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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 781-810 of 1130 questions

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781 Easy
781 of 1130
Understanding +4 | -1
A and B play a game where each is asked to select a number from 1 to 25. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial is
View answer and explanation
Correct option: B
24/2524 / 25
Option B (24/25) is correct because it represents the complement of the probability that they will win a prize, which is 1/25.
782 Easy
782 of 1130
Application +4 | -1
A random variable XX takes values 0, 1, 2, 3,… with probability P(X=x)=k(x+1)(15)x,P\left( {X = x} \right) = k\left( {x + 1} \right){\left( {\frac{1}{5}} \right)^x}, where kk is constant, then P(X=0)P\left( {X = 0} \right) is
View answer and explanation
Correct option: D
1625\frac{{16}}{{25}}
Option D (1625\frac{{16}}{{25}}) is correct because it is the value of P(X=0) that satisfies the condition for a valid probability distribution.
783 Medium
783 of 1130
Application +4 | -1
A fair die is thrown twenty times. The probability that on the tenth throw the fourth six appears, is
View answer and explanation
Correct option: C
84×56610\frac{84 \times 5^6}{6^{10}}
Option C (84×56610\frac{{84 \times 5^6}}{{6^{10}}}) is correct because it calculates the probability of getting the fourth six on the tenth throw.
784 Medium
784 of 1130
Application +4 | -1
When a missile is fired from a ship, the probability that it is intercepted is 13\frac{1}{3} and the probability that the missile hits the target, given that it is not intercepted, is 34\frac{3}{4}. If three missiles are fired independently from the ship, then the probability that all three hit the target, is
View answer and explanation
Correct option: C
18\frac{1}{8}
Option C (1/8) is correct. It involves calculating the probability of a missile hitting the target and then finding the probability of all three missiles hitting the target.
785 Easy
785 of 1130
Understanding +4 | -1
If A\mathrm{A} and B\mathrm{B} are independent events, then which of the following is not true?
View answer and explanation
Correct option: C
P(A/B)=P(B/A)\mathrm{P}(\mathrm{A} / \mathrm{B})=\mathrm{P}(\mathrm{B} / \mathrm{A})
Option C (P(A/B)=P(B/A)\mathrm{P}(\mathrm{A} / \mathrm{B})=\mathrm{P}(\mathrm{B} / \mathrm{A})) is not true because for independent events, P(A/B) = P(A) and P(B/A) = P(B), and in general, P(A) is not equal to P(B).
786 Medium
786 of 1130
Application +4 | -1
India play two matches each with West Indies and Australia. In any match the probabilities of India getting points 0,1 and 2 are 0.45,0.050.45,0.05 and 0.500.50, respectively. Assuming that the outcomes are independent, then the probability of India getting atleast 7 points, is
View answer and explanation
Correct option: B
0.08750.0875
Option B is correct because the probability is calculated by considering the two possible scenarios: all four matches result in 2 points, or three matches result in 2 points and one match results in 1 point.
787 Easy
787 of 1130
Understanding +4 | -1
If AA and BB are two mutually exclusive events, then
View answer and explanation
Correct option: A
P(A)≤P(Bˉ)P(A) \leq P(\bar{B})
Option A is correct because if A and B are mutually exclusive, then A is a subset of the complement of B, meaning P(A) is less than or equal to P(B').
788 Medium
788 of 1130
Application +4 | -1
A die is rolled three times, if pp be the probability of getting a large number than the previous number, then the value of 54p54 p is
View answer and explanation
Correct option: A
5
Option A (5) is correct because it calculates the probability of getting a larger number than the previous number when a die is rolled three times and then multiplies by 54.
789 Easy
789 of 1130
Application +4 | -1
Three numbers are chosen at random without replacement from {1,2,3,…,10}\{1,2,3, \ldots, 10\}. The probability that minimum of the chosen number is 3 or their maximum is 7, is
View answer and explanation
Correct option: B
1140\frac{11}{40}
Option B (1140\frac{{11}}{{40}}) is correct because it represents the probability that the minimum of the chosen numbers is 3 or their maximum is 7, calculated using the principle of inclusion-exclusion.
790 Medium
790 of 1130
Application +4 | -1
If the probability that the random variable XX takes values xx is given by P(X=x)=k(x+1)3−xP(X=x)=k(x+1) 3^{-x}, x=0,1,2,3,…x=0,1,2,3, \ldots. , where kk is a constant, then P(X≥2)P(X \geq 2) is equal to
View answer and explanation
Correct option: A
727\frac{7}{27}
Option A (727\frac{7}{27}) is correct because it is the probability of X≥2X \geq 2 after finding the constant k and using the given probability distribution.
