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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 961-990 of 1130 questions

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961 Easy
961 of 1130
Application +4 | -1
Two events EE and FF are independent. If P(E)=0.3P(E)=0.3, P(E∪F)=0.5P(E \cup F)=0.5, then P(E∣F)−P(F∣E)P(E \mid F)-P(F \mid E) equals
View answer and explanation
Correct option: C
170\frac{1}{70}
Option C (170\frac{1}{70}) is correct because it is the difference between P(E|F) and P(F|E) calculated using the independence of E and F.
962 Easy
962 of 1130
Analyzing +4 | -1
The probability that a randomly chosen one-one function from the set {a,b,c,d}\{a, b, c, d\} to the set {1,2,3\{1,2,3, 4,5}4,5\} satisfies f(a)+2f(b)−f(c)=f(d)f(a)+2 f(b)-f(c)=f(d) is :
View answer and explanation
Correct option: D
120\frac{1}{20}
Option D (120\frac{1}{20}) is correct because it is calculated by dividing the number of favorable one-to-one functions by the total number of one-to-one functions.
963 Easy
963 of 1130
Application +4 | -1
A dice is thrown 100 times, getting an even number is considered a success. The variance of the number of successes is
View answer and explanation
Correct option: B
25
Option B (25) is correct because it correctly calculates the variance for a binomial distribution where getting an even number is considered a success.
964 Easy
964 of 1130
Understanding +4 | -1
A box contains 10 good articles and 6 with defects, one article is chosen at random. What is the probability that it is either good or has a defect?
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Correct option: D
1
Option D (1) is correct because the article chosen must be either good or defective, covering all possibilities.
965 Medium
965 of 1130
Analyzing +4 | -1
Consider the polynomial P(x)=x3+ax2+bx+cP(x)=x^3+a x^2+b x+c where a,b,c∈{1,2,……..,10}a, b, c \in\{1,2, \ldots \ldots . ., 10\} then the probability that P(x)P(x) is divisible by x2+x+1x^2+x+1 is
View answer and explanation
Correct option: B
91000\frac{9}{1000}
Option B (91000\frac{9}{1000}) is correct because the probability is calculated by finding the number of favorable outcomes (where P(x) is divisible by x2+x+1x^2+x+1) and dividing it by the total number of possible outcomes.
966 Easy
966 of 1130
Understanding +4 | -1
Unbiased die is thrown, probability that outcome is greater than 4, is
View answer and explanation
Correct option: C
13\frac{1}{3}
Option C (13\frac{1}{3}) is correct because there are 2 outcomes greater than 4 (5 and 6) out of a total of 6 possible outcomes, so the probability is 2/6 = 1/3.
967 Easy
967 of 1130
Understanding +4 | -1
Four persons A,B,C,DA, B, C, D are to speak at a function along with 6 others. If they all speak in random order, the probability that AA speaks before B,BB, B before C,CC, C before DD is
View answer and explanation
Correct option: C
1/241 / 24
Option C (1/241/24) is correct because there are 4! = 24 ways to arrange A, B, C, and D, and only one of those arrangements has them in the correct order.
968 Easy
968 of 1130
Application +4 | -1
If the probability that AA and BB will die with in a year are pp and qq respectively, then the probability that only one of them will be alive at the end of the year is
View answer and explanation
Correct option: B
p+q−2pqp + q - 2pq
Option B (p+q−2pqp + q - 2pq) is correct because it represents the probability that only one of A or B will die within a year.
969 Easy
969 of 1130
Application +4 | -1
Let AA and BB be two independent events such that P(A)=13P\left( A \right) = \frac{1}{3} and P(B)=16.P\left( B \right) = \frac{1}{6}. Then, which of the following is TRUE?
View answer and explanation
Correct option: C
P(A/B′)=13P\left( {A/B'} \right) = \frac{1}{3}
Option C (P(A/B′)=13P\left( {A/B'} \right) = \frac{1}{3}) is correct because, for independent events, the conditional probability P(A∣B′)P(A|B') is equal to P(A)P(A).
970 Easy
970 of 1130
Application +4 | -1
An unbiased coin is tossed fixed number of times. If the probability of getting 4 heads equals the probability of getting 7 heads, then the probability of getting 2 heads is
View answer and explanation
Correct option: A
55/2048
Option A (55/2048) is correct because it correctly uses the binomial distribution formula and the given condition to find the number of trials and then calculates the probability of getting 2 heads.
