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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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721 Easy
721 of 822
Application +4 | -1
If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is:
View answer and explanation
Correct option: C
33228\frac{33}{2^{28}}
Option C (33228\frac{33}{2^{28}}) is correct because it is the probability calculated from the binomial distribution with n=32 and p=1/2 for one or two successes.
722 Easy
722 of 822
Application +4 | -1
A number is chosen form each of two sets A={1,2,3,4,5,6,7,8}A=\{1,2,3,4,5,6,7,8\} and B={8,7,6,5,4,3,2,1}B=\{8,7,6,5,4,3,2,1\}.
If p1p_1 is the probability that the sum of the numbers is 9 and p2p_2 is the probability that sum of the numbers is 7 , then p1+p2p_1+p_2 is equal to
View answer and explanation
Correct option: D
732\frac{7}{32}
Option D (732\frac{7}{32}) is correct because it represents the sum of the probabilities of getting a sum of 9 and a sum of 7 when choosing one number from each of the given sets.
723 Easy
723 of 822
Analyzing +4 | -1
For three events A,BA, B and C,PC, P (exactly one of the events AA or BB occurs )=P)=P (exactly one of the events BB or CC occurs) =P(=P( exactly one of the events CC or AA occurs )=p)=p and PP (all the three coents occur simultaneously) =p2= {p^2}, where 0<p<1/20 < p < 1/2. Then the probability of atleast one of the three events A,BA, B and CC occurring is
View answer and explanation
Correct option: A
3p+2p22\frac{3 p+2 p^2}{2}
Option A (3p+2p22\frac{{3 p+2 p^2}}{{2}}) is correct because it represents the probability of at least one of the three events occurring, calculated using the given probabilities and the inclusion-exclusion principle.
724 Easy
724 of 822
Application +4 | -1
What is the probability that for S's come consecutively in the woord 'MISSISSIPPI'?
View answer and explanation
Correct option: C
4165\frac{4}{165}
Option C (4165\frac{4}{165}) is correct because it calculates the probability of the four S's appearing consecutively in the word \'MISSISSIPPI\' by considering the arrangements where the S's are treated as a single unit.
725 Easy
725 of 822
Application +4 | -1
At a telephone enquiry system the number of phone calls regarding relevant enquiry follow poisson distribution with an average of 5 phone calls during 10 min time intervals. The probability that there is at the most one phone cell during a 10 min time period, is
View answer and explanation
Correct option: C
6e5\frac{6}{{{e^5}}}
Option C (6e5\frac{6}{{{e^5}}}) is correct because it represents the probability of at most one phone call, calculated using the Poisson distribution formula with a mean of 5.
726 Easy
726 of 822
Application +4 | -1
Let X be a random variable having binomial distribution B(7, p). If P(X = 3) = 5P(X = 4), then the sum of the mean and the variance of X is:
View answer and explanation
Correct option: C
7736\frac{{77}}{{36}}
Option C (77/36) is correct because it is derived by using the binomial distribution properties to find p and then calculating the mean and variance.
727 Easy
727 of 822
Application +4 | -1
A bag XX contains 2 white and 3 black bails 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, the 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 correctly calculates the probability of selecting a white ball by considering both bags and their respective probabilities.
728 Easy
728 of 822
Understanding +4 | -1
Two fair dice are rolled once. Then the mean of sum of the numbers on them is
View answer and explanation
Correct option: A
7
Option A (7) is correct because the mean of the sum of numbers on two fair dice is 7. This is calculated by considering all possible outcomes and their probabilities.
729 Easy
729 of 822
Application +4 | -1
Probability that a student will succeed in I.I.T. entrance test is 0.2 and that he will succeed in Roorkee entrance test is 0.5. If the probability that he will successful at both the places is 0.3, then the probability that he does not succeed at both the places is
View answer and explanation
Correct option: D
0.6
Option D (0.6) is correct because it calculates the probability that the student does not succeed at both places using the probabilities of success at each place and at both places.
730 Easy
730 of 822
Understanding +4 | -1
A die is rolled, let E{E} be the event "die shows 4" and F{F} be the event "die shows even number". Then

I. E{E} and F{F} are mutually exclusive.

II. E{E} and F{F} are not mutually exclusive.
View answer and explanation
Correct option: B
Only II is true
Option B (Only II is true) is correct because the events E and F are not mutually exclusive since they both include the outcome '4'.