791 Easy
791 of 1130
Application +4 | -1
A random variable XX takes the values 0, 1,2, 3, ..., with probability P(X=x)=k(x+1)(15)xP(X=x)=k(x+1)\left(\frac{1}{5}\right)^x, where kk is aa constant, then P(X=0)P(X=0) is
View answer and explanation
Correct option: B
1625\frac{16}{25}
Option B (1625\frac{16}{25}) is correct because it's the probability calculated after finding the constant k using the property that the sum of all probabilities equals 1.
792 Easy
792 of 1130
Application +4 | -1
If XX follows a binomial distribution with parameters n=8n=8 and p=12p=\frac{1}{2}, then p(∣x−4∣≤2)p(|x-4| \leq 2) is equal to
View answer and explanation
Correct option: B
119128\frac{119}{128}
Option B (119128\frac{119}{128}) is correct because it calculates the probability of |x-4| <= 2 for a binomial distribution with n=8 and p=1/2 by summing the probabilities for x=2, 3, 4, 5, and 6.
793 Easy
793 of 1130
Application +4 | -1
Four persons independently solve a certain problem correctly with probabilities 12,34,14\frac{1}{2}, \frac{3}{4}, \frac{1}{4} and 18\frac{1}{8}. Then, the probability that the problem is solved correctly by atleast one of them, is
View answer and explanation
Correct option: A
235256\frac{235}{256}
Option A (235256\frac{{235}}{{256}}) is correct because it calculates the probability that at least one person solves the problem by subtracting the probability that none of them solve it from 1.
794 Easy
794 of 1130
Application +4 | -1
The minimum number of times one has to toss a fair coin so that the probability of observing atleast one head is atleast 90% is
View answer and explanation
Correct option: D
4
Option D (4) is correct because at least 4 tosses are required to have a probability of at least 90% of observing at least one head.
795 Medium
795 of 1130
Application +4 | -1
The probability of India winning a test match against West-Indies is 12\frac{1}{2} assuming independence from match to match the probability that in a match series India’s second win occurs at the third test, is
View answer and explanation
Correct option: B
14\frac{1}{4}
Option B is correct because the probability is calculated by considering the two possible sequences of wins and losses that result in the second win occurring at the third test.
796 Medium
796 of 1130
Application +4 | -1
Four dice are rolled. The probability that the same number will appear on each of them is .
View answer and explanation
Correct option: C
163\frac{1}{6^3}
Option C (163\frac{1}{6^3}) is correct because it represents the probability of all four dice showing the same number, which is 6 favorable outcomes out of 646^4 total outcomes.
797 Medium
797 of 1130
Application +4 | -1
Past records reveal that during a particular was, out of 9 vessels expected to arrive at the Mumbai harbour exactly 7 reached the harbour safely. If 3 vessels were expected to arrive there on a particular data, the probability that exactly two would arrive at the harbour safely, is
View answer and explanation
Correct option: D
None of these
None of the provided options are correct. The probability that exactly two vessels arrive safely is calculated using the binomial distribution formula, resulting in 98/243.
798 Medium
798 of 1130
Application +4 | -1
These are four balls of different colours and four boxes of colours, same as those of the balls. The number of ways in which the balls, one each in a box, could be placed such that a ball does not go to a box of its own colour is
View answer and explanation
Correct option: B
38\frac{3}{8}
The provided answer and explanation are incorrect. The correct approach involves derangements. Option D (None of these) is most likely the correct answer because the probability is not 3/8.
799 Easy
799 of 1130
Application +4 | -1
The probability that a marksman will hit a target is given as 1/5.1 / 5. Then the probablity that at least one hit in 10 shots is
View answer and explanation
Correct option: A
1−(45)101-\left(\frac{4}{5}\right)^{10}
Option A (1−(45)101-(\frac{4}{5})^{10}) is correct because it's the probability of at least one hit, calculated as 1 minus the probability of no hits.
800 Easy
800 of 1130
Application +4 | -1
If P(A)=12,P(B)=35,P(B∩A)=310\mathrm{P}(\mathrm{A})=\frac{1}{2}, \mathrm{P}(\mathrm{B})=\frac{3}{5}, \mathrm{P}(\mathrm{B} \cap \mathrm{A})=\frac{3}{10}, then P(B/A)\mathrm{P}(\mathrm{B} / \mathrm{A}) is equal to
View answer and explanation
Correct option: D
35\frac{3}{5}
Option D (35\frac{3}{5}) is correct because it is the result of applying the conditional probability formula P(B/A) = P(B∩A) / P(A).
