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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 601-630 of 1130 questions

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601 Easy
601 of 1130
Application +4 | -1
An experiment succeeds twice as often as it fails. The probability of at least 5 successes in the six trials of the experiment, is
View answer and explanation
Correct option: B
256729\frac{256}{729}
Option B (256729\frac{256}{729}) is correct because it calculates the probability of at least 5 successes in 6 trials using binomial distribution with p=2/3 and q=1/3.
602 Easy
602 of 1130
Application +4 | -1
There are two balls in an urn. Each ball can be either white or black. If a white ball is put into the urn and there after a ball is drawn at random from the urn, then the probability that it is white is
View answer and explanation
Correct option: B
23\frac{2}{3}
The correct answer is B (2/3) because it is calculated by considering the possible initial states of the urn and then applying conditional probability.
603 Easy
603 of 1130
Application +4 | -1
The mean and variance of binomial distribution are 4 and 3 respectively. Then, the probability of getting exactly six success 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}}
Option A (16C6(14)6(34)10^{16}{C_6}{\left( {\frac{1}{4}} \right)^6}{\left( {\frac{3}{4}} \right)^{10}}) is correct because it correctly applies the binomial distribution formula with the calculated values of n, p, and q.
604 Easy
604 of 1130
Application +4 | -1
When three dice are thrown the probability of getting 4 or 5 on each of the dice simultaneously, is
View answer and explanation
Correct option: D
None of these
The correct probability is 127\frac{1}{{27}}, which is not listed in the options. The explanation shows the calculation of this probability by considering all possible favorable outcomes when throwing three dice.
605 Easy
605 of 1130
Understanding +4 | -1
There is an objective type question with 4 answer choices exactly one of which is correct. A student has not studied the topic on which the question has been set. The probability that the student guesses the correct answer, is
View answer and explanation
Correct option: B
14\frac{1}{4}
Option B (14\frac{1}{4}) is correct because with 4 answer choices, the probability of guessing the correct one is 1/4.
606 Easy
606 of 1130
Application +4 | -1
If AA and BB are independent events such that P(B)=27,P(A∪Bˉ)=0.8P(B)=\frac{2}{7}, P(A \cup \bar{B})=0.8, then P(A)=P(A)=
View answer and explanation
Correct option: C
0.30.3
Option C (0.3) is correct because it correctly uses the formula for the probability of the union of events and the independence of A and B.
607 Easy
607 of 1130
Application +4 | -1
A six faced fair die is thrown until 1 comes. Then, the probability that 1 comes in even number of trials, is
View answer and explanation
Correct option: A
511\frac{5}{11}
Option A (5/11) is correct because the probability of getting 1 in an even number of trials is the sum of an infinite geometric series.
608 Easy
608 of 1130
Application +4 | -1
The mean of the numbers obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
View answer and explanation
Correct option: B
2
Option D (83\frac{8}{3}) is correct because it is the expected value (mean) calculated by summing the product of each outcome and its probability.
609 Easy
609 of 1130
Application +4 | -1
A coin is tossed three times is succession. If E\mathrm{E} is the event that there are at least two heads and F\mathrm{F} is the event in which first throw is a head, then P(E/F)\mathrm{P}(\mathrm{E} / \mathrm{F}) equal to:
View answer and explanation
Correct option: A
34\frac{3}{4}
Option A (34\frac{3}{4}) is correct because it is the conditional probability of having at least two heads given that the first throw is a head.
610 Easy
610 of 1130
Application +4 | -1
In shuffling a pack of playing cards, four are accidently dropped. The probability that missing cards should be one from each suit, is
View answer and explanation
Correct option: C
219720825\frac{{2197}}{{20825}}
Option C is correct because the probability is calculated by dividing the number of ways to choose one card from each suit by the total number of ways to choose four cards.
611 Easy
611 of 1130
Analyzing +4 | -1
A quadratic equation is chosen from the set of all quadratic equations which are unchanged by squaring their roots. The chance that the chosen equation has equal roots, is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because it represents the ratio of quadratic equations with equal roots to the total number of quadratic equations unchanged by squaring their roots.
