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JEE • MATHEMATICS • PROBABILITY

Probability practice · Mixed difficulty

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Practice format and chapter quiz

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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Showing 1-30 of 1130 questions
1 Easy
1 of 1130
Application +4 | -1
Consider the probability distribution

X:12345P(X):k22kk2k5k2\begin{array}{cccccc}X: & 1 & 2 & 3 & 4 & 5 \\ P(X): & k^2 & 2 k & k & 2 k & 5 k^2\end{array}
The value of P(X>2)P(X>2) is
View answer and explanation
Correct option: D
2336\frac{23}{36}
Option D (2336\frac{23}{36}) is correct because it calculates the probability P(X>2) after finding the value of k from the given probability distribution.
2 Easy
2 of 1130
Application +4 | -1
A bag contains (2  n+1)\left( {2\;n + 1} \right) coins. It is known that nn of these coins have a head on both sides, whereas the remaining n+1n + 1 coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is 31/42, then nn is equal to
View answer and explanation
Correct option: A
10
Option A (10) is correct because solving the probability equation derived from the given conditions yields n = 10.
3 Medium
3 of 1130
Application +4 | -1
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05 . What is the probability that out of 5 such bulbs none will fuse after 150 days of use
View answer and explanation
Correct option: B
(1920)5\left(\frac{19}{20}\right)^5
Option B is correct. The probability that none of the 5 bulbs will fuse is calculated using the binomial probability formula with p = 19/20, representing the probability of a bulb not fusing.
4 Easy
4 of 1130
Application +4 | -1
If AA and BB are two independent events such that P(A∩B′)=325P\left( {A \cap B'} \right) = \frac{3}{{25}} and P(A′∩B)=825P\left( {A' \cap B} \right) = \frac{8}{{25}}, then P(A)P\left( A \right) is equal to
View answer and explanation
Correct option: A
15\frac{1}{5}
Option A (1/5) is correct because it is found by using the independence of events A and B to solve for P(A) from the given probabilities.
5 Easy
5 of 1130
Analyzing +4 | -1
In a game two players AA and BB take turns in throwing a pair of fair dice starting with player AA 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 BB throws a total of 7 and BB 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 player wins. The probability of AA winning the game is
View answer and explanation
Correct option: A
3061\frac{30}{61}
Option A (3061\frac{30}{61}) is correct because it calculates the probability of player A winning by summing the probabilities of A winning on their first turn, third turn, fifth turn, and so on.
6 Easy
6 of 1130
Application +4 | -1
A box contains 5 brown and 4 white socks. A man takes out two socks. The probability that they are of the same colour is
View answer and explanation
Correct option: D
49\frac{4}{9}
Option D (49\frac{4}{9}) is correct because it represents the probability of drawing two socks of the same color, calculated by summing the probabilities of drawing two brown socks and two white socks.
7 Medium
7 of 1130
Application +4 | -1
Given that XX is discrete random variable which takes the values 0, 1, 2 and P(X=0)=144169,  P(X=1)=1169,P\left( {X = 0} \right) = \frac{{144}}{{169}},\;P\left( {X = 1} \right) = \frac{1}{{169}}, then the value of P(X=2)P\left( {X = 2} \right) is
View answer and explanation
Correct option: B
24169\frac{{24}}{{169}}
Option B (24169\frac{{24}}{{169}}) is correct because the sum of probabilities for all possible values of a discrete random variable must equal 1.
8 Easy
8 of 1130
Application +4 | -1
If n(A)=4n(A)=4 and n(B)=7n(B)=7, then the difference between maximum and minimum value of n(A∪B)n(A \cup B) is
View answer and explanation
Correct option: A
4
Option A (4) is correct because it represents the difference between the maximum and minimum possible values of n(A∪B)n(A \cup B) based on the given sizes of sets A and B.
9 Easy
9 of 1130
Application +4 | -1
There are 3 works. One is of 3 volumes and one is of 4 volumes and one is of exactly one volume. If the books are placed in a random order on a shelf, the probability that all the volumes of the same work are together is
View answer and explanation
Correct option: B
3140\frac{3}{140}
Option D (3/140) is correct because it calculates the probability that all volumes of the same work are together when the books are placed in a random order.
10 Easy