731 Easy
731 of 822
Application +4 | -1
Out of 30 consecutive integers, 2 are chosen at random. The probability that their sum is odd, is
View answer and explanation
Correct option: C
1529\frac{{15}}{{29}}
Option C (1529\frac{{15}}{{29}}) is correct because it calculates the probability that the sum of two randomly chosen integers from 30 consecutive integers is odd.
732 Easy
732 of 822
Application +4 | -1
The probability that certain electronic conponent fails when first used in 0.10 . If it does not fail immediately, the probability that it lasts for one year is 0.99 . The probability that a new component will last for one year is
View answer and explanation
Correct option: A
0.891
Option A (0.891) is correct because it calculates the probability that a new component will last for one year by considering the probability that it doesn't fail immediately and then lasts for one year.
733 Easy
733 of 822
Application +4 | -1
The probability that a man can hit a target is 3/4.3/4. He tries 5 times. The probability that he will hit the target at least three times is
View answer and explanation
Correct option: D
459/512459/512
Option D (459512\frac{459}{512}) is correct because it calculates the probability of hitting the target at least three times using the binomial probability formula.
734 Easy
734 of 822
Application +4 | -1
A pair of unbiased dice is rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is
View answer and explanation
Correct option: A
25\frac{2}{5}
Option A (25\frac{2}{5}) is correct because it represents the probability of getting a sum of 5 before a sum of 7, calculated using infinite geometric progression.
735 Easy
735 of 822
Application +4 | -1
A die is tossed thrice. If event of getting an even number is a success, then the probability of getting at least two successes is
View answer and explanation
Correct option: D
12\frac{1}{2}
Option D (12\frac{1}{2}) is correct because it calculates the probability of getting at least two successes (even numbers) in three tosses of a die.
736 Easy
736 of 822
Application +4 | -1
A consignment of 15 record players contains 4 defectives. The record players are selected at random, one by one, and examined. The ones examined are not put back. The probability that 9th one examined is the last defective, is
View answer and explanation
Correct option: B
4C3×11C515C8×17\frac{{ }^4 C_3 \times{ }^{11} C_5}{{ }^{15} C_8} \times \frac{1}{7}
Option B is correct because it calculates the probability of the 9th record player being the last defective, considering the conditional probability.
737 Easy
737 of 822
Application +4 | -1
A room has 3 lamps. From a collection of 10 light bulbs of which 6 are not good, a person selects 3 at random and puts them in a socket, the Probability that he will have light, is
View answer and explanation
Correct option: A
56\frac{5}{6}
Option A (56\frac{5}{6}) is correct because it represents the probability that the person will have light, given that 6 out of 10 bulbs are not good.
738 Easy
738 of 822
Application +4 | -1
If three natural numbers from 1 to 100 are selected randomly, then probability that all are divisible by both 2 and 3, is
View answer and explanation
Correct option: D
41155\frac{4}{{1155}}
Option D is correct because the probability is calculated by dividing the number of ways to choose 3 numbers divisible by 6 by the total number of ways to choose 3 numbers from 1 to 100.
739 Easy
739 of 822
Application +4 | -1
The mean and the variance of a binomial distribution are 4 and 2 respectively. Then, the probability of 2 successes is
View answer and explanation
Correct option: A
28/25628 / 256
The correct answer is 28/25628 / 256 because it calculates the probability of 2 successes in a binomial distribution with mean 4 and variance 2.
740 Easy
740 of 822
Analyzing +4 | -1
If AA and BB are independent events of a random experiments such that P(A∩B)=16P\left( {A \cap B} \right) = \frac{1}{6} and P(Aˉ∩Bˉ)=13,  P\left( {\bar A \cap \bar B} \right) = \frac{1}{3},\; then P(A)P\left( A \right) is equal to
View answer and explanation
Correct option: B
13\frac{1}{3}
Option D (23\frac{2}{3}) is incorrect. The correct answer is B (13\frac{1}{3}). The solution involves using the independence of events A and B to find P(A) and P(B) from the given probabilities of their intersection and the intersection of their complements.
741 Easy
741 of 822
Analyzing +4 | -1
If EE and FF are two independent events such that 0<P(E)<10 < P\left( E \right) < 1 and 0<P(F)<1,  0 < P\left( F \right) < 1,\; then
View answer and explanation
Correct option: C
P(EF)+P(EcFc)=1P\left( {\frac{E}{F}} \right) + P\left( {\frac{{{E^c}}}{{{F^c}}}} \right) = 1
Option C, P(EF)+P(EcFc)=1P\left( {\frac{E}{F}} \right) + P\left( {\frac{{{E^c}}}{{{F^c}}}} \right) = 1, is correct because for independent events E and F, the conditional probabilities simplify and sum to 1.
