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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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841 Easy
841 of 1130
Application +4 | -1
Let n(A−B)=25+x,n(B−A)=2xn(A-B)=25+x, n(B-A)=2 x and n(A∩B)=2xn(A \cap B)=2 x. If n(A)=2n(B)n(A)=2 n(B) then ' xx ' is
View answer and explanation
Correct option: D
5
Option D (5) is correct. The value of x is found by setting up equations based on the given information about the number of elements in different sets and solving for x.
842 Easy
842 of 1130
Application +4 | -1
7 white balls and 3 black balls are placed in a row at random. The probability that no two black balls are adjacent is
View answer and explanation
Correct option: B
715\frac{7}{15}
Option B (715\frac{{7}}{{15}}) is correct because it calculates the ratio of favorable arrangements (no two black balls adjacent) to the total possible arrangements of white and black balls.
843 Easy
843 of 1130
Application +4 | -1
If AA and BB are two events such that P(A)=25P(A)=\frac{2}{5} and P(A∩B)=320P(A \cap B)=\frac{3}{20}, then the conditional probability P(A/A′∪B′)P\left(A / A^{\prime} \cup B^{\prime}\right), where A′A^{\prime} denotes the complement of AA, is equal to
View answer and explanation
Correct option: C
517\frac{5}{17}
Option C (517\frac{5}{17}) is correct because it correctly applies the conditional probability formula and De Morgan's law.
844 Easy
844 of 1130
Analyzing +4 | -1
Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is :
View answer and explanation
Correct option: B
518\frac{5}{18}
Option A (1) is correct because the LHS and RHS of the equation are equal only when x=0, resulting in one element in the set.
845 Easy
845 of 1130
Application +4 | -1
The probability that a candidate secures a seat in Engineering through “EAMCET” is 1/10.1/10. 7 candidates are selected at random from a centre. The probability that exactly two will get seats is
View answer and explanation
Correct option: C
21  (0.1)2  (0.9)521\;{\left( {0.1} \right)^2}\;{\left( {0.9} \right)^5}
Option C (21(0.1)2(0.9)521{\left( {0.1} \right)^2}{\left( {0.9} \right)^5}) is correct because it uses the binomial probability formula with n=7, k=2, p=1/10, and q=9/10.
846 Medium
846 of 1130
Application +4 | -1
A sportsman's chance of shooting an animal at a distance r>ar > a ("a" is constant) is given to be a2r2\frac{a^2}{r^2}. He fires at r=2ar=2 a and if he misses, then again fires at r=3ar=3 a. He repeats the same process at r=4a,5ar=4 a, 5 a and 6a6 a. When he misses at r=6ar=6 a, the animal escapes into the jungle. If the odds against the sportsman are p:q{p}: {q}, then q−p{q}-{p} is ____.
View answer and explanation
Correct option: A
5
Option A (5) is correct because it is the difference between q and p, where p:q are the odds against the sportsman, calculated from the probability of the animal escaping.
847 Easy
847 of 1130
Application +4 | -1
The probability that the 6th day of a randomly chosen month of a year is a Sunday, is
View answer and explanation
Correct option: C
184\frac{1}{{84}}
Option C (1/84) is correct. The probability that the 6th day of a randomly chosen month is a Sunday is calculated by considering the probability of choosing a month and the probability that the month starts on a Tuesday.
848 Easy
848 of 1130
Application +4 | -1
Six coins are tossed simultaneously. The probability of getting at least 4 heads is
View answer and explanation
Correct option: B
Nov-32
Option A (Nov-32) is correct because it calculates the probability of getting at least 4 heads by summing the probabilities of getting 4, 5, or 6 heads.
849 Medium
849 of 1130
Analyzing +4 | -1
For two events AA and B,B, if P(A)=P(AB)=14   and  P(BA)=12,P\left( A \right) = P\left( {\frac{A}{B}} \right) = \frac{1}{4}\;\,\mathrm{and}\,\,P\left( {\frac{B}{A}} \right) = \frac{1}{2}, then
View answer and explanation
Correct option: C
P(B′A′)=12P\left( {\frac{{B'}}{{A'}}} \right) = \frac{1}{2}
Option D (P(B′A′)=12P(\frac{{B'}}{{A'}}) = \frac{1}{2}) is correct because after calculating the probabilities of the intersections and conditional probabilities, it satisfies the given conditions and independence.
850 Easy
850 of 1130
Application +4 | -1
A lottery sells n2n^2 tickets and declares nn prizes. If a man purchases nn tickets, the probability of his winning exactly one prize is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the provided options do not match the correct probability calculation for winning exactly one prize.
