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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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211 Easy
211 of 1130
Application +4 | -1
The chances to fail in physics are 20% and the chances to fail in mathematics are 10%. What are the chances to fail in at least one object?
View answer and explanation
Correct option: A
28%
Option A (28%) is correct because it calculates the probability of failing in at least one subject using the formula for the union of two events.
212 Easy
212 of 1130
Application +4 | -1
In a three throw of two dice, the probability throwing doublets not more than twice is
View answer and explanation
Correct option: C
215216\frac{215}{216}
Option C (215/216) is correct because it calculates the probability of throwing doublets not more than twice in three throws of two dice, using the binomial distribution.
213 Easy
213 of 1130
Application +4 | -1
Let AA and BB are two events and P(A′)=0.3,  P(B)=0.4,  P(A∩B′)=0.5,P\left( {A'} \right) = 0.3,\;P\left( B \right) = 0.4,\;P\left( {A \cap B'} \right) = 0.5, then P(A∪B′)P\left( {A \cup B'} \right) is
View answer and explanation
Correct option: B
0.8
Option B (0.8) is correct because it is the probability of A∪B′A \cup B' calculated using the given probabilities and the formula for the probability of the union of two events.
214 Easy
214 of 1130
Application +4 | -1
If pp and qq are chosen from the set {1,2,3\{1,2,3, 4,5,6,7,8,9,10}4,5,6,7,8,9,10\}, with replacement, the probability that the roots of the equation x2+2px+q=0x^2+2 p x+q=0 are real is
View answer and explanation
Correct option: A
0.84
Option A (0.84) is correct because it represents the probability calculated by finding the complement of the probability of non-real roots.
215 Easy
215 of 1130
Analyzing +4 | -1
Let 0<P(A)<1,0<P(B)<10 < P(A) < 1,0 < P(B) < 1 and P(A∪B)=P(A)+P(B)−P(A)P(B)P(A \cup B)=P(A)+P(B)-P(A) P(B). Then,
View answer and explanation
Correct option: B
P(A∪B)c=P(Ac)P(Bc)P(A \cup B)^c=P\left(A^c\right) P\left(B^c\right)
Option B (P(A∪B)c=P(Ac)P(Bc)P(A \cup B)^c=P(A^c) P(B^c)) is correct because it indicates that AA and BB are independent events, which is derived from the given equation P(A∪B)=P(A)+P(B)−P(A)P(B)P(A \cup B)=P(A)+P(B)-P(A) P(B).
216 Easy
216 of 1130
Application +4 | -1
Three identical dice are rolled. The probability that the same number will appear on each of them, is
View answer and explanation
Correct option: B
Jan-36
Option B (Jan-36) is correct because the probability of getting the same number on each of the three dice is 1/36.
217 Medium
217 of 1130
Application +4 | -1
From a group of 8 boys and 3 girls, a committee of 5 members to be formed. Find the probability that 2 particular girls are included in the committee
View answer and explanation
Correct option: B
211\frac{2}{{11}}
Option B (211\frac{2}{{11}}) is correct because it is the probability of including 2 particular girls in a committee of 5, calculated by dividing the number of favorable outcomes by the total number of possible committees.
218 Easy
218 of 1130
Application +4 | -1
A fair coin is tossed nn times. If the probability that heads occurs 6 times is equal to the probability that heads occurs 8 times, then value of nn is
View answer and explanation
Correct option: C
14
Option C (14) is correct because equating the probabilities of getting 6 and 8 heads leads to the equation nC6=nC8{ }^n C_6 = { }^n C_8, which solves for n = 14.
219 Medium
219 of 1130
Application +4 | -1
In a sequence of independent trials, the probability of success is 1/41 / 4. If pp denotes the probability that the second success occurs on the fourth trial or later trial, then 32p−2032 p-20 is
View answer and explanation
Correct option: C
7
Option C (7) is correct because the probability that the second success occurs on the fourth trial or later is 27/32, and 32p - 20 = 7.
220 Easy
220 of 1130
Analyzing +4 | -1
For two events AA and BB it is given that P(A)=P(A/B)=14P(A)=P(A / B)=\frac{1}{4} and P(B/A)=12P(B / A)=\frac{1}{2}. Then,
View answer and explanation
Correct option: B
AA and BB are independent events such that P(Aˉ/B)=3/4P(\bar{A} / B)=3 / 4
Option B (AA and BB are independent events such that P(Aˉ/B)=3/4P(\bar{A} / B)=3 / 4) is correct because the given conditions P(A)=P(A/B)=14P(A)=P(A / B)=\frac{1}{4} and P(B/A)=12P(B / A)=\frac{1}{2} imply that AA and BB are independent, and P(Aˉ/B)P(\bar{A} / B) is indeed 3/4.
