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JEE • MATHEMATICS • PROBABILITY

Probability practice · Easy

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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 31-60 of 822 questions
31 Easy
31 of 822
Application +4 | -1
Three rotten apples are accidently mixed with fifteen good apples. Assuming the random variable xx to be the number of rotten apples in a draw of two apples, the variance of xx is
View answer and explanation
Correct option: D
40153\frac{40}{153}
Option D (40/153) is correct because it is the variance calculated using the probabilities of drawing 0, 1, or 2 rotten apples.
32 Easy
32 of 822
Application +4 | -1
An unbiased coin is tossed nn times. Let XX denote the number of times head occurs. If P(X=4),P(X=5)P\left( {X = 4} \right),P\left( {X = 5} \right) and P(X=6)P\left( {X = 6} \right) are in AP, then the value of nn can be
View answer and explanation
Correct option: A
7, 14
Option A (7, 14) is correct because these values of n satisfy the condition that P(X=4), P(X=5), and P(X=6) are in arithmetic progression.
33 Easy
33 of 822
Application +4 | -1
A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is
View answer and explanation
Correct option: D
8243\frac{8}{{243}}
Option D (8243\frac{8}{{243}}) is correct because it calculates the probability of getting a score of exactly 9 twice in three independent throws of a pair of dice using the binomial probability formula.
34 Easy
34 of 822
Application +4 | -1
Among the workers in a factory only 30% receive bonus and among those receiving bonus only 20% are skilled. The probability that a randomly selected worker is skilled and is receiving bonus is
View answer and explanation
Correct option: C
0.06
Option C (May-24) is correct because it is calculated by considering the probability of choosing each bag and then the probability of drawing a ticket with the number 4 from each bag.
35 Easy
35 of 822
Application +4 | -1
If six students, including two particular students AA and BB, stand in a row, then the probability that AA and BB are separated with one student in between them is
View answer and explanation
Correct option: D
4/154 / 15
Option D (4/154 / 15) is correct because it represents the probability that two particular students A and B are separated by one student when six students stand in a row.
36 Easy
36 of 822
Analyzing +4 | -1
A dice is thrown (2n+1)(2 n+1) times. The probability that faces with even numbers show odd number of times is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because in a binomial distribution with p=1/2p=1/2, the probability of an odd number of successes in 2n+12n+1 trials is always 1/21/2.
37 Easy
37 of 822
Analyzing +4 | -1
Let A,BA, B and CC are pairwise independent events with P(C)>0P(C)>0 and P(A∩B∩C)=0P(A \cap B \cap C)=0. Then, P((Ac∩Bc)C)P\left(\frac{\left(A^c \cap B^c\right)}{C}\right) is
View answer and explanation
Correct option: A
P(Ac)−P(B)P\left(A^c\right)-P(B)
Option A (P(Ac)−P(B)P(A^c)-P(B)) is correct because it results from applying the properties of pairwise independent events and conditional probability.
38 Easy
38 of 822
Application +4 | -1
In a college, 30%30 \% students fail in physics, 25%25 \% fail in Mathematics and 10%10 \% fail in both. One student is chosen at random. The probability that she fails in Physics, if she has failed in mathematics, is
View answer and explanation
Correct option: B
25\frac{2}{5}
Option B (25\frac{2}{5}) is correct because it is the conditional probability of failing in Physics given that the student failed in Mathematics.
39 Easy
39 of 822
Application +4 | -1
The probability that AA can solve a problem is 2/3 and BB can solve it is 3/4. If both attempt the problem, what is the probability that the problem gets solved?
View answer and explanation
Correct option: A
11-Dec
Option A (11/12) is correct because it calculates the probability of the problem being solved by either A or B or both.
40 Easy
40 of 822
Application +4 | -1
If P(A∩B)=1/2,P(A′∩B′)=1/3,P(A)=pP(A \cap B)=1 / 2, P\left(A^{\prime} \cap B^{\prime}\right)=1 / 3, P(A)=p and P(B)=2pP(B)=2 p then value of pp is
View answer and explanation
Correct option: B
718\frac{7}{18}
Option B (718\frac{7}{18}) is correct because it is the value of p that satisfies the given probability conditions using the properties of probability and set theory.