851 Medium
851 of 1130
Application +4 | -1
From a well-shuffled pack of 52 cards, four cards are selected at random. The probability of drawing exactly 2 spades and exactly 2 aces is
View answer and explanation
Correct option: A
1494270725\frac{1494}{270725}
Option A (1494270725\frac{1494}{270725}) is correct because it calculates the probability of drawing exactly 2 spades and 2 aces considering both cases: one spade ace and two non-spade aces.
852 Easy
852 of 1130
Application +4 | -1
A lot consists of 102 good pencils, 6 with minor defects and 2 with major defects. A pencil is choosen at random. The probability that this pencil is not defective is
View answer and explanation
Correct option: A
3/53/5
Option C, 4/54/5 is incorrect. The correct probability is 102/110=51/55102/110 = 51/55. The provided solution and answer are incorrect.
853 Easy
853 of 1130
Application +4 | -1
A five digits number is formed by writing the digits 1, 2, 3, 4, 5 in a random order without repetitions. Then the probability that the number is divisible by 4, is
View answer and explanation
Correct option: C
15\frac{1}{5}
Option C (1/5) is correct because it calculates the probability of forming a number divisible by 4 from the given digits.
854 Medium
854 of 1130
Application +4 | -1
If a∈[−20,0],a \in \left[ { - 20,0} \right], then the probability that the graph of the function y=16  x2+8(a+5)x−7  a−5y = 16\;{x^2} + 8\left( {a + 5} \right)x - 7\;a - 5 is strictly above the xx-axis is
View answer and explanation
Correct option: A
01-Feb
Option A (01-Feb) is correct because the probability is calculated by finding the range of 'a' for which the quadratic is always positive and dividing by the total range of 'a'.
855 Medium
855 of 1130
Application +4 | -1
Let E1,E2,E3E_1, E_2, E_3 be three mutually exclusive events such that P(E1)=2+3p6,P(E2)=2−p8P\left(E_1\right)=\frac{2+3 p}{6}, P\left(E_2\right)=\frac{2-p}{8} and P(E3)=1−p2P\left(E_3\right)=\frac{1-p}{2}. If the maximum and minimum values of pp are p1p_1 and p2p_2, then (p1+p2)\left(p_1+p_2\right) is equal to :
View answer and explanation
Correct option: B
53\frac{5}{3}
Option B (53\frac{{5}}{{3}}) is correct because after finding the range of p for each probability and their sum, the intersection gives the valid range, and p1+p2 = 5/3.
856 Easy
856 of 1130
Application +4 | -1
If the probability that a person is not a swimmer is 0.3 , then the probability that out of 5 persons 4 are swimmers is
View answer and explanation
Correct option: A
5C4(0.7)4(0.3){ }^5 C_4(0.7)^4(0.3)
Option A (5C4(0.7)4(0.3){}^5 C_4(0.7)^4(0.3)) is correct because it represents the probability of exactly 4 swimmers out of 5 people, using the binomial probability formula.
857 Hard
857 of 1130
Application +4 | -1
Nine identical balls are numbers 1, 2, ...9. are put in a bag. A draws a ball and gets the number aa. The ball is put back the bag. Next BB draws a ball gets the number bb. The probability that aa and bb satisfics the inequality a−2b+10>0a-2 b+10 > 0 is
View answer and explanation
Correct option: C
6181\frac{61}{81}
Option C (6181\frac{{61}}{{81}}) is correct because it is the probability calculated by finding the number of pairs (a, b) that satisfy the inequality a−2b+10>0a - 2b + 10 > 0, divided by the total number of possible pairs.
858 Easy
858 of 1130
Application +4 | -1
The probability that a company executive will travel by train is 23\frac{2}{3} and that he will travel by plane is 15.\frac{1}{5}. The probability of his journey by train or plane is
View answer and explanation
Correct option: B
1315\frac{{13}}{{15}}
Option B (13/15) is correct because it calculates the probability of the executive traveling by train or plane by adding their individual probabilities since the events are mutually exclusive.
859 Easy
859 of 1130
Application +4 | -1
If a dice is thrown twice, the probability of occurrence of 4 at least once, is
View answer and explanation
Correct option: A
11/3611/36
Option A (11/3611/36) is correct because it represents the probability of getting at least one 4 when a dice is thrown twice.