221 Easy
221 of 1130
Application +4 | -1
A bag contains 7 red and 2 white balls and another bag contains 5 red and 4 white balls. Two balls are drawn, one from each bag. The probability that both the balls are white, is
View answer and explanation
Correct option: C
8/818/81
Option C (8/818/81) is correct because it is the product of the probabilities of drawing a white ball from each bag.
222 Hard
222 of 1130
Application +4 | -1
A bag contains some white and some
black balls, all combinations of balls being equally likely. The
total number of balls in the bag is 10. If three balls are drawn
at random without replacement and all of them are found to
be black, the probabili ty that the bag contains 1 white and
9 black balls is
View answer and explanation
Correct option: A
14/5514 / 55
Option A (14/55) is correct because it is the conditional probability calculated using Bayes' theorem, given that three black balls were drawn.
223 Medium
223 of 1130
Application +4 | -1
A complete cycle of a traffic light takes 60 s. During each cycle the light is green for 25 s, yellow for 5 s and red for 30 s. At a randomly chosen time, the probability that the light will not be green, is
View answer and explanation
Correct option: D
712\frac{7}{{12}}
Option A (12\frac{1}{2}) is correct because it is found by using the given probabilities and the formula P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B).
224 Medium
224 of 1130
Application +4 | -1
An experiment succeeds twice as often as it fails. Find the probability that in 4 trials there will be at least three success
View answer and explanation
Correct option: C
1627\frac{16}{27}
Option C (1627\frac{16}{27}) is correct because it calculates the probability of at least three successes in four trials, considering both 3 successes and 4 successes.
225 Medium
225 of 1130
Application +4 | -1
There are 10 pairs of shoes in a cupboard, from which 4 shoes are picked at random. If pp is the probability that there is at least one pair, then 323p323 p -97 is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because the probability of at least one pair is 99/323, and 323p - 97 = 2.
226 Easy
226 of 1130
Application +4 | -1
If P(A)=P(B)=x  P\left( A \right) = P\left( B \right) = x\; and P(A∩B)=P(A′∩B′)=13P\left( {A \cap B} \right) = P\left( {A' \cap B'} \right) = \frac{1}{3}, then xx is equal to
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (0.6976) is correct because it is calculated by subtracting the probability of missing the plane with all four shots from 1.
227 Easy
227 of 1130
Application +4 | -1
If P(A)=0.4,P(B)=0.8P(A)=0.4, P(B)=0.8 and P(B/A)=0.6P(B / A)=0.6, then P(A∪B)P(A \cup B) is equal to
View answer and explanation
Correct option: D
0.96
Option D (0.96) is correct because it is the result of calculating P(A∪B)P(A \cup B) using the given probabilities and the formula P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B).
228 Easy
228 of 1130
Application +4 | -1
Four persons can hit a target correctly with probabilities 12, 13, 14\frac{1}{2},\,\frac{1}{3},\,\frac{1}{4} and 18\frac{1}{8} respectively. If all hit at the target independently, then the probability that the target would be hit, is
View answer and explanation
Correct option: B
2532\frac{{25}}{{32}}
Option B (2532\frac{{25}}{{32}}) is correct because it represents the probability that the target is hit, calculated as 1 minus the probability that none of the four persons hit the target.
229 Medium
229 of 1130
Application +4 | -1
If AA and BB are events such that P(A)=0.42,P(B)=0.48P(A)=0.42, P(B)=0.48 and P(A and B)=0.16P(A \text{ and } B)=0.16, then,

I. P(not⁡A)=0.58{P}(\operatorname{not}{A})=0.58

II. P(not B)=0.52{P}(\text{not }{B})=0.52

III. P(A or B)=0.47{P}({A} \text{ or }{B})=0.47
View answer and explanation
Correct option: A
Only I and II are correct
Option A is correct because statements I and II are true based on the given probabilities, while statement III is false as the calculated probability of A or B is 0.74, not 0.47.
230 Easy
230 of 1130
Application +4 | -1
If the probability of hitting a target by a shooter in any shot, is 13\frac{1}{3}, then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than 56\frac{5}{6}, is
View answer and explanation
Correct option: C
5
Option C (5) is correct because it is the minimum number of trials required for the probability of hitting the target at least once to be greater than 5/6, calculated using binomial distribution.