612 Medium
612 of 1130
Analyzing +4 | -1
The probability that two randomly selected subsets of the set {1,2,3,4,5}\left\{ {1,2,3,4,5} \right\} have exactly two elements in their intersection, is
View answer and explanation
Correct option: C
13529\frac{{135}}{{{2^9}}}
Option C (13529\frac{{135}}{{{2^9}}}) is correct because it calculates the probability of two randomly selected subsets having exactly two elements in their intersection by considering the number of ways to choose the intersection and the remaining elements.
613 Easy
613 of 1130
Application +4 | -1
Suppose nn students appear in an examination. Let X=X= the number of students who pass the examination. Suppose P(X=k)=λk2P(X=k)=\lambda k^2, then value of λ\lambda is
View answer and explanation
Correct option: B
6n(n+1)(2n+1)\frac{6}{n(n+1)(2 n+1)}
Option B (6n(n+1)(2n+1)\frac{6}{n(n+1)(2 n+1)}) is correct because it is derived from the condition that the sum of all probabilities must equal 1.
614 Easy
614 of 1130
Application +4 | -1
The probability of forming a three digit numbers with the same digits when three digit numbers are formed out of the digit 0, 2, 4, 6, 8 is
View answer and explanation
Correct option: D
125\frac{1}{{25}}
Option D (125\frac{1}{{25}}) is correct because there are 4 favorable cases (222, 444, 666, 888) out of 100 possible three-digit numbers formed using the digits 0, 2, 4, 6, 8.
615 Easy
615 of 1130
Application +4 | -1
If AA and BB are events of a random experiment such that P(A∪B)=45,P(Aˉ∪Bˉ)=710P\left( {A \cup B} \right) = \frac{4}{5},P\left( {\bar A \cup \bar B} \right) = \frac{7}{{10}} and P(B)=25P\left( B \right) = \frac{2}{5} then P(A)P\left( A \right) equals
View answer and explanation
Correct option: C
710\frac{7}{{10}}
Option C (710\frac{7}{{10}}) is correct because it uses DeMorgan's law and the formula for the probability of the union of two events to find P(A).
616 Easy
616 of 1130
Application +4 | -1
A bag AA contains 4 green and 3 red balls and bag BB contains 4 red and 3 green balls. One bag is taken at random and a ball is drawn and noted to be green. The probability that it comes from bag BB, is
View answer and explanation
Correct option: C
37\frac{3}{7}
Option C (37\frac{3}{7}) is correct because it applies Bayes' theorem to calculate the probability that the green ball came from bag B, given that a green ball was drawn.
617 Easy
617 of 1130
Application +4 | -1
The probability of having at least one tail in 4 throws with a coin, is
View answer and explanation
Correct option: A
1516\frac{{15}}{{16}}
Option A (1516\frac{{15}}{{16}}) is correct because it represents the probability of getting at least one tail, which is the complement of getting no tails in 4 throws.
618 Easy
618 of 1130
Application +4 | -1
Five persons A,   B,   C,   DA,\,\;B,\,\;C,\;\,D and EE are in queue of a shop. The probability that AA and EE always together, is
View answer and explanation
Correct option: C
25\frac{2}{5}
Option C (25\frac{2}{5}) is correct because it represents the probability of A and E being together, calculated by considering them as a single unit.
619 Easy
619 of 1130
Application +4 | -1
Suppose n(≥3)n(\geq 3) persons are arranged in a row. The probability that two particular persons are not together is
View answer and explanation
Correct option: A
1−2n1-\frac{2}{n}
Option A (1−2n1-\frac{2}{n}) is correct because it represents the probability that two particular persons are not together, calculated by subtracting the probability that they are together from 1.
620 Easy
620 of 1130
Application +4 | -1
An article manufactured by a company consits of two parts AA and BB. In the processs of manufacutre of part AA, 9 out of 100 are likely to be defective. Similarly 5 out of 100 are likely to be defective in the manufacture of part BB. The probability that the assembled part will not be defective, is
View answer and explanation
Correct option: A
0.8645
Option A (0.8645) is correct because it represents the probability that both parts A and B are non-defective, resulting in a non-defective assembled article.