971 Medium
971 of 1130
Application +4 | -1
A sample space consists of 3 sample points with associated probabilities given as 2p,p2,4p−12 p, p^2, 4 p-1. Then the value of pp is -
View answer and explanation
Correct option: A
p=11−3\mathrm{p}=\sqrt{11}-3
Option D (p=11−3\mathrm{p}=\sqrt{11}-3) is correct because the sum of probabilities of all sample points in a sample space must equal 1, leading to a quadratic equation whose solution gives the value of p.
972 Easy
972 of 1130
Application +4 | -1
Three numbers are chosen at random from the numbers 1,2,…201,2, \ldots 20. The probability that the arithmetic mean of these numbers is 4 is
View answer and explanation
Correct option: A
7/22807 / 2280
Option A (7/22807 / 2280) is correct because it represents the probability that the arithmetic mean of three randomly chosen numbers from 1 to 20 is 4.
973 Medium
973 of 1130
Application +4 | -1
The probability of event AA occurring is 0.50.5 and B occurring is 0.3. If AA and BB are mutually exclusive events, then the probability of neither AA nor BB occurring is
View answer and explanation
Correct option: D
none of these
The correct answer is D (none of these) because the probability of neither A nor B occurring, when A and B are mutually exclusive with probabilities 0.5 and 0.3 respectively, is 0.2.
974 Easy
974 of 1130
Application +4 | -1
Three persons P,Q\mathrm{P}, \mathrm{Q} and R\mathrm{R} independently try to hit a target. If the probabilities of their hitting the target are 34,12\frac{3}{4}, \frac{1}{2} and 58\frac{5}{8} respectively, then the probability that the target is hit by P\mathrm{P} or Q\mathrm{Q} but not by R\mathrm{R} is :
View answer and explanation
Correct option: A
2164\frac{21}{64}
Option A (21/64) is correct because it calculates the probability of P or Q hitting the target but not R, considering all possible combinations.
975 Easy
975 of 1130
Application +4 | -1
Two cards are drawn one by one at random from a pack of 52 cards. The prohahility that both of them are king, is
View answer and explanation
Correct option: C
1221\frac{1}{221}
Option C (1221\frac{1}{221}) is correct because it is the probability of drawing two kings in a row without replacement from a standard deck of cards.
976 Easy
976 of 1130
Application +4 | -1
Two cards are drawn one by one from a pack of cards. The probability of getting first card an ace and second a coloured one is (before drawing second card first card is not placed again in the pack)
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Correct option: C
5/221
Option C (5/221) is correct because it calculates the probability of drawing an ace first and then a colored card without replacement.
977 Medium
977 of 1130
Application +4 | -1
A fair coin is tossed repeatedly until a head is obtained. If X\mathrm{X} denotes the number of tosses required, then the conditional probability of X≥6X \geq 6, given that X>3X>3 is
View answer and explanation
Correct option: B
14\frac{1}{4}
Option B (14\frac{1}{4}) is correct because it is the conditional probability calculated based on the geometric distribution of coin tosses.
978 Easy
978 of 1130
Application +4 | -1
Three numbers are chosen from 1 to 30. The probability that they are not consecutive is
View answer and explanation
Correct option: A
144145\frac{{144}}{{145}}
Option A (144145\frac{{144}}{{145}}) is correct because it represents the probability of choosing three non-consecutive numbers from 1 to 30, calculated by subtracting the probability of choosing consecutive numbers from 1.
979 Easy
979 of 1130
Application +4 | -1
In a class of 80 students numbered 1 to 80 , all odd numbered students opt of Cricket, students whose numbers are divisible by 5 opt for Football and those whose numbers are divisible by 7 opt for Hockey. If the number of students who do not opt any of the three games is nn, then n4\frac{n}{4} is equal to
View answer and explanation
Correct option: A
7
Option A (7) is correct because it represents the value of n4\frac{{n}}{{4}} where nn is the number of students who do not opt for any of the three games, calculated using set theory principles.
980 Easy
980 of 1130
Application +4 | -1
Three numbers are chosen at random from numbers 1 to 30 . The probability that the minimum of the chosen numbers is 9 and maximum is 25 , is
View answer and explanation
Correct option: C
3812\frac{3}{812}
Option C (3812\frac{3}{812}) is correct because it represents the probability that the minimum of three chosen numbers is 9 and the maximum is 25 when choosing from numbers 1 to 30.