41 Easy
41 of 822
Application +4 | -1
In a test an examinee either guesses, copies or knows the answer to multiple choice question with five choices. The probability that he makes a guess is 14\frac{1}{4} and probability that he copies the answer is 18\frac{1}{8}. The probability that his answer is correct given that he copies is 110\frac{1}{10}. Find the probability that he knows the answer to the question given that he correctly answered it:
View answer and explanation
Correct option: B
1011\frac{10}{11}
Option B (1011\frac{10}{11}) is correct because it is the probability that the examinee knows the answer given that he answered correctly, calculated using Bayes' theorem.
42 Easy
42 of 822
Analyzing +4 | -1
In tossing of a coin (m+n)(m>n)\left( {m + n} \right)(m > n) times, the probability of coming consecutive heads at least mm times is
View answer and explanation
Correct option: A
n+22m+1\frac{{n + 2}}{{{2^{m + 1}}}}
Option A, n+22m+1\frac{{n + 2}}{{{2^{m + 1}}}}, is correct because it calculates the probability of getting at least m consecutive heads in (m+n) coin tosses.
43 Easy
43 of 822
Application +4 | -1
Five numbers, x1,x2,x3,x4,x5{x_1},{x_2},{x_3},{x_4},{x_5} are randomly selected from the numbers 1, 2, 3,….., 18 and are arranged in the increasing order (x1<x2<x3<x4<x5).({x_1} < {x_2} < {x_3} < {x_4} < {x_5}). The probability that x2=7{x_2} = 7 and x4=11{x_4} = 11 is:
View answer and explanation
Correct option: C
168\frac{1}{{68}}
Option C (1/68) is correct because it is calculated by dividing the number of favorable outcomes by the total number of possible outcomes when selecting and arranging the numbers.
44 Easy
44 of 822
Application +4 | -1
If XX and YY are independent binomial variates B(5,1/2)B(5,1 / 2) and B(7,1/2)B(7,1 / 2) then P(X+Y)=3P(X+Y)=3, is
View answer and explanation
Correct option: B
55/102455 / 1024
The correct answer is 55/102455 / 1024 because it calculates the probability of X+Y=3, where X and Y are independent binomial variates.
45 Easy
45 of 822
Application +4 | -1
A fair die is tossed eight times. The probability that a third six is observed on the eight throw, is
View answer and explanation
Correct option: B
  7C2×5568\frac{{{\;^7}{C_2} \times {5^5}}}{{{6^8}}}
Option B (  7C2×5568\frac{{{\;^7}{C_2} \times {5^5}}}{{{6^8}}}) is correct because it represents the probability of getting exactly two sixes in the first seven throws, and then a six on the eighth throw.
46 Easy
46 of 822
Application +4 | -1
Two numbers are chosen, one by one (without replacement) from the set of numbers A={1,2,3,4,5,6}{A}=\{1,2,3,4,5,6\}. Then the probability that minimum value of two numbers chosen is less than 4 , is:
View answer and explanation
Correct option: D
4/54 / 5
Option D (45\frac{4}{5}) is correct because it represents the probability that the minimum of two numbers chosen without replacement from the set {1, 2, 3, 4, 5, 6} is less than 4.
47 Easy
47 of 822
Application +4 | -1
If eight coins are tossed together, then the probability of getting exactly 3 heads is
View answer and explanation
Correct option: B
732\frac{7}{32}
Option B (732\frac{7}{32}) is correct because it is the probability of getting exactly 3 heads when tossing eight coins, calculated using the binomial probability formula.
48 Easy
48 of 822
Application +4 | -1
Three squares of a chess board are chosen at random, the probability that two are of one colour and one of another
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Correct option: A
16/2116/21
Option A (16/21) is correct because it represents the probability of choosing two squares of one color and one of another color from a chessboard.
49 Easy
49 of 822
Application +4 | -1
If the mean and S.D. of a binomial distribution are 20 and 4 respectively, then the number of trials is
View answer and explanation
Correct option: B
25
Option B (25) is correct because the number of trials is calculated using the mean and standard deviation of a binomial distribution.
50 Easy
50 of 822
Application +4 | -1
If P(A)=45P(A)=\frac{4}{5} and P(A∩B)=710P(A \cap B)=\frac{7}{10}, then P(B/A)P(B / A) is equal to
View answer and explanation
Correct option: C
78\frac{7}{8}
Option C (78\frac{7}{8}) is correct because it is the conditional probability of B given A, calculated using the formula P(B/A)=P(A∩B)P(A)P(B/A) = \frac{P(A \cap B)}{P(A)}.
51 Easy
51 of 822
Application +4 | -1
Two numbers are selected at random from set of the first 100 natural numbers. The probability that the product obtained is divisible by 3 is k\mathrm{k} then 150k83\frac{150 \mathrm{k}}{83} is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because the probability that the product is divisible by 3 is 83/150, and (150k)/83 equals 1.
