.2
Option D (0.20.2) is correct because it is calculated using the properties of mutually exclusive events and the given probabilities.
590 Medium
590 of 1130
Application +4 | -1
Four boys and three girls stand in a queue for an interview, probability that they will in alternate position, is
View answer and explanation
Correct option: B
1/351/35
Option B (1/351/35) is correct because it calculates the probability of boys and girls standing in alternate positions by dividing the number of favorable arrangements by the total number of possible arrangements.
591 Medium
591 of 1130
Application +4 | -1
A missile is destroyed before hitting the target by a probability of 13\frac{1}{3}. If the missile is not destroyed probability of hitting the target is 34\frac{3}{4}. Three missiles are fired. The probability of exactly two hits the target is
View answer and explanation
Correct option: C
38\frac{3}{8}
Option B (83\frac{8}{3}) is correct because as x approaches 2 from the right, the greatest integer function evaluates to 0, leaving only the cubic term.
592 Easy
592 of 1130
Application +4 | -1
A four digit number is formed by the digits 1, 2, 3, 4 with no repetition. The probability that the number is odd, is
View answer and explanation
Correct option: A
0.5
Option A (0.5) is correct because it calculates the probability by dividing the number of odd numbers formed by the total number of possible numbers.
593 Easy
593 of 1130
Understanding +4 | -1
Assertion: Probability of getting a head in a toss of an unbiased coin is 12\frac{1}{2}.

Reason: In a simultaneous toss of two coins, the probability of getting 'no tails' is 14\frac{1}{4}.
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a correct explanation for assertion
Option B (Assertion is correct, reason is correct; reason is not a correct explanation for assertion) is correct because both the probability of getting a head is 1/2 and the probability of getting no tails in two coin tosses is 1/4 are correct, but they are unrelated.
594 Easy
594 of 1130
Application +4 | -1
AA and BB are two independent events. The probability that both AA and BB occur is 1/61/6 and the probability that neither of them occurs is 1/31/3. Then,
View answer and explanation
Correct option: A
P(A)=1/2,P(B)=1/3P\left( A \right) = 1/2,P\left( B \right) = 1/3
Option A (P(A)=1/2,P(B)=1/3P(A) = 1/2, P(B) = 1/3) is correct because it satisfies the conditions for independent events and the given probabilities.
595 Medium
595 of 1130
Application +4 | -1
AA is a set containing nn elements. Two subsets PP and QQ of AA are chosen at random. (PP and QQ may have elements in common). The probability that P∩Q=P \cap Q= ϕ\phi is
View answer and explanation
Correct option: A
(34)n\left(\frac{3}{4}\right)^n
Option A ((34)n\left(\frac{3}{4}\right)^n) is correct because it represents the probability that the intersection of two randomly chosen subsets P and Q of A is empty.
596 Easy
596 of 1130
Application +4 | -1
A bag contains 16 coins of which two are counterfeit with heads on both sides. The rest are fair coins. One is selected at random from the bag and tossed. The probability of getting a head is
View answer and explanation
Correct option: A
9/169 / 16
Option A (9/169 / 16) is correct because it uses the law of total probability to calculate the probability of getting a head.
597 Easy
597 of 1130
Analyzing +4 | -1
If AA and BB are independent events of aa random experiment such that P(A∩B)=16P(A \cap B)=\frac{1}{6} and P(Aˉ∩Bˉ)=13P(\bar{A} \cap \bar{B})=\frac{1}{3}, then P(A)P(A) is equal to
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{1}{3}) is correct because solving the system of equations derived from the independence and complement properties yields P(A)=13P(A) = \frac{1}{3}.
598 Easy
598 of 1130
Application +4 | -1
A bag XX contains 2 white and 3 black balls and another bag YY contains 4 white and 2 black balls. One bag is selected at random and a ball is drawn from it. Then, probability for the ball chosen be white, is
View answer and explanation
Correct option: C
815\frac{8}{{15}}
Option C (8/15) is correct because it calculates the probability of drawing a white ball by considering the probabilities of selecting each bag and then drawing a white ball from that bag.
599 Easy
599 of 1130
Application +4 | -1
On each evening a boy either watches DOORDARSHAN channel or TENSPORTS. The probability that he watches TEN SPORTS is 45\frac{4}{5}. If he watches DOORDARSHAN, there is a chance of 34\frac{3}{4} that he will be asleep, while it is 14\frac{1}{4} when he watches TEN SPORTS. On one day, the boy is found to be asleep. The probability that the boy watched DOORDARSHAN is
View answer and explanation
Correct option: C
37\frac{3}{7}
Option C (37\frac{3}{7}) is correct because it is the conditional probability that the boy watched DOORDARSHAN given that he was found asleep.
600 Easy
600 of 1130
Analyzing +4 | -1
Five different objects A1, A2, A3, A4, A5{A_1},\,{A_2},\,{A_3},\,{A_4},\,{A_5} are distributed randomly in 5 places marked 1, 2, 3, 4, 5. One arrangement is picked at random. The probability that in the selected arrangement, none of the object occupies the place corresponding to its number, is
View answer and explanation
Correct option: C
Nov-30
Option D (Nov-30) is correct because it represents the probability that none of the objects occupies the place corresponding to its number, which is a derangement problem.
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