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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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421 Easy
421 of 822
Analyzing +4 | -1
A biased die is marked with numbers 2, 4, 8, 16, 32, 32 on its faces and the probability of getting a face with mark n is 1/n. If the die is thrown thrice, then the probability, that the sum of the numbers obtained is 48, is:
View answer and explanation
Correct option: D
13212\frac{{13}}{{{2^{12}}}}
Option D (13212\frac{{13}}{{{{2^{12}}}}}) is correct because it represents the probability of obtaining a sum of 48 when the biased die is thrown thrice, considering the two possible combinations and their probabilities.
422 Easy
422 of 822
Analyzing +4 | -1
Let S={1,2,3,…,2022}S=\{1,2,3, \ldots, 2022\}. Then the probability, that a randomly chosen number nn from the set SS such that HCF⁡(n,2022)=1\operatorname{HCF}(n, 2022)=1, is
View answer and explanation
Correct option: D
112337\frac{112}{337}
Option D (112337\frac{112}{337}) is correct because it represents the probability of choosing a number from S that is relatively prime to 2022, calculated using the principle of inclusion-exclusion.
423 Easy
423 of 822
Application +4 | -1
A student takes his examination in four subjects α,β\alpha, \beta, γ,δ\gamma, \delta. He estimates his chance of passing in α\alpha as 45\frac{4}{5}, in β\beta as 34\frac{3}{4}, in γ\gamma as 56\frac{5}{6} and in δ\delta as 23\frac{2}{3}. To qualify he must pass in α\alpha and at least two other subjects. The Probability that he qualifies is
View answer and explanation
Correct option: B
6190\frac{61}{90}
Option B (6190\frac{61}{90}) is correct because it sums the probabilities of all qualifying scenarios where the student passes in alpha and at least two other subjects.
424 Easy
424 of 822
Application +4 | -1
Four positive integers are taken at random and are multiplied together. Then the probability that the product ends in an odd digit other than 5, is
View answer and explanation
Correct option: B
16/625
Option C (17\frac{1}{7}) is correct because it calculates the probability of drawing balls of alternating colors by considering both possible sequences (WBWB and BWBW) and summing their probabilities.
425 Easy
425 of 822
Application +4 | -1
If mm rupee coins and nn ten paise coins are placed in a line, then the probability that the extreme coins are ten paise coins, is
View answer and explanation
Correct option: B
n(n−1)(m+n)(m+n−1)\frac{{n\left( {n - 1} \right)}}{{\left( {m + n} \right)\left( {m + n - 1} \right)}}
Option B (n(n−1)(m+n)(m+n−1)\frac{{n(n - 1)}}{{(m + n)(m + n - 1)}}) is correct because it represents the probability of selecting a ten paise coin for both extreme positions out of the total possible arrangements.
426 Easy
426 of 822
Application +4 | -1
Two cards are drawn successively with replacement from a well shuffled deck of 52 cards, then the mean of the number of aces is
View answer and explanation
Correct option: C
213\frac{2}{{13}}
Option C (213\frac{2}{{13}}) is correct because it calculates the mean number of aces by considering the probability of drawing an ace in each draw with replacement.
427 Easy
427 of 822
Application +4 | -1
Let E1,E2{E_1},{E_2} be two mutually exclusive events of an experiment with PP(not E2{E_2}) =0.6=P(E1∪E2)= 0.6 = P\left( {{E_1} \cup {E_2}} \right).Then, P(E1)P\left( {{E_1}} \right) is equal to
View answer and explanation
Correct option: D
0.20.2
Option D (0.20.2) is correct because it is calculated using the properties of mutually exclusive events and the given probabilities.
428 Easy
428 of 822
Application +4 | -1
A four digit number is formed by the digits 1, 2, 3, 4 with no repetition. The probability that the number is odd, is
View answer and explanation
Correct option: A
0.5
Option A (0.5) is correct because it calculates the probability by dividing the number of odd numbers formed by the total number of possible numbers.
