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JEE • MATHEMATICS • PROBABILITY

Probability practice · Easy

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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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631 Easy
631 of 822
Application +4 | -1
If the random variable XX takes the values x1,   x2, x10{x_1},\;\,{x_2},\,{x_{10}} with probability P(X=x1)=ki,P\left( {X = {x_1}} \right) = ki, then the value of kk is equal to
View answer and explanation
Correct option: C
155\frac{1}{{55}}
Option C (1/55) is correct. The value of k is found by using the fact that the sum of probabilities for all possible values of a random variable must equal 1.
632 Easy
632 of 822
Application +4 | -1
The probability that a leap year will have 53 Fridays or 53 Saturdays, is
View answer and explanation
Correct option: B
37\frac{3}{7}
Option B (37\frac{3}{7}) is correct because it calculates the probability of a leap year having 53 Fridays or 53 Saturdays by considering the possible combinations of the extra two days.
633 Easy
633 of 822
Application +4 | -1
For a party 8 guests are invited by a husband and his wife. They sit for a dinner around a round table. The probability that the husband and his wife sit together, is
View answer and explanation
Correct option: B
29\frac{2}{9}
Option B (29\frac{2}{9}) is correct because it calculates the probability of the husband and wife sitting together around a round table.
634 Easy
634 of 822
Application +4 | -1
If (  1+a3)   and   (  1−a4)\left( {\;\frac{{1 + a}}{3}} \right)\;\,\mathrm{and}\,\;\left( {\;\frac{{1 - a}}{4}} \right) are probability of two mutually exclusive events, then set of all values of aa is
View answer and explanation
Correct option: A
−1≤a≤1- 1 \le a \le 1
Option A (−1≤a≤1- 1 \le a \le 1) is correct because it satisfies the conditions for probabilities of mutually exclusive events, ensuring each probability is between 0 and 1, and their sum is also between 0 and 1.
635 Easy
635 of 822
Application +4 | -1
A bag contains 3 same balls and 3 different balls of three different colours. Two balls are drawn randomly with replacement. The probability they have same colour is mm. Again four balls are drawn one by one with replacement, then probability of getting three same balls is nn. The value of mm. nn is
View answer and explanation
Correct option: D
881\frac{8}{81}
Option D (8/81) is correct because it is the product of the probabilities m and n, calculated based on the given conditions of drawing balls with replacement.
636 Easy
636 of 822
Analyzing +4 | -1
There is a five-volume dictionary among 50 books arranged on a shelf in a random order. If the volumes are not necessarily kept side-by side, the probability that they occur in increasing order from left to right is
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the probability of the dictionary volumes being in increasing order is 15!=1120\frac{1}{5!}=\frac{1}{120}, which is not among the given options.
637 Easy
637 of 822
Application +4 | -1
The chances of defective screws in three boxes A, B, C are 15,16,17\frac{1}{5}, \frac{1}{6}, \frac{1}{7} respectively. A box is selected at random and a screw drawn from it at random is found to be defective. Then the probability that it came from box AA is
View answer and explanation
Correct option: D
42/10742 / 107
Option D (42/10742 / 107) is correct because it uses Bayes' theorem to find the probability that the defective screw came from box A.
638 Easy
638 of 822
Application +4 | -1
A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to the probability of getting 9 heads, then the probability of getting 2 heads is
View answer and explanation
Correct option: A
15213\frac{15}{2^{13}}
Option A (15213\frac{15}{2^{13}}) is correct because it's the probability of getting 2 heads after determining the number of tosses based on the given probabilities of 7 and 9 heads.
639 Easy
639 of 822
Analyzing +4 | -1
If the independent events AA and BB are such that 0<P(A)<10 < P(A) < 1 and 0<P(B)<10 < P(B) < 1, then which of the following alternatives is not correct?
View answer and explanation
Correct option: A
AA and BB are mutually exclusive
Option A (AA and BB are mutually exclusive) is not correct because if AA and BB are independent events with 0<P(A)<10 < P(A) < 1 and 0<P(B)<10 < P(B) < 1, then P(A∩B)=P(A)P(B)≠0P(A \cap B) = P(A)P(B) \neq 0, which means they cannot be mutually exclusive.
640 Easy
640 of 822
Application +4 | -1
In a class 60%60 \% passed their Physics examination and 58%58 \% passed in Mathematics. Atleast what percentage of students passed both their Physics and Mathematics examination?
View answer and explanation
Correct option: A
18%18 \%
Option A (18%18 \%) is correct because it represents the minimum percentage of students who passed both exams, calculated using set theory principles.
