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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 631-660 of 1130 questions

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631 Medium
631 of 1130
Application +4 | -1
A dice is rolled 4 times, the numbers appearing are listed. The number of different throws, such that the largest number appearing in the list is 4 , is
View answer and explanation
Correct option: D
175
Option C (1540) is correct because it is the number of ways to form the committee under the given constraint.
632 Hard
632 of 1130
Application +4 | -1
In a bag there are three tickets numbered 1, 2, 3. A ticket is drawn at random and put back, and this is done four times. The probability that the sum of the numbers is even, is
View answer and explanation
Correct option: C
40/81
Option C (40/81) is correct because it represents the probability that the sum of four tickets drawn with replacement is even.
633 Easy
633 of 1130
Application +4 | -1
′X′'X' speaks truth in 60%60 \% and ' YY ' in 50%50 \% of the cases. The probability that they contradict each other while narrating the some incident, is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because it calculates the probability of them contradicting each other by considering the cases where one speaks the truth and the other lies.
634 Easy
634 of 1130
Application +4 | -1
An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn; if the drawn ball is red, then a green ball is added to the urn the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is
View answer and explanation
Correct option: B
3249\frac{32}{49}
Option B (3249\frac{{32}}{{49}}) is correct because it calculates the probability of drawing a red ball in the second draw considering the two possible scenarios of the first draw.
635 Easy
635 of 1130
Application +4 | -1
A box contains 10 mangoes out of which 4 are rotten. 2 mangoes are taken out together. If one of them is found to be good, the probability that the other is also good is
View answer and explanation
Correct option: C
5/135/13
Option C (5/135/13) is the correct probability, calculated using conditional probability considering that one mango is already known to be good.
636 Medium
636 of 1130
Application +4 | -1
Twelve tickets are numbered from 1 to 12. One ticket is drawn at random, then the probability of the number to be divisible by 2 or 3, is
View answer and explanation
Correct option: A
23\frac{2}{3}
Option A (2/3) is correct because it correctly calculates the probability of a number being divisible by 2 or 3 from a set of 12 numbers.
637 Easy
637 of 1130
Application +4 | -1
2n2 n boys are randomly divided into two subgroups containing nn boys each. The probability that the two tallest boys are in different groups is
View answer and explanation
Correct option: A
n2n−1\frac{n}{2 n-1}
Option A (n2n−1\frac{{n}}{{2n-1}}) is correct because it represents the probability that the two tallest boys are in different groups when 2n boys are divided into two groups of n.
638 Easy
638 of 1130
Application +4 | -1
If the standard deviation of the binomial distribution (q+p)16(q+p)^{16} is 2, then mean of the distribution is
View answer and explanation
Correct option: B
8
Option B (8) is correct. Given the standard deviation, we can find p and then calculate the mean using the formula mean = np.
639 Easy
639 of 1130
Application +4 | -1
If the mean and variance of a binomial variate XX are 2 and 1 respectively, then the probability that XX takes a value greater than 1, is
View answer and explanation
Correct option: D
15/1615 / 16
Option D (15/1615 / 16) is correct because it calculates the probability that X is greater than 1, given the mean and variance of a binomial variate.
640 Easy
640 of 1130
Application +4 | -1
While shuffling a pack of 52 playing cards. Unfortunately 4 cards are dropped on the ground it is known that 2 out of 4 are spades. What is the probability that all dropped 4 are spades?
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the provided options do not match the correct conditional probability calculation.
641 Medium
641 of 1130
Application +4 | -1
If 3 numbers are selected from the set (1,2,3,.........,24)(1,2,3,........., 24) without replacement, then what is the probability that the selected numbers do not form an A.P.?
View answer and explanation
Correct option: C
43/4643 / 46
Option C (43/46) is correct because it represents the probability that the three selected numbers do not form an arithmetic progression.
