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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 1051-1080 of 1130 questions
1051
Easy
1051 of 1130
Application
+4 | -1
For a binomial variate with , if , then its variance is
View answer and explanation
Correct option: C
Option C (9/8) is correct because the variance is calculated using the binomial distribution formula npq after finding p and q.
1052
Easy
1052 of 1130
Application
+4 | -1
If has a binomial distribution, with parameters and such that then the mean of variable is
View answer and explanation
Correct option: B
Option B () is correct because, given , we can derive the relationship for a binomial distribution, and .
1053
Medium
1053 of 1130
Application
+4 | -1
Fifteen coupons are numbered 1 to 15. Seven coupons are selected at random, one at a time with replacement. The probability that the largest number appearing on a selected coupon be 9, is
View answer and explanation
Correct option: C
Option D is incorrect. Option C is correct because it represents the probability of selecting a number less than or equal to 9 in each of the 7 trials with replacement.
1054
Easy
1054 of 1130
Application
+4 | -1
If and are two events associated with a random experiment for which and , then is
View answer and explanation
Correct option: B
0.67
Option B (0.67) is correct because it is the probability of event F, calculated using the formula for the probability of the union of two events.
1055
Easy
1055 of 1130
Application
+4 | -1
To open a lock, a key is taken out from a collection of keys at random. If the lock is not opened with this key, it is put back into the collection and another key is tried. The process is repeated again and again. If it is given that with only one key in the collection, the lock can be opened, then the probability that the lock will open in trials, is
View answer and explanation
Correct option: C
Option C () is correct because it calculates the probability of opening the lock within n trials by subtracting the probability of not opening the lock in any of the n trials from 1.
1056
Easy
1056 of 1130
Application
+4 | -1
The probability that in the toss of two dice, we obtain the sum 7 or 11, is
View answer and explanation
Correct option: C
Option C () is correct because it is the probability of obtaining a sum of 7 or 11 when tossing two dice, calculated by dividing the number of favorable outcomes by the total number of possible outcomes.
1057
Hard
1057 of 1130
Application
+4 | -1
A four digit number (numbered from 0000 to 9999) is said to be lucky if the sum of its first two digits is equal to the sum of its last two digits. If a four digit number is picked up at random, then the probability that it is lucky is
View answer and explanation
Correct option: D
0.067
Option D (0.067) is correct because it represents the probability of a four-digit number being lucky, calculated by dividing the number of favorable outcomes by the total number of possible outcomes.
1058
Easy
1058 of 1130
Application
+4 | -1
The probability that randomly selected calculator from a store is of brand is proportional to , . Further, the probability of a calculator of brand being defective is . Then the probability that a calculator randomly selected from the store being defective is
View answer and explanation
Correct option: A
Option A (8/63) is correct because it calculates the probability of a randomly selected calculator being defective, considering the probability of each brand and its defect rate.
1059
Medium
1059 of 1130
Application
+4 | -1
A group of 6 boys and 6 girls is randomly divided into two equal groups. The probability that each group contains 3 boys and 3 girls is
View answer and explanation
Correct option: D
Option D (100 / 231) is correct because it is calculated by dividing the number of ways to form a group with 3 boys and 3 girls by the total number of ways to form a group of 6 from 12 people.
1060
Easy
1060 of 1130
Understanding
+4 | -1
There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is
View answer and explanation
Correct option: B
Option B (1/6) is correct because there is only one way to choose the two faulty machines in the first two tests out of the six possible combinations.
1061
Easy
1061 of 1130
Application
+4 | -1
A random variable has the following probability distribution
Then, the mean of is
Then, the mean of is
View answer and explanation
Correct option: A
3
Option A (3) is correct because it is the mean of the random variable X calculated using the given probability distribution.
1062
Easy
1062 of 1130
Application
+4 | -1
From a box containing 20 tickets marked with numbers 1 to 20, four tickets are drawn one by one. After each draw, the ticket is replaced. The probability that the largest value of tickets drawn is 15 is
View answer and explanation
Correct option: B
The correct answer is because it represents the probability that the largest value of four tickets drawn with replacement is 15.
1063
Easy
1063 of 1130
Application
+4 | -1
If and , then equals to
View answer and explanation
Correct option: A
Option A () is correct because it is the conditional probability of A given B, calculated using the formula .
1064
Easy
1064 of 1130
Application
+4 | -1
In a maths paper there are 3 sections A, B and C. Section A is compulsory. Out of sections and a student has to attempt any one. Passing in the paper means passing in and passing in or . The probability of the student passing in and are and respectively. If the probability that the student is successful is then
View answer and explanation
Correct option: D
Option D (p=q=1/2) is correct because it satisfies the probability condition for the student being successful.
1065
Easy
1065 of 1130
Application
+4 | -1
Probability that in the toss of two dice we obtain an even sum or a sum less than 5, is
View answer and explanation
Correct option: D
Option D () is correct because it's calculated by finding the probability of an even sum, the probability of a sum less than 5, and subtracting the probability of their intersection.
