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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 931-960 of 1130 questions

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931 Easy
931 of 1130
Application +4 | -1
In a binomial distribution B(n,p=1/4)B(n, p=1 / 4) If the probability of at least one success is greater than or equal to 9/109 / 10, then nn is greater than
View answer and explanation
Correct option: A
1log⁡104−log⁡103\frac{1}{\log _{10} 4-\log _{10} 3}
The correct answer is 1log⁡104−log⁡103\frac{1}{\log _{10} 4-\log _{10} 3} because it represents the lower bound for n, given the probability condition.
932 Easy
932 of 1130
Application +4 | -1
nn persons are seated on nn chairs at a round table. The probability that two specified persons are sitting next to each other, is
View answer and explanation
Correct option: B
2n−1\frac{2}{n-1}
Option B (2n−1\frac{2}{n-1}) is correct because it represents the probability of two specific people sitting next to each other at a round table.
933 Easy
933 of 1130
Application +4 | -1
Two cards are drawn successively with replacement from a well shuffled deck of 52 cards. Let XX denote the random variable of number of aces obtained in the two drawn cards. Then, P(X=1)+P(X=2)P(X = 1) + P(X = 2) equals
View answer and explanation
Correct option: A
25169\frac{{25}}{{169}}
Option A (25/169) is correct because it is the sum of probabilities of getting exactly one ace and two aces in two draws with replacement.
934 Medium
934 of 1130
Application +4 | -1
Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house, is
View answer and explanation
Correct option: C
19\frac{1}{9}
Option C (19\frac{1}{9}) is correct because it represents the probability that all three persons apply for the same house out of the three available houses.
935 Medium
935 of 1130
Application +4 | -1
Two marbles are drawn in succession from a box
containing 10 red, 30 white, 20 blue and 15 orange
marbles, with replacement being made after each
drawing. Then the probability, that first drawn
marble is red and second drawn marble is white, is
View answer and explanation
Correct option: A
475\frac{4}{75}
Option A (475\frac{4}{75}) is correct because it is the probability of drawing a red marble first and a white marble second, with replacement, calculated as the product of the individual probabilities.
936 Easy
936 of 1130
Application +4 | -1
Three integers are chosen at random without replacement from the first 20 integers. The probability that their product is even is
View answer and explanation
Correct option: C
17/1917 / 19
Option C (17 / 19) is correct because it is calculated by finding the probability that the product is odd and subtracting it from 1.
937 Medium
937 of 1130
Application +4 | -1
If from each of three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn, is
View answer and explanation
Correct option: A
1332\frac{{13}}{{32}}
Option A (1332\frac{{13}}{{32}}) is correct because it calculates the probability of drawing 2 white and 1 black ball by considering all possible combinations and their respective probabilities.
938 Easy
938 of 1130
Application +4 | -1
An experiment succeeds twice as often as it fails. The probability that in the next six trials there are at least 4 successes, is
View answer and explanation
Correct option: A
496729\frac{496}{729}
Option A (496729\frac{496}{729}) is correct because it's the probability of at least 4 successes, calculated by summing the probabilities of 4, 5, and 6 successes.
939 Easy
939 of 1130
Application +4 | -1
In a series of three trials the probability of exactly two successes in nine times as large as the probability of three successes. Then, the probability of success in each trial is
View answer and explanation
Correct option: C
01-Apr
Option D (01-Apr) is correct because the probability of success is calculated by setting up an equation based on the given relationship between the probabilities of two and three successes.
940 Easy
940 of 1130
Understanding +4 | -1
A,B,CA,B,C are any three events. If P(S)P\left( S \right) denotes the probability of SS happening, then P(A∩(B∪C))=P\left( {A \cap \left( {B \cup C} \right)} \right) =
View answer and explanation
Correct option: C
P(A∩B)+P(A∩C)−P(A∩B∩C)P\left( {A \cap B} \right) + P\left( {A \cap C} \right) - P\left( {A \cap B \cap C} \right)
Option C (P(A∩B)+P(A∩C)−P(A∩B∩C)P\left( {A \cap B} \right) + P\left( {A \cap C} \right) - P\left( {A \cap B \cap C} \right)) is correct because it correctly applies the distributive property and the inclusion-exclusion principle for probabilities.