641 Easy
641 of 822
Application +4 | -1
In a binomial distribution, mean is 3 and standard deviation is 32\frac{3}{2}, then the probability distribution is
View answer and explanation
Correct option: A
(34+14)12{\left( {\frac{3}{4} + \frac{1}{4}} \right)^{12}}
Option A ((34+14)12{\left( {\frac{{3}}{{4}} + \frac{{1}}{{4}}} \right)^{{12}}}) is correct because the binomial distribution is determined by finding n and p from the given mean and standard deviation.
642 Easy
642 of 822
Application +4 | -1
If ten objects are distributed at random among ten persons, the probability that at least one of them will not get any thing is
View answer and explanation
Correct option: B
1010−10 ! 1010\frac{10^{10}-10 \text { ! }}{10^{10}}
Option B (1010−10!1010\frac{{10^{10}-10 ! }}{{10^{10}}}) is correct because it calculates the probability that at least one person doesn't receive an object when distributing ten objects among ten persons.
643 Easy
643 of 822
Analyzing +4 | -1
AA box B1B_1 contains 1 white ball, 3 red balls and 2 black balls. Another box B2B_2 contains 2 white balls, 3 red balls and 4 black balls. A third box B3B_3 contains 3 white balls, 4 red balls and 5 black balls. If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B2B_2 is
View answer and explanation
Correct option: D
55181\frac{55}{181}
Option C (P(X/X2)=516P(X / X_2)=\frac{{5}}{{16}}) is incorrect because the correct calculation yields P(X/X2)=58P(X / X_2) = \frac{{5}}{{8}}.
644 Easy
644 of 822
Application +4 | -1
A number nn is chosen at random from {1, 2, 3, 4, ……….,1000}. The probability that nn is a number that leaves remainder 1 when divided by 7, is
View answer and explanation
Correct option: B
1431000\frac{{143}}{{1000}}
Option B (1431000\frac{{143}}{{1000}}) is correct because it calculates the probability of n leaving a remainder of 1 when divided by 7 by finding the number of favorable outcomes and dividing by the total number of outcomes.
645 Easy
645 of 822
Application +4 | -1
In a market research project, 20%20 \% opted for 'Nirma' detergent whereas 60%60 \% opted for 'Surf blue' detergent. The remaining individuals were not certain. If the difference between those who opted for 'Surf blue' and those who were uncertain was 720 , how many respondents were covered in the survey
View answer and explanation
Correct option: C
1800
Option C (1800) is correct because it is the total number of respondents calculated based on the given percentages and the difference between 'Surf blue' and 'uncertain' preferences.
646 Easy
646 of 822
Analyzing +4 | -1
A natural number is selected from 1 to 1000 at random, then the probability that a particular non-zero digit appears at most once, is
View answer and explanation
Correct option: A
243250\frac{{243}}{{250}}
Option A (243250\frac{{243}}{{250}}) is correct because it calculates the probability by subtracting the probability of numbers with a digit appearing more than once from 1.
647 Easy
647 of 822
Application +4 | -1
Four fair dice are thrown simultaneously. If the probability that the highest number obtained is 4 is 25a1296\frac{25 a}{1296} then &#39; aa &#39; is equal to
View answer and explanation
Correct option: A
7
The correct answer is A (7) because the probability that the highest number obtained is 4 is calculated by considering all outcomes where the numbers are less than or equal to 4, and subtracting the outcomes where the numbers are less than or equal to 3.
648 Easy
648 of 822
Application +4 | -1
A fair coin is tossed repeatedly. The probability of getting a result in fifth toss different from those obtained in the first four tosses is
View answer and explanation
Correct option: D
116\frac{1}{16}
Option A (116\frac{1}{16}) is correct because the probability of the fifth toss being different from the first four is the sum of the probabilities of getting four heads followed by a tail and four tails followed by a head.
649 Easy
649 of 822
Application +4 | -1
Box AA contains 2 black and 3 red balls. While box BB contains 3 black and 4 red balls. Out of these two boxes one is selected at random; and the probability of choosing box AA is double that of box BB. If a red ball is drawn from the selected box, then the probability that it has come from box BB. is
View answer and explanation
Correct option: B
1031\frac{{10}}{{31}}
Option B (1031\frac{{10}}{{31}}) is correct because it's the conditional probability calculated using Bayes' theorem, considering the probabilities of selecting each box and drawing a red ball.
650 Easy
650 of 822
Application +4 | -1
One ticket is selected at random from 100 tickets numbered 00,01,02,…,9900,01,02, \ldots, 99.