621 Medium
621 of 1130
Application +4 | -1
The mean and variance of 5 observations of an experiment are 4 and 5.2 respectively. If from these observations three are 1,2 and 6 , then the mean of remaining will be-
View answer and explanation
Correct option: A
5.5
Option A (5.5) is correct because after calculating the sum of the remaining two observations, their mean is found to be 5.5.
622 Easy
622 of 1130
Application +4 | -1
If a triangle is formed by choosing the vertices of 10 sided regular polygon. Then probability that triangle has exactly one side in common with the polygon, is
View answer and explanation
Correct option: D
12\frac{1}{2}
The correct answer is D (12\frac{{1}}{{2}}) because it is obtained by calculating the probability of forming a triangle with exactly one side common with the polygon.
623 Easy
623 of 1130
Application +4 | -1
A fair die is tossed repeatedly until a six is obtained. Let XX denote the number of tosses required. The probability that X≥3X \geq 3 equals
View answer and explanation
Correct option: D
2536\frac{25}{36}
Option D (2536\frac{{25}}{{36}}) is correct because it represents the probability that at least 3 tosses are required to get a six, which is the probability of not getting a six in the first two tosses.
624 Easy
624 of 1130
Application +4 | -1
A rifleman is firing at a distant target and has only 10%10 \% chance of hitting it. The least number of rounds, he must fire in order to have more than 50%50 \% chance of hitting it at least once, is
View answer and explanation
Correct option: C
7
Option C (7) is correct because it finds the minimum number of rounds needed for the probability of hitting the target at least once to be greater than 50%.
625 Medium
625 of 1130
Application +4 | -1
Two different families AA and BB are blessed with equal number of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket.
If the probability that all the tickets go to the children of the family BB is 112\frac{1}{12}, then the number of children in each family is?
View answer and explanation
Correct option: D
5
The correct answer is D (5) because the probability equation is set up correctly, and solving for the number of children in each family yields 5.
626 Easy
626 of 1130
Application +4 | -1
A fair die is rolled. The probability that the first time 1 occurs at the even throw is
View answer and explanation
Correct option: B
511\frac{5}{{11}}
Option B (5/11) is correct because it represents the sum of an infinite geometric series where the first 1 occurs on an even throw.
627 Easy
627 of 1130
Analyzing +4 | -1
If AA and BB are two independent events such that P(Bˉ)=αP(Aˉ)=710,P(Bˉ)=αP(\bar{B})=\alpha P(\bar{A})=\frac{7}{10}, P(\bar{B})=\alpha and P(A∪B)=810P(A \cup B)=\frac{8}{10}, then α\alpha, is
View answer and explanation
Correct option: A
2/72 / 7
Option A (2/7) is correct because it is the value of α\alpha that satisfies the given probability conditions for independent events A and B.
628 Medium
628 of 1130
Application +4 | -1
The probability of a man hitting a target is 25\frac{2}{5}. He fires at the target kk times ( kk, a given number). Then the minimum kk, so that the probability of hitting the target at least once is more than 710\frac{7}{10}, is :
View answer and explanation
Correct option: A
3
Option A (3) is correct because it is the minimum number of times the man must fire to have a probability greater than 7/10 of hitting the target at least once.
629 Hard
629 of 1130
Analyzing +4 | -1
Fifteen football players of a club-team are given 15 T-shirts with their names written on the backside. If the players pick up the T-shirts randomly, then the probability that at least 3 players pick the correct T- shirt is
View answer and explanation
Correct option: A
16\frac{1}{6}
Option A (16\frac{1}{6}) is the correct answer because it represents the calculated probability of at least 3 players picking the correct T-shirt using derangements.
630 Easy
630 of 1130
Analyzing +4 | -1
Let A,B,CA, B, C be pairwise independent events with P(C)>0P(C)>0 and P(A∩B∩C)=0P(A \cap B \cap C)=0. Then, P(Ac∩Bc/C)P\left(A^c \cap B^c / C\right) is equal to
View answer and explanation
Correct option: C
P(Ac)−P(B)P\left(A^c\right)-P(B)
Option C (P(Ac)−P(B)P(A^c)-P(B)) is the correct answer because it correctly applies the properties of pairwise independent events and conditional probability to simplify the expression for P(Ac∩Bc/C)P(A^c \cap B^c / C).
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