231 Easy
231 of 1130
Application +4 | -1
Three natural numbers are taken at random one by one with replacement from the set A={x∣1≤x≤100,x∈N}\mathrm{A}=\{x \mid 1 \leq x \leq 100, x \in N\}. The probability that the AM\mathrm{AM} of the number is 25 is
View answer and explanation
Correct option: C
74C721003\frac{{ }^{74} C_{72}}{100^3}
Option C (74C721003\frac{{ }^{74} C_{72}}{100^3}) is correct because it represents the number of favorable outcomes (combinations of three numbers summing to 75) divided by the total possible outcomes (1003).
232 Medium
232 of 1130
Application +4 | -1
Seven chits are numbered 1 to 7. Four chits are drawn one by one with replacement. The probability that the least number appearing on any selected chit is 5, is
View answer and explanation
Correct option: A
(37)4{\left( {\frac{3}{7}} \right)^4}
Option A ((37)4{\left( {\frac{3}{7}} \right)^4}) is correct because it represents the probability that each of the four drawn chits has a number greater than or equal to 5.
233 Easy
233 of 1130
Application +4 | -1
In a school 80 students like chocolate, 40 like coffee if the number of students doesn't like any of them is equal to the number of students who like both of them then what is the total number of students in the school?
View answer and explanation
Correct option: C
120
Option C (120) is correct because it represents the total number of students calculated using the principle of inclusion-exclusion and the given conditions.
234 Easy
234 of 1130
Application +4 | -1
In a game called "odd man out" each person in a group of n≥2n \geq 2 persons tosses a coin to determine who will buy refreshment for the entire group. The odd man out is the one whose outcome is different from the rest of the persons. The probability that there is a loser in the game is
View answer and explanation
Correct option: A
n2n−1\frac{n}{2^{n-1}}
Option A (n2n−1\frac{n}{2^{n-1}}) is correct because it represents the probability of having exactly one person with a different outcome than the others.
235 Easy
235 of 1130
Understanding +4 | -1
If AA and BB are any two events, then P(A∩B′)P\left( {A \cap B'} \right) is equal to
View answer and explanation
Correct option: D
P(A)−P(A∩B)P\left( A \right) - P\left( {A \cap B} \right)
Option D (P(A)−P(A∩B)P\left( A \right) - P\left( {A \cap B} \right)) is correct because it represents the probability of A occurring without B occurring.
236 Easy
236 of 1130
Application +4 | -1
The Probability of solving a problem by three students A,B,CA, B, C is 12,13,14\frac{1}{2}, \frac{1}{3}, \frac{1}{4} respectively. The probability that the problem will be solved is
View answer and explanation
Correct option: C
34\frac{3}{4}
Option C (34\frac{3}{4}) is correct because it calculates the probability that the problem will be solved by finding the complement, which is the probability that none of them solve it.
237 Easy
237 of 1130
Application +4 | -1
For a random variable X,E(X)=3X,E\left( X \right) = 3 and E(X2)=11.E\left( {{X^2}} \right) = 11. Then, variable of XX is
View answer and explanation
Correct option: C
2
Option C (2) is correct because it is calculated using the formula for variance: Var(X) = E(X²) - [E(X)]².
238 Medium
238 of 1130
Analyzing +4 | -1
Let MM be the maximum value of the product of two positive integers when their sum is 66 . Let the sample space S={x∈Z:x(66−x)≥59M}S=\left\{x \in \mathbb{Z}: x(66-x) \geq \frac{5}{9} M\right\} and the event A={x∈S:xA=\{x \in S: x is a multiple of 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 we first find the maximum product M, then determine the sample space S and event A, and finally calculate the probability P(A).
239 Easy
239 of 1130
Understanding +4 | -1
The event AA is independent of itself if and only if P(A)P(A) is
View answer and explanation
Correct option: A
0 or 1
Option A (0 or 1) is correct because for an event A to be independent of itself, P(A) must be either 0 or 1.
240 Easy
240 of 1130
Analyzing +4 | -1
Let AA and BB be two events such that P(A∪B‾)=16,P(A∩B)=14P(\overline{A \cup B})=\frac{1}{6}, P(A \cap B)=\frac{1}{4} and P(Aˉ)=14P(\bar{A})=\frac{1}{4}, where Aˉ\bar{A} stands for the complement of the event AA. Then the events AA and BB are
View answer and explanation
Correct option: A
independent but not equally likely
Option A (independent but not equally likely) is correct because P(A∩B)=P(A)P(B)P(A \cap B) = P(A)P(B) and P(A)≠P(B)P(A) \neq P(B).
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