742 Easy
742 of 822
Application +4 | -1
If P(A)=14,P(Bˉ)=12P(A)=\frac{1}{4}, P(\bar{B})=\frac{1}{2} and P(A∪B)=59P(A \cup B)=\frac{5}{9} then P(A/B)P(A / B) is
View answer and explanation
Correct option: C
718\frac{7}{18}
Option C (718\frac{7}{18}) is correct because it is the conditional probability of A given B, calculated using the formula P(A/B) = P(A ∩ B) / P(B).
743 Easy
743 of 822
Analyzing +4 | -1
If P(A)=1/3,P(B)=1/2P\left( A \right) = 1/3,P\left( B \right) = 1/2 and P(A∪B)=5/6,P\left( {A \cup B} \right) = 5/6, then events AA and BB are
View answer and explanation
Correct option: A
Mutually exclusive
The correct answer is A (Mutually exclusive) because the probability of the intersection of events A and B is zero, indicating they cannot occur simultaneously.
744 Easy
744 of 822
Application +4 | -1
A bag contains 6 white and 6 black socks. A man randomly takes out two socks from the bag. The probability that socks are the same colour is:
View answer and explanation
Correct option: B
511\frac{5}{11}
Option B (511\frac{5}{11}) is correct because the probability is calculated by dividing the number of ways to choose two socks of the same color by the total number of ways to choose two socks.
745 Easy
745 of 822
Application +4 | -1
Six dice are thrown simultaneously. The probability that all of them show the same face, is
View answer and explanation
Correct option: B
165\frac{1}{6^5}
Option B (165\frac{1}{6^5}) is correct because it represents the probability of all six dice showing the same face, calculated as the number of favorable outcomes divided by the total possible outcomes.
746 Easy
746 of 822
Application +4 | -1
A number xx is chosen at random from the set {1,2,3,…,100}\{1,2,3, \ldots, 100\}. Let p=p= probability that xx is divisible by 19 and q=q= probability that xx is divisible by 31, then p+qp+q is equal to
View answer and explanation
Correct option: D
0.8
Option D (0.08) is correct because it represents the sum of the probabilities that x is divisible by 19 and 31, calculated by dividing the number of multiples of each number within the set by the total number of elements in the set.
747 Easy
747 of 822
Analyzing +4 | -1
Three-digit numbers are formed using the digits 0,0 , 1,2,3,4,51,2,3,4,5 without repetition of digits. If a number is chosen at random, then the probability that the digits either increase or decrease, is
View answer and explanation
Correct option: C
310\frac{3}{10}
Option C (310\frac{3}{10}) is correct because it represents the ratio of numbers with increasing or decreasing digits to the total number of three-digit numbers formed.
748 Easy
748 of 822
Application +4 | -1
If n(A)=8n\left( A \right) = 8 and n(A∩B)=2,n\left( {A \cap B} \right) = 2, then n[(A∩B)′∩A]n\left[ {{{\left( {A \cap B} \right)}^\prime } \cap A} \right] is equal to
View answer and explanation
Correct option: A
6
Option A (6) is correct because n[(A∩B)′∩A]=n(A)−n(A∩B)=8−2=6n[ (A ∩ B)' ∩ A ] = n(A) - n(A ∩ B) = 8 - 2 = 6.
749 Easy
749 of 822
Application +4 | -1
Each of the persons AA and BB independently tosses three fair coins. The probability that both of them get the same number of heads is
View answer and explanation
Correct option: C
516\frac{5}{{16}}
Option C (5/16) is correct because it calculates the probability of both persons getting the same number of heads by considering all possible cases (0, 1, 2, or 3 heads) and summing their probabilities.
750 Easy
750 of 822
Analyzing +4 | -1
For the three events A, BA,\,B and CC, PP (exactly one of the events AA or BB occurs) = P= \,P (exactly one of the events BB or CC occurs) −P- P (exactly one of the events CC or AA occurs) = p= \,p and PP ( all the three events occur simultaneously) =p2,  = {p^2},\; where 0<p<120 < p < \frac{1}{2}. Then, the probability of at least one of the three events AA, BB and CC occuring, is
View answer and explanation
Correct option: A
3p+2p22\frac{{3p + 2{p^2}}}{2}
Option A (3p+2p22\frac{{3p + 2{p^2}}}{2}) is correct because it represents the probability of at least one of the three events occurring, derived from the given probabilities of exactly one event occurring and all three events occurring simultaneously.
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