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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 541-570 of 1130 questions

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541 Easy
541 of 1130
Application +4 | -1
A positive integer NN is selected so as to be 100<N<200100 < N < 200. Then, the probability that it is divisible by 4 or 7 , is
View answer and explanation
Correct option: D
3499\frac{34}{99}
Option D (3499\frac{34}{99}) is correct because it calculates the probability of a number being divisible by 4 or 7 within the given range using the principle of inclusion-exclusion.
542 Easy
542 of 1130
Application +4 | -1
If P(A)=1/4,P(B)=1/2,P(A∪B)=5/8P(A)=1 / 4, P(B)=1 / 2, P(A \cup B)=5 / 8 then P(A∩B)P(A \cap B) is
View answer and explanation
Correct option: B
1/81 / 8
Option B (18\frac{1}{8}) is correct because it correctly applies the formula for the probability of the intersection of two events given their individual probabilities and the probability of their union.
543 Easy
543 of 1130
Analyzing +4 | -1
A binary operation is chosen at random from the set of all binary operations on a set AA containing nn elements. The probability that the binary operation is commutative, is
View answer and explanation
Correct option: C
nn/2nn2/2\frac{{{n^{n/2}}}}{{{n^{{n^2}/2}}}}
Option C (nn/2nn2/2\frac{{{n^{n/2}}}}{{{n^{{n^2}/2}}}}) is correct because it represents the probability of a binary operation being commutative, calculated by dividing the number of commutative binary operations by the total number of binary operations.
544 Medium
544 of 1130
Application +4 | -1
If AA and BB are events such that P(Aˉ∪Bˉ)=34,P(Aˉ∩Bˉ)=14P(\bar{A} \cup \bar{B})=\frac{3}{4}, \quad P(\bar{A} \cap \bar{B})=\frac{1}{4} and P(Aˉ)=23P(\bar{A})=\frac{2}{3}, then P(Aˉ∩B)P(\bar{A} \cap B) is
View answer and explanation
Correct option: A
512\frac{5}{12}
Option A (5/12) is correct because using De Morgan's laws and probability rules, we can find P(A ∪ B) and then calculate P(¬A ∩ B).
545 Medium
545 of 1130
Understanding +4 | -1
A,B{A}, {B} are two events in a random experiment such that 0<p(A)<10 < p(A) < 1 and P(B)≠1P(B) \neq 1
Assertion (A): P(ABc)=P(A)−P(A∩B)1−P(B)P\left(\frac{A}{B^c}\right)=\frac{P(A)-P(A \cap B)}{1-P(B)}
Reason (R): P(AB)=P(A∩B)P(B)P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}
View answer and explanation
Correct option: A
Both A and R are true and R is the correct explanation of A
Option A (Both A and R are true and R is the correct explanation of A) is correct because both the assertion and the reason are true, and the reason correctly explains the assertion.
546 Easy
546 of 1130
Application +4 | -1
Given a throw of three dice shows different face, the probability that at least one face shows 6 is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (1/2) is correct because it calculates the conditional probability of at least one face showing 6, given that three dice show different faces.
547 Easy
547 of 1130
Application +4 | -1
The mean and variance of a binomial distribution are 4 and 3 respectively. Then the probability of getting exactly six successes in this distribution, is
View answer and explanation
Correct option: A
16C6(14)6(34)10{ }^{16} C_6\left(\frac{1}{4}\right)^6\left(\frac{3}{4}\right)^{10}
The correct answer is 16C6(14)6(34)10{ }^{16} C_6\left(\frac{1}{4}\right)^6\left(\frac{3}{4}\right)^{10} because it calculates the probability of getting exactly six successes in the binomial distribution.
548 Easy
548 of 1130
Application +4 | -1
A seven digit number is formed using digits 3, 3, 4, 4, 4, 5, 5. The probability, that number so formed is divisible by 2, is
View answer and explanation
Correct option: C
37\frac{3}{7}
Option C (37\frac{3}{7}) is correct because it accurately calculates the probability of forming a seven-digit number divisible by 2 using the given digits.
549 Easy
549 of 1130
Application +4 | -1
If P(E)=0.40,P(F)=0.35,P(E∪F)=0.55P(E)=0.40, P(F)=0.35, P(E \cup F)=0.55, then P(F/E)P(F / E) is
View answer and explanation
Correct option: A
0.5
Option A (0.5) is correct because it is calculated by using the formula for conditional probability P(F/E)=P(E∩F)P(E)P(F/E) = \frac{{P(E \cap F)}}{{P(E)}} after finding P(E∩F)P(E \cap F).