981 Easy
981 of 1130
Application +4 | -1
Two squares are chosen at random on a chessboard. If pp denotes the probability that they have exactly one side in common, then 144p144 p is
View answer and explanation
Correct option: C
7
Option C (7) is correct because the probability that two randomly chosen squares on a chessboard have exactly one side in common is 7/144, and 144p = 7.
982 Medium
982 of 1130
Application +4 | -1
A group consists 2n2 n students including you and your friend. If the group is split into two different sections AA and BB, each containing nn students, then the probability that you and your friend are in the different sections 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 the probability that you and your friend are in different sections is calculated by dividing the number of favorable arrangements by the total number of arrangements.
983 Easy
983 of 1130
Application +4 | -1
If A={x:x=4n+1,n≤5,n∈N}A=\{x: x=4 n+1, n \leq 5, n \in N\} and B{3n:n≤8B\{3 n: n \leq 8, n∈N}n \in N\}, then find (A−(A−B))(A-(A-B)).
View answer and explanation
Correct option: A
{9,21}\{9,21\}
The correct answer is A ({9,21}) because A - (A - B) simplifies to A ∩\cap B, which is the set of elements common to both A and B.
984 Medium
984 of 1130
Analyzing +4 | -1
10 different books and 2 different pens are given to 3 boys so that each gets equal number of things. The probability that the some boy does not receive both the pens is
View answer and explanation
Correct option: A
5/115/11
Option A (5/11) is correct because it calculates the probability that no single boy receives both pens by subtracting the probability that one boy receives both pens from 1.
985 Medium
985 of 1130
Application +4 | -1
A five digit mumber is formed by the digits 1, 2, 3, 4, 5,6 and 8. The probability that the number has even digit at both ends is
View answer and explanation
Correct option: C
27\frac{2}{7}
Option C (27\frac{2}{7}) is correct because the probability is calculated by finding the ratio of favorable outcomes (numbers with even digits at both ends) to the total possible outcomes.
986 Easy
986 of 1130
Application +4 | -1
Events A,B,CA, B, C are mutually exclusive events such that P(A)=3x+13,P(B)=1−x4P(A)=\frac{3 x+1}{3}, P(B)=\frac{1-x}{4} and P(C)=1−2x2P(C)=\frac{1-2 x}{2}. The set of all possible values of xx are in the interval
View answer and explanation
Correct option: B
[−1/3,1/2][-1 / 3,1 / 2]
Option B ([−13,12]\left[-\frac{{1}}{{3}}, \frac{{1}}{{2}}\right]) is correct because it determines the range of x based on the conditions that probabilities must be non-negative and their sum must be less than or equal to 1.
987 Easy
987 of 1130
Application +4 | -1
One hundred identical coins, each with probability pp of showing heads are tossed once. If 0<p<10 < p < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, the value of pp, is
View answer and explanation
Correct option: B
51/10151 / 101
Option B (51/10151 / 101) is correct because it uses the binomial distribution and the condition P(X=50) = P(X=51) to solve for p.
988 Easy
988 of 1130
Application +4 | -1
A random variable XX has poisson distribution with mean 2. Then, P(X>1.5)P(X > 1.5) equals
View answer and explanation
Correct option: B
1−3e21 - \frac{3}{{{e^2}}}
Option B (1−3e21 - \frac{3}{{{{e^2}}}}) is correct because it calculates the probability of X > 1.5 using the Poisson distribution with a mean of 2.
989 Easy
989 of 1130
Application +4 | -1
If the sum and the product of mean and variance of a binomial distribution are 24 and 128 respectively, then the probability of one or two successes is:
View answer and explanation
Correct option: C
33228\frac{33}{2^{28}}
Option C (33228\frac{33}{2^{28}}) is correct because it is the probability calculated from the binomial distribution with n=32 and p=1/2 for one or two successes.
990 Easy
990 of 1130
Application +4 | -1
A number is chosen form each of two sets A={1,2,3,4,5,6,7,8}A=\{1,2,3,4,5,6,7,8\} and B={8,7,6,5,4,3,2,1}B=\{8,7,6,5,4,3,2,1\}.
If p1p_1 is the probability that the sum of the numbers is 9 and p2p_2 is the probability that sum of the numbers is 7 , then p1+p2p_1+p_2 is equal to
View answer and explanation
Correct option: D
732\frac{7}{32}
Option D (732\frac{7}{32}) is correct because it represents the sum of the probabilities of getting a sum of 9 and a sum of 7 when choosing one number from each of the given sets.
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