10 of 1130
Application +4 | -1
The probability that a leap year selected at random contains either 53 Sundays or 53 Mondays, is
View answer and explanation
Correct option: C
37\frac{3}{7}
Option C (37\frac{3}{7}) is correct because it uses the principle of inclusion-exclusion to find the probability of either 53 Sundays or 53 Mondays in a leap year.
11 Easy
11 of 1130
Application +4 | -1
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
View answer and explanation
Correct option: B
1736\frac{{17}}{{36}}
Option B (1736\frac{{17}}{{36}}) is correct because it represents the probability of the sum of two dice being less than or equal to 8, given the specific numbers on the dice faces.
12 Easy
12 of 1130
Application +4 | -1
A binary number is made up of 8 digits. Suppose that the probability of an incorrect digit appearing in pp and that of errors in different digits are independent of each other. The the probability of forming an incorrect number, is
View answer and explanation
Correct option: D
1−(1−p)81-(1-p)^8
Option D (1−(1−p)81-(1-p)^8) is correct because it calculates the probability of at least one incorrect digit by subtracting the probability of all digits being correct from 1.
13 Easy
13 of 1130
Application +4 | -1
A purse contains 4 copper and 3 silver coins. Another purse contains 6 copper and 2 silver coins. A coin is taken out from any purse, the probability that it is a silver coin, is
View answer and explanation
Correct option: B
1956\frac{{19}}{{56}}
Option B (1956\frac{{19}}{{56}}) is correct because it calculates the probability of selecting a silver coin from either of the two purses.
14 Easy
14 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 0.30.3 and 0.20.2 respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by second plane is
View answer and explanation
Correct option: D
0.140.14
Option D (0.14) is correct because it calculates the probability of the second plane hitting the target given the first plane misses.
15 Easy
15 of 1130
Application +4 | -1
Four numbers are chosen at random from {1, 2, 3, …, 40}. The probability that they are not consecutive, is
View answer and explanation
Correct option: C
24692470\frac{{2469}}{{2470}}
Option C (24692470\frac{{2469}}{{2470}}) is correct because it calculates the probability that four numbers chosen at random from the set are not consecutive.
16 Easy
16 of 1130
Application +4 | -1
A coin is tossed three times. The probability of getting a head once and a tail twice is
View answer and explanation
Correct option: C
38\frac{3}{8}
Option C (38\frac{3}{8}) is correct because there are 3 favorable outcomes (HTT, THT, TTH) out of 8 total possible outcomes.
17 Easy
17 of 1130
Application +4 | -1
A bag contains 2 white and 4 black balls. A ball is drawn 5 times with replacement. The probability that at least 4 of the balls drawn are white, is
View answer and explanation
Correct option: C
11243\frac{{11}}{{243}}
Option C (11243\frac{{11}}{{243}}) is correct because it calculates the probability of drawing at least 4 white balls out of 5 draws with replacement using binomial probability.
18 Easy
18 of 1130
Application +4 | -1
A certain player, say XX, is known to win with probability 0.3 if the track is fast and 0.4 if the track is slow. For Monday, there is a 0.7 probability of a fast track and 0.3 probability of a slow track. The probability that player X{X} will win a Monday, is
View answer and explanation
Correct option: C
0.33
Option B (2n−1\frac{2}{n-1}) is correct because it calculates the probability that two specified persons are sitting next to each other at a round table.
19 Medium
19 of 1130
Analyzing +4 | -1
If an unbiased die, marked with −2,−1,0,1,2,3-2,-1,0,1,2,3 on its faces, is thrown five times, then the probability that product of the outcomes is positive is:
View answer and explanation
Correct option: D
5212592\frac{521}{2592}
Option D (5212592\frac{{521}}{{2592}}) is correct because it is the probability calculated by considering the cases where the product of the outcomes is positive.
20 Easy
20 of 1130
Analyzing +4 | -1
Two numbers aa and bb are chosen at random from the set {1,2,3,…,3n}\{1,2,3, \ldots, 3 n\}. The probability that a3+b3a^3+b^3 is divisible by 3, is
View answer and explanation
Correct option: D
13\frac{1}{3}
Option D (13\frac{{1}}{{3}}) is correct because it represents the probability calculated by dividing the number of ways to choose two numbers such that a3+b3a^3 + b^3 is divisible by 3 by the total number of ways to choose two numbers.