550 Easy
550 of 1130
Application +4 | -1
If P(B)=34,P(A∩B∩Cˉ)=13P(B)=\frac{3}{4}, P(A \cap B \cap \bar{C})=\frac{1}{3} and P(Aˉ∩B∩Cˉ)=13P(\bar{A} \cap B \cap \bar{C})=\frac{1}{3}, then P(B∩C)P(B \cap C) is
View answer and explanation
Correct option: A
112\frac{1}{12}
Option A (112\frac{{1}}{{12}}) is correct because it is derived by using the given probabilities and the properties of conditional probability to find P(B∩C)P(B \cap C).
551 Easy
551 of 1130
Analyzing +4 | -1
A coin is tossed nn times. The probability that head will turn up an 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 the probability of getting an odd number of heads in n tosses is always 1/2.
552 Medium
552 of 1130
Analyzing +4 | -1
Let XX and YY are two events such that P(X∪Y)=P(X∩Y)P(X \cup Y)=P(X \cap Y).
Statement 1: P(X∩Y′)=P(X′∩Y)=0P\left(X \cap Y^{\prime}\right)=P\left(X^{\prime} \cap Y\right)=0
Statement 2: P(X)+P(Y)=2P(X∩Y)P(X)+P(Y)=2 P(X \cap Y)
View answer and explanation
Correct option: B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1 .
Option B (Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1) is correct because both statements are true, but Statement 2 doesn't directly explain Statement 1.
553 Easy
553 of 1130
Analyzing +4 | -1
A\mathrm{A} and B\mathrm{B} are two independent events. The probability that both A\mathrm{A} and B\mathrm{B} occur is 1/61 / 6 and the probability that neither of them occurs is 1/31 / 3. The probability of occurrence of A is:
View answer and explanation
Correct option: D
None of these
The correct answer is None of these because the probability of A is either 1/2 or 1/3, neither of which is provided as an option.
554 Easy
554 of 1130
Understanding +4 | -1
If MM and NN are any two events. The probability, that exactly one of them occurs, is
View answer and explanation
Correct option: D
P(M)+P(N)−2P(M∩N)P\left( M \right) + P\left( N \right) - 2P\left( {M \cap N} \right)
Option D (P(M)+P(N)−2P(M∩N)P\left( M \right) + P\left( N \right) - 2P\left( {M \cap N} \right)) is correct because it accounts for the probability of M occurring without N and N occurring without M.
555 Easy
555 of 1130
Application +4 | -1
Four digit numbers are formed using each of the digits 1,2,…,81,2, \ldots, 8 only once. One number from them is picked up at random. The probability that the selected number contains unity is
View answer and explanation
Correct option: C
1/21 / 2
Option C (1/21 / 2) is correct because the number of four-digit numbers containing unity is half the total number of four-digit numbers that can be formed.
556 Medium
556 of 1130
Application +4 | -1
A coin is biased so that a head is twice as likely to
occur as a tail. If the coin is tossed 3 times, the
probability of getting two tails and one head is
View answer and explanation
Correct option: A
29\frac{2}{9}
Option A (2/9) is correct because it is the probability of getting two tails and one head when the coin is tossed 3 times.
557 Easy
557 of 1130
Application +4 | -1
A bag contains ww white and bb black balls. A ball is drawn at random and put back in the bag with kk additional balls of the same colour. A ball is again drawn at random from the bag. The probability that it is black is
View answer and explanation
Correct option: C
b/(w+b)b /(w+b)
Option C (b/(w+b)) is correct because the probability of drawing a black ball in the second draw is the same as the probability of drawing a black ball in the first draw, due to the ball being replaced with additional balls of the same color.
558 Medium
558 of 1130
Application +4 | -1
Three six-faced dice are thrown together. The probability that the sum of the numbers appearing on the dice is k(3≤k≤8)k(3 \leq k \leq 8), is
View answer and explanation
Correct option: C
(k−1)(k−2)432\frac{(k-1)(k-2)}{432}
Option C ((k−1)(k−2)432\frac{{(k-1)(k-2)}}{{432}}) is correct because it represents the probability calculated by finding the coefficient of xkx^k in the expansion of (x+x2+...+x6)3(x+x^2+...+x^6)^3 and dividing by the total possible outcomes.
559 Easy
559 of 1130
Understanding +4 | -1
If CC and DD are two events, such that C⊂DC \subset D and P(D)≠0P(D) \neq 0, then the correct statement among the following, is
View answer and explanation
Correct option: A
P(CD)≥P(C)P\left(\frac{C}{D}\right) \geq P(C)
Option A (P(CD)≥P(C)P(\frac{{C}}{{D}}) \geq P(C)) is correct because since C⊂DC \subset D, P(CD)=P(C)P(D)P(\frac{{C}}{{D}}) = \frac{{P(C)}}{{P(D)}} and P(D)≤1P(D) \le 1, thus P(CD)≥P(C)P(\frac{{C}}{{D}}) \geq P(C).