941 Hard
941 of 1130
Application +4 | -1
A dice is weighted such that the probability of rolling n\mathrm{n} is proportional to n2(n=1,2,3,4,5,6)\mathrm{n}^2(\mathrm{n}=1,2,3,4,5,6). The dice is rolled twice, yielding the numbers a and b\mathrm{b}. The probability that a<b\mathrm{a}<\mathrm{b} in lowest form is pq\frac{\mathrm{p}}{\mathrm{q}} where 620(p+q)=\frac{620}{(\mathrm{p}+\mathrm{q})}=
View answer and explanation
Correct option: A
5
Option A (5) is correct because the probability a < b is calculated by finding the probability of a=b and using symmetry to find P(a < b).
942 Medium
942 of 1130
Analyzing +4 | -1
Let S={w1,w2,……}S=\left\{w_1, w_2, \ldots \ldots\right\} be the sample space associated to a random experiment. Let P(wn)=P(wn−1)2,n≥2P\left(w_n\right)=\frac{P\left(w_{n-1}\right)}{2}, n \geq 2. Let A={2k+3l:k,l∈N}A=\{2 k+3 l: k, l \in N\} and B={wn:n∈A}B=\left\{w_n: n \in A\right\}. Then P(B)P(B) is equal to
View answer and explanation
Correct option: C
364\frac{3}{64}
Option C (364\frac{3}{64}) is correct because it is the probability of the event B, calculated by finding the probabilities of the individual events and summing them.
943 Easy
943 of 1130
Application +4 | -1
Let XX have a binomial distribution B(n,p)B(n, p) such that the sum and the product of the mean and variance of XX are 24 and 128 respectively. If P(X>n−3)=P(X>n-3)= k2n\frac{k}{2^n}, then kk is equal to :
View answer and explanation
Correct option: B
529
Option B (529) is correct because it is the sum of the probabilities P(X=n-2), P(X=n-1), and P(X=n) for a binomial distribution with n=32 and p=1/2.
944 Easy
944 of 1130
Application +4 | -1
In a group of people, if 4 are selected at a random, the probability that the any two of the four do not have same month of birth is pp then 96p11\frac{96 p}{11} is equal to
View answer and explanation
Correct option: A
5
Option A (5) is correct because the probability that no two of the four selected people have the same month of birth is 55/96, and (96p)/11 equals 5.
945 Easy
945 of 1130
Application +4 | -1
A bag contains 100 tickets numbered 1 to 100. If one ticket is picked up at random, then the probability that the number is divisible by 3 or 5 is
View answer and explanation
Correct option: A
0.47
Option A (0.47) is correct because it is calculated by finding the probability of a number being divisible by 3, the probability of being divisible by 5, and subtracting the probability of being divisible by both (15).
946 Easy
946 of 1130
Application +4 | -1
A 7-digit number is formed using the digit 4 three times, digit 5 two times and digit 3 two times. A number is chosen at random, the probability that the chosen mumber is even, is
View answer and explanation
Correct option: C
37\frac{3}{7}
Option D (37\frac{3}{7}) is correct because it calculates the probability of the chosen number being even by considering the arrangements where the last digit is 4.
947 Medium
947 of 1130
Application +4 | -1
Three persons A,B,CA, B, C throw a die in succession. The one getting 'six' wins. If AA starts then the probability of BB winning is
View answer and explanation
Correct option: D
3091\frac{30}{91}
Option D (3091\frac{30}{91}) is correct because it represents the probability of B winning the game, calculated by considering the infinite series of probabilities where B wins on their turn.
948 Easy
948 of 1130
Application +4 | -1
If the mean and variance of a variate XX having a binomial distribution are 6 and 4 respectively. then the number of values of the variate in the distribution is
View answer and explanation
Correct option: D
18
Option D (18) is correct because the number of trials n in a binomial distribution can be found using the mean (np) and variance (npq).
949 Easy
949 of 1130
Application +4 | -1
If X{X} follows a binomial distribution with parameters nn =8=8 and p=1/2p=1 / 2, then P(∣X−4∣≤2)={P}(|{X}-4| \leq 2)=
View answer and explanation
Correct option: A
119128\frac{119}{128}
Option A (119/128) is correct because it calculates the probability of X being between 2 and 6 inclusive, given a binomial distribution with n=8 and p=1/2.