52 Easy
52 of 822
Understanding +4 | -1
If EE and FF are two events such that 0<P(F)<10 < P(F) < 1, then
View answer and explanation
Correct option: A
P(E∣Fˉ)+P(Eˉ∣Fˉ)=1P(E \mid \bar{F})+P(\bar{E} \mid \bar{F})=1
Option A (P(E∣Fˉ)+P(Eˉ∣Fˉ)=1P(E \mid \bar{{F}})+P(\bar{{E}} \mid \bar{{F}})=1) is correct because the conditional probabilities of an event and its complement given the same condition always sum to 1.
53 Easy
53 of 822
Application +4 | -1
If the mean and variance of a binomial distribution are 4 and 3 respectively, then P(X=1)P(X=1) is
View answer and explanation
Correct option: B
4(34)154\left(\frac{3}{4}\right)^{15}
Option B (4(34)154(\frac{{3}}{{4}})^{{15}}) is correct because it is calculated by first finding n, p, and q from the given mean and variance, and then using the binomial probability formula.
54 Easy
54 of 822
Application +4 | -1
A coffee connoisseur claims that he can distinguish between a cup of instant coffee and a cup of percolator coffee 75%75 \% of time. It is agreed that his claim will be accepted if he correctly identifies at least 5 of the 6 cups. His chances of having the claim accepted is
View answer and explanation
Correct option: A
0.534
Option A (0.534) is correct because it is the sum of probabilities of correctly identifying 5 or 6 cups, calculated using the binomial distribution.
55 Easy
55 of 822
Application +4 | -1
If A={2,3,4,8,10},B={3,4,5,10,12},C={4,5,6,12,14}A=\{2,3,4,8,10\}, B=\{3,4,5,10,12\}, C=\{4,5, 6,12,14\} then find (A∩B)∪(A∩C)(A \cap B) \cup(A \cap C).
View answer and explanation
Correct option: A
{3,4,10}\{3,4,10\}
The correct answer is A ({3,4,10}) because (A∩B)∪(A∩C)(A \cap B) \cup (A \cap C) involves finding the intersection of A with B and C separately, and then taking the union of the results.
56 Easy
56 of 822
Application +4 | -1
The numbers 1,2,3,…,n1,2,3, \ldots, n are arranged in a random order. Then, the probability that the digits 1,2,3,…,k(k<n)1,2,3, \ldots ,k(k < n) appears as neighbours, is
View answer and explanation
Correct option: B
(n−k+1)nCk\frac{(n-k+1)}{{ }^n C_k}
Option B is correct because the probability is calculated by dividing the number of favorable arrangements by the total number of possible arrangements, resulting in (n-k+1)/{}^n C_k.
57 Easy
57 of 822
Application +4 | -1
A box contains 100 bulbs out of which 10 are defective. A sample of 5 bulbs is drawn. The probability that none is defective, is
View answer and explanation
Correct option: C
(910)5{\left( {\frac{9}{{10}}} \right)^5}
Option C ((910)5{\left( {\frac{9}{{10}}} \right)^5}) is correct because it is the probability of selecting 5 non-defective bulbs out of 100, where 10 are defective.
58 Easy
58 of 822
Application +4 | -1
Let EE and FF be two independent events. The probability that exactly one of them occurs is 1125\frac{11}{25} and the probability of none of them occurring is 225\frac{2}{25}. If P(T)P(T) denotes the probability of occurrence of the event TT, then
View answer and explanation
Correct option: A
P(E)=45,P(F)=35P(E)=\frac{4}{5}, P(F)=\frac{3}{5}
Option A (P(E)=4/5, P(F)=3/5) is correct because it satisfies the given conditions for independent events and the probabilities provided.
59 Easy
59 of 822
Analyzing +4 | -1
Three different numbers are selected at random from the set A={1,2,…,10}A=\{1,2, \ldots, 10\}. The probability that the product of two of the numbers is equal to third is
View answer and explanation
Correct option: B
140\frac{1}{40}
Option B (140\frac{{1}}{{40}}) is correct because it represents the probability that the product of two of the three selected numbers from the set A equals the third number.
60 Easy
60 of 822
Application +4 | -1
The least number of times a fair coin must be tossed so that the probability of getting at least one head is at least 0.95
View answer and explanation
Correct option: A
5
Option A (5) is correct because it's the least number of tosses needed for the probability of at least one head to be at least 0.95.
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