560 Easy
560 of 1130
Understanding +4 | -1
A single letter is selected at random from the word ‘PROBABILITY’. The probability that it is a vowel is
View answer and explanation
Correct option: B
4/114/11
Option B (4/114/11) is correct because there are 4 vowels out of 11 letters in the word 'PROBABILITY'.
561 Easy
561 of 1130
Application +4 | -1
Suppose A,B,CA, B, C are three independent events such that P(A)=P(B)=P(C)=pP(A)=P(B)=P(C)=p, then PP (at least two of A,B,CA, B, C) equals
View answer and explanation
Correct option: B
3p2−2p33 p^2-2 p^3
Option B (3p2−2p33 p^2-2 p^3) is correct because it represents the probability of at least two of the independent events A, B, C occurring, given that P(A)=P(B)=P(C)=p.
562 Easy
562 of 1130
Application +4 | -1
The probability that an electric bulb will last 150 days
or more is 0.7 and that it will last at most 160 days
is 0.8. The probability that the bulb will last 150 to
160 days is
View answer and explanation
Correct option: A
0.5
Option A (0.5) is correct because it represents the probability that the bulb will last between 150 and 160 days, given the probabilities of lasting at least 150 days and at most 160 days.
563 Easy
563 of 1130
Application +4 | -1
One ticket is selected at random from 50 tickets numbered 00,01,02,...,49. Then the probability that the sum of the digits on the selected ticket is 8 , given that the product of these digits is zero, equals
View answer and explanation
Correct option: A
114\frac{1}{14}
Option A (114\frac{1}{14}) is correct because it correctly calculates the conditional probability P(A/B) where A is the sum of digits being 8 and B is the product of digits being zero.
564 Easy
564 of 1130
Application +4 | -1
Two balls are drawn from an urn containing 2 white, 3 red and 4 black balls one by one without replacement. The probability that at least one ball is red, is
View answer and explanation
Correct option: A
712\frac{7}{12}
Option A (712\frac{{7}}{{12}}) is correct because it calculates the probability of at least one red ball being drawn by subtracting the probability of no red balls being drawn from 1.
565 Hard
565 of 1130
Application +4 | -1
A cricket club has 15 members, of whom only 5 can bowl. If the names of 15 members are put into a box and 11 are drawn at random, then the probability of obtaining an eleven containing at least 3 bowlers is
View answer and explanation
Correct option: D
Dec-13
Option D (12/13) is correct because it calculates the probability of obtaining an eleven containing at least 3 bowlers by considering all possible combinations of bowlers and other players.
566 Easy
566 of 1130
Application +4 | -1
If two natural number x,y\mathrm{x}, \mathrm{y} are selected at randomly and probability that x2+y2x^2+y^2 is multiply of 5 is pp, then 25p25 \mathrm{p} is
View answer and explanation
Correct option: A
9
Option A (9) is correct because the probability that x2+y2x^2 + y^2 is a multiple of 5 can be calculated by considering the last digits of x and y.
567 Medium
567 of 1130
Application +4 | -1
A determinant of the second order is made with the elements 0 and 1. If mn\frac{{m}}{{n}} be the probability that the determinant made is non-negative, where m{m} and n{n} are relative primes, then the value of n−m{n}-{m} is
View answer and explanation
Correct option: A
3
Option A (3) is correct because it represents the value of n-m, where m/n is the probability that the determinant is non-negative.
568 Easy
568 of 1130
Application +4 | -1
A, B, C try to hit a target simultaneously but independently. Their respective probabilities of hitting the targets are 34,12,58\frac{3}{4}, \frac{1}{2}, \frac{5}{8}. The probability that the target is hit by A\mathrm{A} or B but not by C\mathrm{C} is :
View answer and explanation
Correct option: A
21/6421/64
Option A (21/64) is correct because it calculates the probability of A or B hitting the target, but C not hitting the target, using independent probabilities.
569 Hard
569 of 1130
Analyzing +4 | -1
Three dice are thrown. The probability of getting a sum which is a perfect square, is
View answer and explanation
Correct option: D
none of these.
Option D (none of these) is correct because the calculated probability of getting a sum that is a perfect square is 17108\frac{17}{108}, which is not among the given options.
570 Medium
570 of 1130
Application +4 | -1
Out of 400 people, 160 are smokers and non-vegetarian, 100 are vegetarian smokers, 140 are non-smokers and vegetarian. If a person is smoker and non-vegetarian, then the chances of his death is 35%35 \%, if he is vegetarian and smoker then chances of his death is 15%15 \%. If he is non-smoker and vegetarian then chances of his death is 10%10 \%. If a persons is found to be dead, then the probability that he is smoker and non-vegetarian, is
View answer and explanation
Correct option: A
5685\frac{56}{85}
Option A (5685\frac{56}{85}) is correct because it's the probability calculated using Bayes' theorem, considering the probabilities of death given each lifestyle choice.
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