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1066
Easy
1066 of 1130
Application
+4 | -1
Assuming that for a husband-wife couple the chances of their child being a boy or a girl are the same, the probability of their two children being a boy and a girl is
View answer and explanation
Correct option: C
Option C () is correct because there are two equally likely possibilities: boy then girl, or girl then boy, out of the four possible outcomes (BB, GG, BG, GB).
1067
Easy
1067 of 1130
Application
+4 | -1
For the box contains red balls and white balls,
Let and
A box is selected at random and a ball is drawn from it. If a red ball is drawn, then the probability that it has come from box , is
Let and
A box is selected at random and a ball is drawn from it. If a red ball is drawn, then the probability that it has come from box , is
View answer and explanation
Correct option: B
Option B (14/39) is correct because it is calculated using Bayes' theorem to find the probability that the red ball came from box B2 given that a red ball was drawn.
1068
Medium
1068 of 1130
Application
+4 | -1
is a set containing n elements. A subset of is chosen at random. The set is reconstructed by replacing the elements of . A subset is again chosen at random. The probability that and are disjoint sets, is
View answer and explanation
Correct option: D
Option D () is correct because it calculates the probability of P and Q being disjoint by considering the choices for each element in A.
1069
Easy
1069 of 1130
Application
+4 | -1
A natural number is chosen at random from the first 100 natural numbers. The probability that
View answer and explanation
Correct option: C
Option C () is the correct probability, calculated by finding the range of x values that satisfy the inequality and dividing by the total number of natural numbers.
1070
Easy
1070 of 1130
Application
+4 | -1
If is a poisson variate such that , then is equal to
View answer and explanation
Correct option: C
Option C (2/3e²) is correct because by equating the probabilities P(X=1) and P(X=2), we find λ=2, and then calculate P(X=4) using the Poisson distribution formula.
1071
Easy
1071 of 1130
Application
+4 | -1
Four tickets are marked 00,01, 10 and 11 respectively, are placed in a bag. A ticket is drawn at random five times, being replaced each time. The probability that the sum of the numbers on the five tickets drawn is 24, is
View answer and explanation
Correct option: B
Option B () is correct because it is the probability calculated by dividing the number of favorable outcomes (sum of 24) by the total possible outcomes when drawing the tickets.
1072
Easy
1072 of 1130
Application
+4 | -1
There are 12 white and 12 red balls in a bag. Balls are drawn one by one with replacement from the bag. The probability that 7 th drawn ball is 4 th white, is
View answer and explanation
Correct option: B
The correct answer is because it calculates the probability that the 7th drawn ball is the 4th white ball.
1073
Easy
1073 of 1130
Analyzing
+4 | -1
If and , then is equal to
View answer and explanation
Correct option: D
1
Option D (1) is correct because the expression simplifies to 1 using the given probabilities and conditional probability formula.
1074
Easy
1074 of 1130
Analyzing
+4 | -1
Two numbers and are chosen at random from the set . The probability that is divisible by 5, is
View answer and explanation
Correct option: C
Option C () is correct because it represents the probability calculated by dividing the number of ways to choose two numbers such that is divisible by 5 by the total number of ways to choose two numbers.
1075
Easy
1075 of 1130
Application
+4 | -1
In tossing 10 coins, the probability of getting exactly 5 heads is
View answer and explanation
Correct option: B
Option B () is correct because it is calculated using the binomial probability formula for exactly 5 heads out of 10 coin tosses.
1076
Easy
1076 of 1130
Understanding
+4 | -1
Suppose and are two events such that and , then equals
View answer and explanation
Correct option: A
Option A () is correct because it correctly applies the definition of conditional probability and De Morgan's law.
1077
Easy
1077 of 1130
Application
+4 | -1
Let be a non-zero real number. A determinant is chosen from the set of all determinants of order 2 with entries or only. The probability that the value of the determinant is non-zero is
View answer and explanation
Correct option: C
Option C () is correct because it represents the probability that the value of the determinant is non-zero, given the entries are either x or -x.
1078
Easy
1078 of 1130
Application
+4 | -1
A bag contains white and black balls. Two players and alternately draw a ball from the bag replacing the ball each time after the draw till one of them draws a white ball and wins the game. begins the game. If the probability of winning the game is three times that of , then the ratio is
View answer and explanation
Correct option: C
Option C () is correct because the probability of A winning is three times that of B, which leads to the ratio a:b = 2:1 after solving the probabilities of A and B winning.
1079
Medium
1079 of 1130
Application
+4 | -1
The probabilities that a student will obtain grades are 0.30, 0.35, 0.20 and 0.15 respectively. The probability that he will receive atleast grade, is
View answer and explanation
Correct option: D
0.2
Option B (0.85) is correct because it is the sum of the probabilities of receiving a grade of A, B, or C.
1080
Easy
1080 of 1130
Understanding
+4 | -1
Which one is not a requirement of a binomial distribution?
View answer and explanation
Correct option: C
The outcomes must be dependent on each other
Option C (The outcomes must be dependent on each other) is correct because one of the requirements of a binomial distribution is that the outcomes of each trial must be independent.
Revision notes and key takeaways
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