642 Easy
642 of 1130
Application +4 | -1
If four positive integers are taken at random and multiplied together, then the probability that the last digit is 1, 3, 7 or 9 is
View answer and explanation
Correct option: D
16625\frac{16}{625}
Option D (16625\frac{16}{625}) is correct because it represents the probability that the last digit of the product of four random positive integers is 1, 3, 7, or 9.
643 Medium
643 of 1130
Application +4 | -1
If P(A)=25,P(B)=310\mathrm{P}(\mathrm{A})=\frac{2}{5}, \mathrm{P}(\mathrm{B})=\frac{3}{10} and P(A∩B)=15\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{5}, then P(A′∣B′)⋅P(B′∣A′)\mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{B}^{\prime} \mid \mathrm{A}^{\prime}\right) is equal to
View answer and explanation
Correct option: C
2542\frac{25}{42}
Option C is correct because it is obtained by using the conditional probability formula and the properties of complementary events.
644 Easy
644 of 1130
Application +4 | -1
The probability that the birth days of six different persons will fall in exactly two calendar months is
View answer and explanation
Correct option: D
341125\frac{341}{12^5}
Option D (341125\frac{341}{12^5}) is correct because it calculates the probability of six birthdays falling in exactly two months, accounting for the combinations of months and excluding cases where all birthdays are in one month.
645 Easy
645 of 1130
Analyzing +4 | -1
A three digit number, which is a multiple of 11, is chosen at random. Probability that the number so chosen is also a multiple of 9, is equal to
View answer and explanation
Correct option: A
19\frac{1}{9}
Option A (1/9) is correct because it represents the probability that a randomly chosen three-digit multiple of 11 is also a multiple of 9.
646 Medium
646 of 1130
Application +4 | -1
There are mm persons sitting in a row. Two of them are selected at random. The probability that the two selected persons are not together is
View answer and explanation
Correct option: D
1−2m1-\frac{2}{m}.
Option D (1−2m1-\frac{2}{m}) is correct because it is the probability calculated by subtracting the probability of selecting two adjacent people from 1.
647 Easy
647 of 1130
Application +4 | -1
An integer is chosen at random from the first 200 positive integers. The probability that the integer chosen is divisible by 6 or 8 is
View answer and explanation
Correct option: B
1/41 / 4
Option B (14\frac{1}{4}) is correct because it calculates the probability of an integer being divisible by 6 or 8 by considering the individual probabilities and subtracting the probability of being divisible by both.
648 Easy
648 of 1130
Application +4 | -1
Let SS be the sample space of all five digit numbers. It pp is the probability that a randomly selected number from SS, is multiple of 7 but not divisible by 5 , then 9p9 p is equal to
View answer and explanation
Correct option: C
1.02851.0285
Option C (1.02851.0285) is correct because it is the result of calculating the probability of a 5-digit number being a multiple of 7 but not divisible by 5, and then multiplying by 9.
649 Medium
649 of 1130
Application +4 | -1
There are 10 pairs of shoes in a cupboard from which 4 shoes are picked at radndom. The probability that there is at least one pair is
View answer and explanation
Correct option: B
99323\frac{99}{323}
Option B (99/323) is correct because it represents the probability of selecting at least one pair of shoes when 4 shoes are randomly picked from 10 pairs.
650 Easy
650 of 1130
Application +4 | -1
If n( A)=1000,n( B)=500n(\mathrm{~A})=1000, n(\mathrm{~B})=500 and if n( A∩B)≥1n(\mathrm{~A} \cap \mathrm{B}) \geq 1 and n( A∪B)=pn(\mathrm{~A} \cup \mathrm{B})=p, then
View answer and explanation
Correct option: D
1000≤p≤14991000 \leq p \leq 1499
Option D (1000≤p≤14991000 \leq p \leq 1499) is correct because it represents the range of possible values for n(A∪B)n(A \cup B) given the constraints on n(A∩B)n(A \cap B).
651 Easy
651 of 1130
Application +4 | -1
A die is thrown. Let AA be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then, P(A∪B)P(A \cup B) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the union of the events A and B covers all possible outcomes when throwing a die.