Suppose, XX and YY are the sum and product of the digit found on the ticket P(X=7/Y=0)P(X=7 / Y=0) is given by
View answer and explanation
Correct option: B
219\frac{2}{19}
Option B is correct because the conditional probability P(X=7 / Y=0) is calculated by finding the intersection of the two events and dividing by the probability of the condition Y=0.
651 Easy
651 of 822
Application +4 | -1
Two unbiased dice are thrown simultaneously. The probability to get a sum more than 8 is
View answer and explanation
Correct option: B
518\frac{5}{{18}}
Option B (518\frac{5}{{18}}) is correct because it calculates the probability of getting a sum more than 8 by counting favorable outcomes and dividing by the total possible outcomes.
652 Easy
652 of 822
Application +4 | -1
Let the sum of two positive integers be 24. If the probability, that their product is not less than 34\frac{3}{4} times their greatest positive product, is mn\frac{\mathrm{m}}{\mathrm{n}}, where gcd⁡(m,n)=1\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1, then n−m\mathrm{n}-\mathrm{m} equals:
View answer and explanation
Correct option: D
10
Option D (10) is correct because by finding the pairs of numbers that satisfy the given condition and calculating the probability, we can determine that n-m = 10.
653 Easy
653 of 822
Application +4 | -1
Two dice are thrown simultaneously. The probability of getting the sum 2 or 8 or 12 is
View answer and explanation
Correct option: B
736\frac{7}{36}
Option B (736\frac{7}{36}) is correct. The possible combinations that sum to 2, 8, or 12 are (1,1), (2,6), (3,5), (4,4), (5,3), (6,2), and (6,6), giving a probability of 7/36.
654 Easy
654 of 822
Analyzing +4 | -1
Numbers 1,2,3,..., 100 are written down on each of the cards A, B and C. One number is selected at random from each of the cards. The probability that the numbers so selected can be the measures (in cm) of three sides of a right-angled triangle is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated probability considering Pythagorean triplets within the range 1-100 does not match any of the given options.
655 Easy
655 of 822
Application +4 | -1
A pair of fair dice is thrown independently 4 times. The probability of getting a sum of exactly 7 twice is
View answer and explanation
Correct option: C
25216\frac{{25}}{{216}}
Option A (3190\frac{3}{{190}}) is correct because it calculates the probability of selecting three consecutive numbers by dividing the number of favorable outcomes (18) by the total number of possible outcomes (1140).
656 Easy
656 of 822
Remembering +4 | -1
If the events AA and BB are independent, then P(A∩B)P(A \cap B) is equal to
View answer and explanation
Correct option: C
P(A)×P(B)P(A) \times P(B)
Option C (P(A)×P(B)P(A) \times P(B)) is correct because for independent events, the probability of their intersection is the product of their individual probabilities.
657 Easy
657 of 822
Application +4 | -1
Let AA and BB be two events such that P(A)P(A) =3/7,P(B)=4/7=3 / 7, P(B)=4 / 7 and P(A∪B)=5/7P(A \cup B)=5 / 7,
then P(A∣B)P(A′∣B)\frac{P(A \mid B)}{P\left(A^{\prime} \mid B\right)} equals
View answer and explanation
Correct option: C
1/21 / 2
Option C (1/21 / 2) is correct because it represents the ratio of conditional probabilities, calculated using the given probabilities and the formula for conditional probability.
658 Easy
658 of 822
Application +4 | -1
If P(A)=112,  P(B)=512P\left( A \right) = \frac{1}{{12}},\;P\left( B \right) = \frac{5}{{12}} and P(BA)=115,P\left( {\frac{B}{A}} \right) = \frac{1}{{15}}, then P(A∪B)P\left( {A \cup B} \right) is equal to
View answer and explanation
Correct option: A
89180\frac{{89}}{{180}}
Option A (89180\frac{{89}}{{180}}) is correct because it is calculated using the formula for the probability of the union of two events and the given conditional probability.
659 Easy
659 of 822
Application +4 | -1
A sum of money is rounded off to the nearest rupee. The probability that round off error is at least ten paise is
View answer and explanation
Correct option: A
81100\frac{81}{100}
Option A (81100\frac{81}{100}) is correct because it represents the probability that the round off error is at least ten paise when a sum of money is rounded off to the nearest rupee.
660 Easy
660 of 822
Application +4 | -1
The probability that A speaks truth is 45\frac{4}{5}, while this probability for BB is 34\frac{3}{4}. The probability that they contradict each other when asked to speak on a fact is
View answer and explanation
Correct option: C
720\frac{7}{20}
Option C (720\frac{{7}}{{20}}) is correct because it calculates the probability that A and B contradict each other by considering the cases where A speaks the truth and B lies, and vice versa.
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