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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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Showing 991-1020 of 1130 questions

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991 Easy
991 of 1130
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.
992 Medium
992 of 1130
Analyzing +4 | -1
There are 7 seats in a row. Three persons take seats at random. The probability that the middle seat is always occupied and no two persons are consecutive is
View answer and explanation
Correct option: C
4/354/35
Option C (4/35) is correct because it calculates the probability of the middle seat being occupied and no two persons sitting consecutively by considering favorable arrangements and total possible arrangements.
993 Easy
993 of 1130
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.
994 Easy
994 of 1130
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.
995 Easy
995 of 1130
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.
996 Easy
996 of 1130
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.
997 Easy
997 of 1130
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.
998 Easy
998 of 1130
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.
999 Easy
999 of 1130
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'.
1000 Easy
1000 of 1130
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.
1001 Hard
1001 of 1130
Application +4 | -1
Three dice are thrown. The probability that the sum of the number appearing is 15, is
View answer and explanation
Correct option: C
5/1085/108
Option C (5/108) is correct because it represents the probability of the sum of three dice being 15, calculated using combinatorial methods.
1002 Medium
1002 of 1130
Application +4 | -1
The number of ways in which a pack of 52 cards of four different suits can be distributed equally among four players so that each player gets the Ace, King, Queen and Knave of the same suit is
View answer and explanation
Correct option: A
36!4!(9!)4\frac{36 ! 4 !}{(9 !)^4}
Option A (36!4!(9!)4\frac{36 ! 4 !}{(9 !)^4}) is correct because it represents the number of ways to distribute the cards according to the given conditions.
1003 Easy
1003 of 1130
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.
1004 Easy
1004 of 1130
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.
1005 Easy
1005 of 1130
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.
1006 Easy
1006 of 1130
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.
1007 Easy
1007 of 1130
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.
1008 Easy
1008 of 1130
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.
1009 Medium
1009 of 1130
Application +4 | -1
Two cards are drawn from a well shuffled deck of 52 cards. The probability that one is red card and the other is a queen is
View answer and explanation
Correct option: C
50/66350/663
Option C (50663\frac{50}{663}) is correct because it calculates the probability of drawing one red card and one queen by considering the two mutually exclusive cases.
1010 Easy
1010 of 1130
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.
1011 Easy
1011 of 1130
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.
1012 Medium
1012 of 1130
Application +4 | -1
If the papers of 4 students can be checked by any one of the 7 teachers, then the probability that all the 4 papers are checked by exactly 2 teachers is
View answer and explanation
Correct option: B
6/496 / 49
Option B (649\frac{{6}}{{49}}) is correct because it is the probability calculated by finding the number of ways the 4 papers can be checked by exactly 2 teachers, divided by the total number of ways the 4 papers can be checked by any of the 7 teachers.
1013 Easy
1013 of 1130
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.
1014 Medium
1014 of 1130
Analyzing +4 | -1
In the above question the probability that the binary operation is non-commutative is
View answer and explanation
Correct option: B
nn2−nn(n+1)2nn2\frac{{{n^{{n^2}}} - {n^{\frac{{n\left( {n + 1} \right)}}{2}}}}}{{{n^{{n^2}}}}}
Option D (nn2−nn(n+1)2nn2\frac{{{n^{{n^2}}} - {n^{\frac{{n\left( {n + 1} \right)}}{2}}}}}{{{n^{{n^2}}}}}) is correct because it represents the probability of a binary operation being non-commutative.
1015 Medium
1015 of 1130
Application +4 | -1
If EE and FF are events such that P(E)=14,P(F)=12{P}({E})=\frac{1}{4}, {P}({F})=\frac{1}{2} and P(EP(E and F)=18F)=\frac{1}{8}, then,
I. P(E or F)=58{P}({E}\text{ or }{F})=\frac{5}{8}
II. P(not E and not F)=38{P}(\text{not }{E} \text{ and not } {F})=\frac{3}{8}
View answer and explanation
Correct option: C
Both I and II are true
Option C (Both I and II are true) is correct because both the probability of E or F and the probability of not E and not F are calculated correctly using the given probabilities.
1016 Easy
1016 of 1130
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.
1017 Easy
1017 of 1130
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).
1018 Easy
1018 of 1130
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.
1019 Easy
1019 of 1130
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.
1020 Easy
1020 of 1130
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.
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