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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 XX with n=6n = 6, if P(X=2)=9P(X=4)P\left( {X = 2} \right) = 9P\left( {X = 4} \right), then its variance is
View answer and explanation
Correct option: C
98\frac{9}{8}
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 XX has a binomial distribution, B(n,p)B(n,p)with parameters nn and pp such that P(X=2)=P(X=3),P\left( {X = 2} \right) = P\left( {X = 3} \right), then E(X),E\left( X \right), the mean of variable X,X, is
View answer and explanation
Correct option: B
3−p3 - p
Option B (3−p3 - p) is correct because, given P(X=2)=P(X=3)P(X=2) = P(X=3), we can derive the relationship np=3−pnp = 3 - p for a binomial distribution, and E(X)=npE(X) = np.
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
(35)7{\left( {\frac{3}{5}} \right)^7}
Option D is incorrect. Option C (35)7{\left( {\frac{{3}}{{5}}} \right)^7} 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 EE and FF are two events associated with a random experiment for which P(E)=0.6,P(E∪F)P(E)=0.6, P(E \cup F) =0.85=0.85 and P(E∩F)=0.42P(E \cap F)=0.42, then P(F)P(F) 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 nn 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 nn trials, is
View answer and explanation
Correct option: C
1−(n−1n)n1 - {\left( {\frac{{n - 1}}{n}} \right)^n}
Option C (1−(n−1n)n1 - {(\frac{{n - 1}}{n})^n}) 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
29\frac{2}{9}
Option C (29\frac{2}{9}) 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 rr is proportional to rr, r=1,2,…,6r=1,2, \ldots, 6. Further, the probability of a calculator of brand rr being defective is 7−r21,r=1,2,…,6\frac{7-r}{21}, r=1,2, \ldots, 6. Then the probability that a calculator randomly selected from the store being defective is
View answer and explanation
Correct option: A
863\frac{8}{63}
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
100/231100 / 231
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
1/61 / 6
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 XX has the following probability distribution
X            1       2        3        4P(X)   k      2k     3k     4k\begin{array}{l} X\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,4\\ P(X)\,\,\,k\,\,\,\,\,\,2k\,\,\,\,\,3k\,\,\,\,\,4k \end{array}
Then, the mean of XX 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
27320\frac{27}{320}
The correct answer is 27320\frac{27}{320} 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 P(A∩B)=710P(A \cap B)=\frac{7}{10} and P(B)=1720P(B)=\frac{17}{20}, then P(A/B)P(A / B) equals to
View answer and explanation
Correct option: A
1417\frac{14}{17}
Option A (1417\frac{14}{17}) is correct because it is the conditional probability of A given B, calculated using the formula P(A/B)=P(A∩B)P(B)P(A/B) = \frac{P(A \cap B)}{P(B)}.
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 BB and CC a student has to attempt any one. Passing in the paper means passing in A\mathrm{A} and passing in B\mathrm{B} or C\mathrm{C}. The probability of the student passing in A,B\mathrm{A}, \mathrm{B} and CC are p,qp, q and 1/21 / 2 respectively. If the probability that the student is successful is 1/21 / 2 then
View answer and explanation
Correct option: D
p=q=1/2\mathrm{p}=\mathrm{q}=1 / 2
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
59\frac{5}{9}
Option D (59\frac{5}{9}) 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.
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
12\frac{1}{2}
Option C (12\frac{1}{2}) 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 k=1,2,3k = 1,2,3 the box Bk{B_k} contains kk red balls and (k+1)\left( {k + 1} \right) white balls,
Let P(B1)=12, P(B2)=13P\left( {{B_1}} \right) = \frac{1}{2},\,P\left( {{B_2}} \right) = \frac{1}{3} and P(B3)=16P\left( {{B_3}} \right) = \frac{1}{6}
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 B2{B_2}, is
View answer and explanation
Correct option: B
1439\frac{{14}}{{39}}
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
AA is a set containing n elements. A subset PP of AA is chosen at random. The set AA is reconstructed by replacing the elements of PP. A subset QQ is again chosen at random. The probability that PP and QQ are disjoint sets, is
View answer and explanation
Correct option: D
(34)n\left(\frac{3}{4}\right)^n
Option D ((34)n\left(\frac{{3}}{{4}}\right)^n) 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 xx is chosen at random from the first 100 natural numbers. The probability that x+100x>50x + \frac{{100}}{x} > 50
View answer and explanation
Correct option: C
1120\frac{{11}}{{20}}
Option C (1120\frac{{11}}{{20}}) 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 XX is a poisson variate such that P(X=1)=P(X=2)P\left( {X = 1} \right) = P\left( {X = 2} \right), then P(X=4)P\left( {X = 4} \right) is equal to
View answer and explanation
Correct option: C
23e2\frac{2}{{3{e^2}}}
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
25512\frac{25}{512}
Option B (25512\frac{25}{512}) 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
5/325 / 32
The correct answer is 5/325 / 32 because it calculates the probability that the 7th drawn ball is the 4th white ball.
1073 Easy
1073 of 1130
Analyzing +4 | -1
If P(B)=35,P(A/B)=12P(B)=\frac{3}{5}, P(A / B)=\frac{1}{2} and P(A∪B)=45P(A \cup B)=\frac{4}{5}, then P(A∪B)′+P(A′∪B)P(A \cup B)^{\prime}+P\left(A^{\prime} \cup B\right) 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 aa and bb are chosen at random from the set {1,2,3,…,5n}\{1,2,3, \ldots, 5 n\}. The probability that a4−b4a^4-b^4 is divisible by 5, is
View answer and explanation
Correct option: C
17n−55(5n−1)\frac{17 n-5}{5(5 n-1)}
Option C (17n−55(5n−1)\frac{{17 n-5}}{{5(5 n-1)}}) is correct because it represents the probability calculated by dividing the number of ways to choose two numbers such that a4−b4a^4 - b^4 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
63256\frac{63}{256}
Option B (63256\frac{{63}}{{256}}) 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 AA and BB are two events such that P(A)>0P(A)>0 and P(B)≠1P(B) \neq 1, then P(A′∣B′)P\left(A^{\prime} \mid B^{\prime}\right) equals
View answer and explanation
Correct option: A
[1−P(A∪B)]/P(B′)[1-P(A \cup B)] / P\left(B^{\prime}\right)
Option A ([1−P(A∪B)]/P(B′)[1-P(A \cup B)] / P(B')) 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 xx be a non-zero real number. A determinant is chosen from the set of all determinants of order 2 with entries xx or −x-x only. The probability that the value of the determinant is non-zero is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) 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 aa white and bb black balls. Two players AA and BB 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. AA begins the game. If the probability of AA winning the game is three times that of BB, then the ratio a:ba:b is
View answer and explanation
Correct option: C
2:12:1
Option C (2:12:1) 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 A, B, C   or   DA,\,B,\,C\;\,\mathrm{or}\;\,D are 0.30, 0.35, 0.20 and 0.15 respectively. The probability that he will receive atleast CC 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.
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