21 Easy
21 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(n>k)1,2,3, \ldots, k(n>k) appear as neighbors is
View answer and explanation
Correct option: B
n−k+1nCk\frac{n-k+1}{{ }^n C_k}
Option B (136\frac{{1}}{{36}}) is correct because it calculates the probability of all three dice showing the same number.
22 Easy
22 of 1130
Application +4 | -1
Anexperiment has 10 equally likely outcomes. Let A 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
The correct answer is 5 or 10 because for A and B to be independent, P(A ∩ B) must equal P(A)P(B), leading to n = 5 or 10.
23 Easy
23 of 1130
Application +4 | -1
Three letters, to each of which corresponds an envelope, are placed in the envelopes at random. The probability that all the letters are not placed in the right envelopes, is
View answer and explanation
Correct option: B
05-Jun
Option B (05-Jun) is correct because it represents the probability that not all letters are placed in the correct envelopes, which is the complement of all letters being placed correctly.
24 Easy
24 of 1130
Application +4 | -1
Let n(U)=700,n(a)=200,n(b)=300n(U)=700, n(a)=200, n(b)=300 and n(A∩B)=100n(A \cap B)=100, then find n(A′∩B′)n\left(A^{\prime} \cap B^{\prime}\right)
View answer and explanation
Correct option: A
300
The correct answer is A (300) because n(A′∩B′)n(A' \cap B') can be found using De Morgan's law: n(A′∩B′)=n((A∪B)′)=n(U)−n(A∪B)n(A' \cap B') = n((A \cup B)') = n(U) - n(A \cup B).
25 Easy
25 of 1130
Application +4 | -1
If P(A∩B)=12,P(Aˉ∩Bˉ)=12P(A \cap B)=\frac{1}{2}, P(\bar{A} \cap \bar{B})=\frac{1}{2} and 2P(A)=P(B)=p2 P(A)=P(B)=p, then the value of pp is equal to
View answer and explanation
Correct option: B
23\frac{2}{3}
Option B (23\frac{{2}}{{3}}) is correct because it calculates the value of p using the given probabilities and the properties of probability.
26 Easy
26 of 1130
Application +4 | -1
If in a binomial distribution n=4,P(X=0)=1681n=4, P(X=0)=\frac{16}{81}, then P(X=4)P(X=4) equals
View answer and explanation
Correct option: B
181\frac{1}{81}
Option B (181\frac{{1}}{{81}}) is correct because it calculates the probability of all 4 trials being successes in a binomial distribution.
27 Medium
27 of 1130
Application +4 | -1
If nn positive integers are taken at random and multiplied together, and pnp_n is the probability that the last digit of the product is 2,4,62,4,6 or 8 then 125p3−50125 p_3-50 is equal to
View answer and explanation
Correct option: B
6
Option C (6) is correct because the probability that the last digit of the product is 2, 4, 6, or 8 is (4/5)^n - (2/5)^n, and for n=3, 125p_3 - 50 = 6.
28 Medium
28 of 1130
Application +4 | -1
A bag contains some white and some black balls, all combinations being equally likely. Total balls are 10. If three are drawn and found black then find probability that the bag contains 1 white and 9 black balls :
View answer and explanation
Correct option: A
1455\frac{14}{55}
Option A (1455\frac{14}{55}) is correct because it is calculated using Bayes' theorem, considering the probability of drawing three black balls given 1 white and 9 black balls, divided by the total probability of drawing three black balls.
29 Medium
29 of 1130
Application +4 | -1
The probability that a teacher will give an unannounced test during any class meeting is 1/5. If a student is absent twice, the probability that he will miss atleast one test, is
View answer and explanation
Correct option: B
9/259/25
Option B (9/25) is correct because it calculates the probability of missing at least one test by subtracting the probability of missing no tests from 1.
30 Easy
30 of 1130
Application +4 | -1
In a binomial distribution B(n,P=14),B\left( {n,P = \frac{1}{4}} \right), if the probability of at least one success is greater than or equal to 910,\frac{9}{{10}}, then nn is greater than
View answer and explanation
Correct option: A
1log⁡104−log⁡103\frac{1}{{{{\log }_{10}}4 - {{\log }_{10}}3}}
Option A (1log⁡104−log⁡103\frac{1}{{{{\log }_{10}}4 - {{\log }_{10}}3}}) is correct because it is derived from the inequality 1−(34)n≥9101 - {(\frac{3}{4})^n} \ge \frac{9}{{10}}, which represents the probability of at least one success being greater than or equal to 910\frac{9}{{10}}.
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Key Takeaways

  • ✓ This set contains 1130 single-correct-answer MCQs on PROBABILITY.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
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