801 Easy
801 of 1130
Application +4 | -1
An unbiased coin is tossed. If the outcome is head then a pair of unbiased dice is rolled and the sum of the numbers obtained on them is noticed. If the coin results in a tail then a card from a well shuffled pack of nine cards numbered 1,2,3,…,91,2,3, \ldots, 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 , is
View answer and explanation
Correct option: A
1972\frac{19}{72}
Option A (1972\frac{{19}}{{72}}) is correct because it calculates the probability of noting a 7 or 8 by considering both cases: getting a head (rolling dice) and getting a tail (picking a card), and then summing the probabilities.
802 Easy
802 of 1130
Application +4 | -1
The probability that a man will hit a target in shooting practice is 0.3. If he shoots 10 times, the probability that he hits the target is
View answer and explanation
Correct option: B
1−(0.7)101 - {\left( {0.7} \right)^{10}}
Option B (1−(0.7)101 - {\left( {0.7} \right)^{10}}) is correct because it represents the complement of the probability that he never hits the target in 10 trials.
803 Easy
803 of 1130
Application +4 | -1
If the probability of hitting a target by a shooter, in any shot, is 1/31 / 3, then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than 5/65 / 6, is
View answer and explanation
Correct option: B
5
Option B (5) is correct because it finds the minimum number of shots required such that the probability of hitting the target at least once is greater than 5/6.
804 Easy
804 of 1130
Application +4 | -1
Fifteen persons, among whom are AA and BB, sit down at random at a round table. The probability that there are 4 persons between AA and BB is
View answer and explanation
Correct option: D
none of these
Option B (59\frac{{5}}{{9}}) is correct because it calculates the probability that all three dice show distinct numbers.
805 Easy
805 of 1130
Application +4 | -1
Two unbiased die are rolled together.

Let A={(a,b):a+b=11}A=\{(a, b): a+b=11\} and B={(a,b):a≠5}B=\{(a, b) : a \neq 5\}, then P(A∣B)P(A \mid B) equals
View answer and explanation
Correct option: A
1/301 / 30
Option A (1/30) is correct because it calculates the conditional probability P(A|B) using the formula P(A∩B) / P(B), where A is the event that the sum is 11 and B is the event that a ≠ 5.
806 Easy
806 of 1130
Application +4 | -1
The probability that a number nn chosen at random from 1 to 30, to satisfy n+(50/n)>27n + \left( {50/n} \right) > 27 is
View answer and explanation
Correct option: D
15\frac{1}{5}
Option D (15\frac{1}{5}) is correct because it calculates the probability of n satisfying the given inequality by finding the ratio of favorable outcomes to the total number of outcomes.
807 Easy
807 of 1130
Application +4 | -1
The outcome of each of 30 items was observed ; 10 items gave an outcome 12−d\frac{1}{2} - d each, 10 items gave outcome 12\frac{1}{2} each and the remaining 10 items gave outcome 12+d\frac{1}{2} + d each. If the variance of this outcome data is 43\frac{4}{3}, then ∣d∣\left| d \right| equals
View answer and explanation
Correct option: C
2\sqrt 2
Option C (2\sqrt 2) is correct because the variance equation is used to solve for |d|, resulting in the square root of 2.
808 Easy
808 of 1130
Application +4 | -1
Two integers are selected at random from integers 1 to 11. If the sum is even, the probability that both the numbers are odd, is
View answer and explanation
Correct option: A
35\frac{3}{5}
Option A (35\frac{{3}}{{5}}) is correct because it represents the conditional probability of both numbers being odd given that their sum is even.
809 Easy
809 of 1130
Application +4 | -1
Three numbers are chosen at random from 1 to 20. The probability that they are consecutive, is
View answer and explanation
Correct option: C
3190\frac{3}{{190}}
Option A is incorrect. Option C (3/190) is correct because it calculates the probability of selecting three consecutive numbers from 1 to 20.
810 Medium
810 of 1130
Application +4 | -1
Two cards are drawn from a pack of 52 cards. What is the probability that at least one of the cards drawn is an ace
View answer and explanation
Correct option: A
33221\frac{33}{221}
Option A (33221\frac{33}{221}) is correct. It calculates the probability of drawing at least one ace by finding the complement (probability of drawing no aces) and subtracting it from 1.
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