950 Easy
950 of 1130
Application +4 | -1
The probability that out of 10 persons, all born in June, at least two have the same birth day is
View answer and explanation
Correct option: C
3010−30C10(30)10\frac{30^{10}-{ }^{30} C_{10}}{(30)^{10}}
Option C (3010−30C10(30)10\frac{30^{10}-{ }^{30} C_{10}}{(30)^{10}}) is correct because it represents the probability that at least two out of 10 persons born in June have the same birthday.
951 Medium
951 of 1130
Application +4 | -1
An unbiased die is tossed until a number greater than 4 appears. The probability that an even number of tosses is needed, is
View answer and explanation
Correct option: B
25\frac{2}{5}
Option B (25\frac{2}{5}) is correct because it calculates the probability of needing an even number of tosses using an infinite geometric series.
952 Easy
952 of 1130
Application +4 | -1
Three numbers are chosen at random (without replacement) from the numbers 1,2,…201,2, \ldots 20. The probability that the numbers are not consecutive are
View answer and explanation
Correct option: D
187190\frac{187}{190}
Option D (187/190) is correct because it calculates the probability that the three numbers chosen are not consecutive by subtracting the probability of choosing consecutive numbers from 1.
953 Easy
953 of 1130
Application +4 | -1
If two coins are tossed five times, then the probability of getting 5 heads and 5 tails is
View answer and explanation
Correct option: A
63/25663 / 256
The correct answer is 63/25663 / 256 because it calculates the probability of getting 5 heads and 5 tails when tossing two coins five times.
954 Easy
954 of 1130
Application +4 | -1
The mode of the binomial distribution for which mean and standard deviation are 10 and 5  \sqrt 5 \; respectively, is
View answer and explanation
Correct option: D
10
Option D (10) is correct because the mode of a binomial distribution lies within the interval (np - q, np + p), and in this case, it equals 10.
955 Medium
955 of 1130
Application +4 | -1
Two persons AA and BB take turns in throwing a pair of dice. The first person to through 9 from both dice will be awarded the prize. If AA throws first, then the probability that BB wins the game, is
View answer and explanation
Correct option: B
817\frac{8}{{17}}
Option B (8/17) is correct because it calculates the probability of B winning based on A not throwing 9 and B then throwing 9, considering all possible turns.
956 Easy
956 of 1130
Application +4 | -1
If the letters of the word PROBABILITY are written down at random in a row, the probability that two Bs are together is
View answer and explanation
Correct option: A
211\frac{2}{11}
Option A (211\frac{2}{11}) is correct because it represents the probability of the two Bs being together when arranging the letters of the word PROBABILITY.
957 Easy
957 of 1130
Application +4 | -1
If 1−3p2,1+4p3,1+p6\frac{1-3 p}{2}, \frac{1+4 p}{3}, \frac{1+p}{6} are the probabilities of three mutually exclusive and exhaustive events, then the set of all values of pp is
View answer and explanation
Correct option: B
[−1/4,1/3][-1 / 4,1 / 3]
Option B ([−14,13]\left[-\frac{{1}}{{4}}, \frac{{1}}{{3}}\right]) is correct because it determines the range of p based on the conditions that probabilities must be non-negative and sum to 1.
958 Easy
958 of 1130
Application +4 | -1
A card is drawn from a pack of cards. The probability that the card will be a queen or a heart, is
View answer and explanation
Correct option: C
413\frac{4}{{13}}
Option C (413\frac{4}{{13}}) is correct because it calculates the probability of drawing a queen or a heart using the formula for the union of two events.
959 Easy
959 of 1130
Analyzing +4 | -1
Which one of the following is not correct?
View answer and explanation
Correct option: C
The probability guessing at least 8 out 10 answers in a true false examination, is 764\frac{7}{{64}}
Option C (The probability guessing at least 8 out 10 answers in a true false examination, is 764\frac{7}{{64}}) is not correct because the correct probability is 11512\frac{11}{512}
960 Easy
960 of 1130
Application +4 | -1
All the heart cards are taken out from a pack of playing cards. From these cards, cards are drawn one by one without replacement till the ace of hearts comes. The probability that ace comes in the 4th draw is
View answer and explanation
Correct option: A
113\frac{1}{13}
Option A (113\frac{1}{13}) is correct because it calculates the probability of drawing 3 non-ace heart cards followed by the ace of hearts.
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