652 Medium
652 of 1130
Application +4 | -1
A letter is taken at random from the letters of the word 'STATISTICS' and another letter is taken at random from the letters of the word 'ASSISTANT'. The probability that they are the same letter is
View answer and explanation
Correct option: C
19/9019 / 90
Option C (19 / 90) is correct because it is calculated by summing the probabilities of choosing the same letter (A, I, S, or T) from both words.
653 Easy
653 of 1130
Application +4 | -1
A,B,CA, B, C are three mutually exclusive and exhaustive events associated with a random experiment. If P(B){P}({B}) =32P(A)=\frac{3}{2} {P}({A}) and P(C)=12P(B){P}({C})=\frac{1}{2} {P}({B}), then P(A)={P}({A})=
View answer and explanation
Correct option: A
413\frac{4}{13}
Option A (413\frac{{4}}{{13}}) is correct because it's the probability of event A, calculated using the mutually exclusive and exhaustive nature of events A, B, and C.
654 Easy
654 of 1130
Analyzing +4 | -1
Six dice are thrown simultaneously. The probability that exactly three of them show the same face and remaining three show different faces, is
View answer and explanation
Correct option: C
(5!)22(66)\frac{(5 !)^2}{2\left(6^6\right)}
Option C ((5!)22(66)\frac{{(5 !)^2}}{{2(6^6)}}) is correct because it represents the probability of exactly three dice showing the same face and the remaining three showing different faces.
655 Medium
655 of 1130
Application +4 | -1
For the married couple living in Jammu, the probability that a husband will vote in an election is 0.5 and the probability that his wife will vote is 0.4 . The probability that the husband votes, given that his wife also votes is 0.7 . Then the probability that husband and wife both will vote is
View answer and explanation
Correct option: A
0.28
Option A (0.28) is correct because the probability of both events occurring is calculated using conditional probability: P(A ∩ B) = P(B) * P(A|B).
656 Medium
656 of 1130
Application +4 | -1
Three are six verities of a regular hexagon are chosen at random, then the possibility that the triangle with three vertices is equilateral, is equal to
View answer and explanation
Correct option: C
110\frac{1}{10}
Option C (1/10) is correct because it calculates the probability of choosing three vertices that form an equilateral triangle from a regular hexagon.
657 Easy
657 of 1130
Application +4 | -1
A class consists of 80 students, 25 of them are girls. If 10 of the students are rich and 20 of the students are fair complexioned, then the probability of selecting a fair complexioned rich girl from the class (assuming three traits as independent) is
View answer and explanation
Correct option: C
5512\frac{5}{512}
Option C (5512\frac{5}{512}) is correct because it correctly calculates the probability of selecting a fair complexioned rich girl assuming independence.
658 Easy
658 of 1130
Application +4 | -1
If AA and BB each toss three coins. The probability that both get the same number of heads is
View answer and explanation
Correct option: C
516\frac{5}{{16}}
Option C (5/16) is correct because it calculates the probability that both A and B get the same number of heads by summing the probabilities of each getting 0, 1, 2, or 3 heads.
659 Hard
659 of 1130
Analyzing +4 | -1
In a convex hexagon two diagonals are drawn at random. The probability that the diagonals intersect at an interior point of the hexagon, is
View answer and explanation
Correct option: A
512\frac{5}{12}
Option A (512\frac{5}{12}) is correct because it represents the ratio of favorable outcomes (intersecting diagonals) to the total possible outcomes when selecting two diagonals.
660 Easy
660 of 1130
Application +4 | -1
The probability that when 12 balls are distributed among three boxes, the first will contain three balls is,
View answer and explanation
Correct option: B
12C3×29312\frac{{ }^{12} C_3 \times 2^9}{3^{12}}
Option B (12C3×29312\frac{{ }^{12} C_3 \times 2^9}{3^{12}}) is correct because it calculates the probability of the first box containing three balls by considering the combinations of choosing the three balls and the distribution of the remaining balls into the other two boxes.
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