429 Easy
429 of 822
Understanding +4 | -1
Assertion: Probability of getting a head in a toss of an unbiased coin is 12\frac{1}{2}.

Reason: In a simultaneous toss of two coins, the probability of getting 'no tails' is 14\frac{1}{4}.
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a correct explanation for assertion
Option B (Assertion is correct, reason is correct; reason is not a correct explanation for assertion) is correct because both the probability of getting a head is 1/2 and the probability of getting no tails in two coin tosses is 1/4 are correct, but they are unrelated.
430 Easy
430 of 822
Application +4 | -1
AA and BB are two independent events. The probability that both AA and BB occur is 1/61/6 and the probability that neither of them occurs is 1/31/3. Then,
View answer and explanation
Correct option: A
P(A)=1/2,P(B)=1/3P\left( A \right) = 1/2,P\left( B \right) = 1/3
Option A (P(A)=1/2,P(B)=1/3P(A) = 1/2, P(B) = 1/3) is correct because it satisfies the conditions for independent events and the given probabilities.
431 Easy
431 of 822
Application +4 | -1
A bag contains 16 coins of which two are counterfeit with heads on both sides. The rest are fair coins. One is selected at random from the bag and tossed. The probability of getting a head is
View answer and explanation
Correct option: A
9/169 / 16
Option A (9/169 / 16) is correct because it uses the law of total probability to calculate the probability of getting a head.
432 Easy
432 of 822
Analyzing +4 | -1
If AA and BB are independent events of aa random experiment such that P(A∩B)=16P(A \cap B)=\frac{1}{6} and P(Aˉ∩Bˉ)=13P(\bar{A} \cap \bar{B})=\frac{1}{3}, then P(A)P(A) is equal to
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{1}{3}) is correct because solving the system of equations derived from the independence and complement properties yields P(A)=13P(A) = \frac{1}{3}.
433 Easy
433 of 822
Application +4 | -1
A bag XX contains 2 white and 3 black balls and another bag YY contains 4 white and 2 black balls. One bag is selected at random and a ball is drawn from it. Then, probability for the ball chosen be white, is
View answer and explanation
Correct option: C
815\frac{8}{{15}}
Option C (8/15) is correct because it calculates the probability of drawing a white ball by considering the probabilities of selecting each bag and then drawing a white ball from that bag.
434 Easy
434 of 822
Application +4 | -1
On each evening a boy either watches DOORDARSHAN channel or TENSPORTS. The probability that he watches TEN SPORTS is 45\frac{4}{5}. If he watches DOORDARSHAN, there is a chance of 34\frac{3}{4} that he will be asleep, while it is 14\frac{1}{4} when he watches TEN SPORTS. On one day, the boy is found to be asleep. The probability that the boy watched DOORDARSHAN is
View answer and explanation
Correct option: C
37\frac{3}{7}
Option C (37\frac{3}{7}) is correct because it is the conditional probability that the boy watched DOORDARSHAN given that he was found asleep.
435 Easy
435 of 822
Analyzing +4 | -1
Five different objects A1, A2, A3, A4, A5{A_1},\,{A_2},\,{A_3},\,{A_4},\,{A_5} are distributed randomly in 5 places marked 1, 2, 3, 4, 5. One arrangement is picked at random. The probability that in the selected arrangement, none of the object occupies the place corresponding to its number, is
View answer and explanation
Correct option: C
Nov-30
Option D (Nov-30) is correct because it represents the probability that none of the objects occupies the place corresponding to its number, which is a derangement problem.
436 Easy
436 of 822
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.
437 Easy
437 of 822
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.
438 Easy
438 of 822
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.
439 Easy
439 of 822
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.
440 Easy
440 of 822
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.
441 Easy
441 of 822
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.
442 Easy
442 of 822
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.
443 Easy
443 of 822
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.
444 Easy
444 of 822
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:
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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.
445 Easy
445 of 822
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.
446 Easy
446 of 822
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.
447 Easy
447 of 822
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.
448 Easy
448 of 822
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.
449 Easy
449 of 822
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).
450 Easy
450 of 822
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.
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