860 Easy
860 of 1130
Application +4 | -1
You are given a box with 20 cards in it. 10 of these cards have letter II printed on them. The other ten have the letter T{T} printed on them. If you pick up 3 cards at random and keep them in same order, the probability of making the word IITIIT, is
View answer and explanation
Correct option: D
538\frac{5}{38}
Option D (5/38) is correct because it is the probability of picking two I cards followed by a T card from the given set of cards.
861 Easy
861 of 1130
Understanding +4 | -1
If AA and BB are arbitrary events, then
View answer and explanation
Correct option: B
P(A∪B)≤P(A)+P(B)P(A \cup B) \leq P(A)+P(B)
Option B (P(A∪B)≤P(A)+P(B)P(A \cup B) \leq P(A)+P(B)) is correct because the probability of the union of two events is always less than or equal to the sum of their individual probabilities.
862 Easy
862 of 1130
Application +4 | -1
If a dice is thrown twice, the probability of occurrence of 4 at least once is
View answer and explanation
Correct option: A
1136\frac{{11}}{{36}}
Option A (1136\frac{{11}}{{36}}) is correct because it represents the probability of getting a 4 at least once in two throws of a dice.
863 Easy
863 of 1130
Application +4 | -1
If AA and BB are two independent events such that P(Aˉ∩B)=2/15P(\bar{A} \cap B)=2 / 15 and P(A∩Bˉ)=1/6P(A \cap \bar{B})=1 / 6, then P(B)P(B), is
View answer and explanation
Correct option: C
45\frac{4}{5} or, 16\frac{1}{6}
Option C (45\frac{4}{5} or, 16\frac{1}{6}) is correct because it is the solution to the probability equations derived from the given independent events.
864 Easy
864 of 1130
Application +4 | -1
If AA and BB are two events such that P(A∪B)=56,P\left( {A \cup B} \right) = \frac{5}{6}, P(A∩B)=13P\left( {A \cap B} \right) = \frac{1}{3} and P(Bˉ)=13,P\left( {\bar B} \right) = \frac{1}{3}, then the value of P(A)P\left( A \right) is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because it is the value of P(A) that satisfies the given probability conditions.
865 Easy
865 of 1130
Application +4 | -1
If the random variable XX takes the values x1,   x2, x10{x_1},\;\,{x_2},\,{x_{10}} with probability P(X=x1)=ki,P\left( {X = {x_1}} \right) = ki, then the value of kk is equal to
View answer and explanation
Correct option: C
155\frac{1}{{55}}
Option C (1/55) is correct. The value of k is found by using the fact that the sum of probabilities for all possible values of a random variable must equal 1.
866 Easy
866 of 1130
Application +4 | -1
The probability that a leap year will have 53 Fridays or 53 Saturdays, is
View answer and explanation
Correct option: B
37\frac{3}{7}
Option B (37\frac{3}{7}) is correct because it calculates the probability of a leap year having 53 Fridays or 53 Saturdays by considering the possible combinations of the extra two days.
867 Easy
867 of 1130
Application +4 | -1
For a party 8 guests are invited by a husband and his wife. They sit for a dinner around a round table. The probability that the husband and his wife sit together, is
View answer and explanation
Correct option: B
29\frac{2}{9}
Option B (29\frac{2}{9}) is correct because it calculates the probability of the husband and wife sitting together around a round table.
868 Medium
868 of 1130
Application +4 | -1
An unbiased die is tossed until a number greater than 4 appears. The probability that an even number of tosses needed, is
View answer and explanation
Correct option: B
25\frac{2}{5}
Option B (2/5) is correct because it calculates the probability that an even number of tosses is needed by considering the geometric series of probabilities.
869 Easy
869 of 1130
Application +4 | -1
If (  1+a3)   and   (  1−a4)\left( {\;\frac{{1 + a}}{3}} \right)\;\,\mathrm{and}\,\;\left( {\;\frac{{1 - a}}{4}} \right) are probability of two mutually exclusive events, then set of all values of aa is
View answer and explanation
Correct option: A
−1≤a≤1- 1 \le a \le 1
Option A (−1≤a≤1- 1 \le a \le 1) is correct because it satisfies the conditions for probabilities of mutually exclusive events, ensuring each probability is between 0 and 1, and their sum is also between 0 and 1.
870 Easy
870 of 1130
Application +4 | -1
A bag contains 3 same balls and 3 different balls of three different colours. Two balls are drawn randomly with replacement. The probability they have same colour is mm. Again four balls are drawn one by one with replacement, then probability of getting three same balls is nn. The value of mm. nn is
View answer and explanation
Correct option: D
881\frac{8}{81}
Option D (8/81) is correct because it is the product of the probabilities m and n, calculated based on the given conditions of drawing balls with replacement.
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