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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 871-900 of 1130 questions

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871 Easy
871 of 1130
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.
872 Easy
872 of 1130
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.
873 Easy
873 of 1130
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.
874 Medium
874 of 1130
Application +4 | -1
Two non-negative integers are chosen at random. The probability that the sum of the square is divisible by 10 , is
View answer and explanation
Correct option: B
950\frac{9}{50}
Option B (9/50) is correct because there are 18 pairs of digits (0-9) whose squares sum to a multiple of 10, out of 100 possible pairs.
875 Easy
875 of 1130
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.
876 Medium
876 of 1130
Application +4 | -1
Two dice are rolled together and the numbers x,yx, y on them form complex number z=x+iyz=x+i y. The probability that ∣z∣≤3|z| \leq 3 is
View answer and explanation
Correct option: C
112\frac{1}{12}
Option C (112\frac{1}{12}) is correct because it represents the probability of the complex number's magnitude being less than or equal to 3, calculated by finding the favorable outcomes and dividing by the total possible outcomes when rolling two dice.
877 Easy
877 of 1130
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.
878 Easy
878 of 1130
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.
879 Easy
879 of 1130
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.
880 Easy
880 of 1130
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}}.
881 Easy
881 of 1130
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.
882 Easy
882 of 1130
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.
883 Easy
883 of 1130
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.
884 Easy
884 of 1130
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.
885 Easy
885 of 1130
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.
886 Medium
886 of 1130
Application +4 | -1
A box contains tickets numbered 1 to 20. 3 tickets are drawn from the box with replacement. The probability that the largest number on the tickets is 7 , is
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the probability is calculated by subtracting the probability of the largest number being less than or equal to 6 from the probability of the largest number being less than or equal to 7.
887 Easy
887 of 1130
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.
888 Medium
888 of 1130
Application +4 | -1
If a fair coin is tossed 20 times and let we get head nn times, then probability that nn is odd, 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 20 tosses of a fair coin is exactly one-half.
889 Medium
889 of 1130
Application +4 | -1
The probability that an event A occurs in a single trial of an experiment is 0.6 . In the first three independent trials of the experiment, the probability that A occurs atleast once is
View answer and explanation
Correct option: B
0.936
Option B (0.936) is correct because it calculates the probability of event A occurring at least once in three trials by subtracting the probability of A not occurring in any of the trials from 1.
890 Easy
890 of 1130
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.
891 Easy
891 of 1130
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.
892 Easy
892 of 1130
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.
893 Easy
893 of 1130
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.
894 Easy
894 of 1130
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.
895 Medium
895 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 7th drawn ball is 4th white, is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because with replacement, the probability of drawing a white ball remains constant at 12\frac{1}{2} for each draw.
896 Easy
896 of 1130
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).
897 Medium
897 of 1130
Application +4 | -1
India plays two ODI matches each with Australia and Pakistan. The probability of India getting 0, 1, 2 are 0.45, 0.05, 0.50. The probability of India getting at least 7 points in the series is
View answer and explanation
Correct option: C
0.08750.0875
Option C (0.0875) is correct because it calculates the probability of India getting at least 7 points in the series based on the given probabilities.
898 Easy
898 of 1130
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.
899 Easy
899 of 1130
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.
900 Easy
900 of 1130
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.
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