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JEE • MATHEMATICS • PROBABILITY

Probability practice · Easy

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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 691-720 of 822 questions

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691 Easy
691 of 822
Application +4 | -1
The probability that out of 10 persons, all born in June, at least two have the same birth day is
View answer and explanation
Correct option: C
3010−30C10(30)10\frac{30^{10}-{ }^{30} C_{10}}{(30)^{10}}
Option C (3010−30C10(30)10\frac{30^{10}-{ }^{30} C_{10}}{(30)^{10}}) is correct because it represents the probability that at least two out of 10 persons born in June have the same birthday.
692 Easy
692 of 822
Application +4 | -1
Three numbers are chosen at random (without replacement) from the numbers 1,2,…201,2, \ldots 20. The probability that the numbers are not consecutive are
View answer and explanation
Correct option: D
187190\frac{187}{190}
Option D (187/190) is correct because it calculates the probability that the three numbers chosen are not consecutive by subtracting the probability of choosing consecutive numbers from 1.
693 Easy
693 of 822
Application +4 | -1
If two coins are tossed five times, then the probability of getting 5 heads and 5 tails is
View answer and explanation
Correct option: A
63/25663 / 256
The correct answer is 63/25663 / 256 because it calculates the probability of getting 5 heads and 5 tails when tossing two coins five times.
694 Easy
694 of 822
Application +4 | -1
The mode of the binomial distribution for which mean and standard deviation are 10 and 5  \sqrt 5 \; respectively, is
View answer and explanation
Correct option: D
10
Option D (10) is correct because the mode of a binomial distribution lies within the interval (np - q, np + p), and in this case, it equals 10.
695 Easy
695 of 822
Application +4 | -1
If the letters of the word PROBABILITY are written down at random in a row, the probability that two Bs are together is
View answer and explanation
Correct option: A
211\frac{2}{11}
Option A (211\frac{2}{11}) is correct because it represents the probability of the two Bs being together when arranging the letters of the word PROBABILITY.
696 Easy
696 of 822
Application +4 | -1
If 1−3p2,1+4p3,1+p6\frac{1-3 p}{2}, \frac{1+4 p}{3}, \frac{1+p}{6} are the probabilities of three mutually exclusive and exhaustive events, then the set of all values of pp is
View answer and explanation
Correct option: B
[−1/4,1/3][-1 / 4,1 / 3]
Option B ([−14,13]\left[-\frac{{1}}{{4}}, \frac{{1}}{{3}}\right]) is correct because it determines the range of p based on the conditions that probabilities must be non-negative and sum to 1.
697 Easy
697 of 822
Application +4 | -1
A card is drawn from a pack of cards. The probability that the card will be a queen or a heart, is
View answer and explanation
Correct option: C
413\frac{4}{{13}}
Option C (413\frac{4}{{13}}) is correct because it calculates the probability of drawing a queen or a heart using the formula for the union of two events.
698 Easy
698 of 822
Analyzing +4 | -1
Which one of the following is not correct?
View answer and explanation
Correct option: C
The probability guessing at least 8 out 10 answers in a true false examination, is 764\frac{7}{{64}}
Option C (The probability guessing at least 8 out 10 answers in a true false examination, is 764\frac{7}{{64}}) is not correct because the correct probability is 11512\frac{11}{512}
699 Easy
699 of 822
Application +4 | -1
All the heart cards are taken out from a pack of playing cards. From these cards, cards are drawn one by one without replacement till the ace of hearts comes. The probability that ace comes in the 4th draw is
View answer and explanation
Correct option: A
113\frac{1}{13}
Option A (113\frac{1}{13}) is correct because it calculates the probability of drawing 3 non-ace heart cards followed by the ace of hearts.
700 Easy
700 of 822
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.
701 Easy
701 of 822
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.
702 Easy
702 of 822
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.
703 Easy
703 of 822
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?
View answer and explanation
Correct option: D
1
Option D (1) is correct because the article chosen must be either good or defective, covering all possibilities.
704 Easy
704 of 822
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.
705 Easy
705 of 822
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.
706 Easy
706 of 822
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.
707 Easy
707 of 822
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).
708 Easy
708 of 822
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.
709 Easy
709 of 822
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.
710 Easy
710 of 822
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.
711 Easy
711 of 822
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.
712 Easy
712 of 822
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)
View answer and explanation
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.
713 Easy
713 of 822
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.
714 Easy
714 of 822
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.
715 Easy
715 of 822
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.
716 Easy
716 of 822
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.
717 Easy
717 of 822
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.
718 Easy
718 of 822
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.
719 Easy
719 of 822
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.
720 Easy
720 of 822
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.
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