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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 1081-1110 of 1130 questions
1081 Easy
1081 of 1130
Application +4 | -1
If P(A)=0.4P(A)=0.4, and P(A∩B)=0.15P(A \cap B)=0.15, then P(A∣A′∪B′)P\left(A \mid A^{\prime} \cup B^{\prime}\right) 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 uses the conditional probability formula and De Morgan's Law.
1082 Easy
1082 of 1130
Analyzing +4 | -1
Let E1,E2E_1, E_2 and E3E_3 be any pairwise independent events with P(E1)>0P\left(E_1\right)>0 and P(E1∩E2∩E3)=0P\left(E_1 \cap E_2 \cap E_3\right)=0, then P(Eˉ2∩Eˉ3/E1)P\left(\bar{E}_2 \cap \bar{E}_3 / E_1\right) is equal to
View answer and explanation
Correct option: A
P(Eˉ3)−P(E2)P\left(\bar{E}_3\right)-P\left(E_2\right)
Option A (P(Eˉ3)−P(E2)P(\bar{E}_3)-P(E_2)) is correct because it simplifies the conditional probability using pairwise independence and the given condition P(E1∩E2∩E3)=0P(E_1 \cap E_2 \cap E_3) = 0.
1083 Easy
1083 of 1130
Application +4 | -1
Two cards are drawn without replacement from a well-shuffled pack. The probability that one of them is an ace of heart, is
View answer and explanation
Correct option: B
126\frac{1}{{26}}
Option B, 126\frac{1}{{26}} is correct because it considers the two possible scenarios: the ace of hearts is drawn first or second.
1084 Easy
1084 of 1130
Application +4 | -1
The mean and variance of a binomial distribution are α\alpha and α3\frac{\alpha}{3} respectively. If P(X=1)=4243P(X=1)=\frac{4}{243}, then P(X=4P(X=4 or 5)) is equal to :
View answer and explanation
Correct option: C
1627\frac{16}{27}
Option C (1627\frac{{16}}{{27}}) is correct because after finding n=6, the probability P(X=4 or 5) is calculated correctly using the binomial distribution formula.
1085 Easy
1085 of 1130
Analyzing +4 | -1
Four dice are thrown simultaneously and the numbers shown on these dice are recorded in 2×22 \times 2 matrices. The probability that such formed matrices have all different entries and are non-singular, is
View answer and explanation
Correct option: D
43162\frac{{43}}{{162}}
Option D (43162\frac{{43}}{{162}}) is correct because it involves calculating the total possible matrices, identifying singular matrices with distinct entries, and subtracting to find the non-singular matrices before calculating the probability.
1086 Easy
1086 of 1130
Understanding +4 | -1
A die is thrown.

I. The probability of a prime number will appear is 12\frac{1}{2}.

II. The probability of a number more than 6 will appear is 1.1.
View answer and explanation
Correct option: A
Only I is correct
Option A (Only I is correct) is correct because the probability of a prime number appearing on a die is 1/2, while the probability of a number more than 6 is 0.
1087 Easy
1087 of 1130
Application +4 | -1
In an experiment the success is twice that of failure. If the experiment is repeated 6 times, the probability that at least 4 times favourable is
View answer and explanation
Correct option: D
496729\frac{{496}}{{729}}
Option D (496/729) is correct because it calculates the probability of at least 4 successes in 6 trials using the binomial probability formula.
1088 Easy
1088 of 1130
Application +4 | -1
If the probability distribution of a random variable X\mathrm{X} is as given below:
X=x:−2−10123P(X=x):110k152k310k\begin{array}{cccccccccccccccccccc} {X = x:}&{ - 2}&{ - 1}&0&1&2&3 \\ {P(X = x):}&{\frac{1}{{10}}}&k&{\frac{1}{5}}&{2k}&{\frac{3}{{10}}}&k \end{array}
Then the value of kk, is
View answer and explanation
Correct option: A
110\frac{1}{10}
Option A (110\frac{1}{10}) is correct because the sum of probabilities in a probability distribution must equal 1, allowing us to solve for k.
1089 Easy
1089 of 1130
Application +4 | -1
One mapping is selected at random from all mappings of the set S={1,2,….,n}S=\{1,2, \ldots ., n\} to itself. If the probability that the mapping is one-one is 5324\frac{5}{324} then the value of nn equals (n∈N)(n \in N)
View answer and explanation
Correct option: A
6
The correct answer is A (6) because the probability that a mapping from a set of size n to itself is one-to-one is n!nn\frac{n!}{n^n}, and solving n!nn=5324\frac{n!}{n^n} = \frac{5}{324} gives n = 6.
1090 Medium
1090 of 1130
Application +4 | -1
India plays two matches each with the West Indies and Australia. In any match, the probabilities of India getting 0,1 and 2 points are 0.45,0.05 and 0.50, respectively. Assuming that the outcomes are independent, the probability of India getting at least 7 points, is
View answer and explanation
Correct option: B
780\frac{7}{80}
Option B (780\frac{7}{80}) is correct because it calculates the probability of India getting at least 7 points by considering the cases of getting exactly 7 or 8 points in the matches.
1091 Easy
1091 of 1130
Application +4 | -1
A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn, is
View answer and explanation
Correct option: A
125\frac{12}{5}
Option A (125\frac{12}{5}) is correct because it is the variance calculated using the formula npq with the given parameters n=10 and p=3/5.
1092 Medium
1092 of 1130
Application +4 | -1
Two small squares on a chess board are chosen at random. Then, the probability that they have a common side, is
View answer and explanation
Correct option: C
118\frac{1}{18}
Option C is correct because the probability is calculated by dividing the number of pairs of squares sharing a side by the total number of ways to choose two squares.
1093 Easy
1093 of 1130
Application +4 | -1
A and B throw a dice. The probability that A's throw is not greater than B's, is
View answer and explanation
Correct option: B
712\frac{7}{12}
Option B (712\frac{7}{12}) is correct because it represents the probability that A's throw is less than or equal to B's throw, calculated by summing the probabilities for each possible outcome of B's throw.
1094 Easy
1094 of 1130
Application +4 | -1
The probability that a year chosen at random has 53 Sundays, is
View answer and explanation
Correct option: D
528\frac{5}{28}
Option D (528\frac{5}{28}) is correct because it calculates the weighted average of the probability of 53 Sundays in a leap year and a non-leap year.
1095 Easy
1095 of 1130
Application +4 | -1
If P(A)=310,P(B)=25P(A)=\frac{3}{10}, P(B)=\frac{2}{5} and P(A∪B)=35P(A \cup B)=\frac{3}{5}, then P(B/A)+P(A/B)P(B / A)+P(A / B) equals to
View answer and explanation
Correct option: D
712\frac{7}{12}
Option D (712\frac{7}{12}) is correct because it is the result of calculating P(B/A)+P(A/B)P(B / A)+P(A / B) using the given probabilities and the formula for conditional probability.
1096 Medium
1096 of 1130
Analyzing +4 | -1
If the numbers 1,2,3,4,5,61,2,3,4,5,6 are arranged in a circle, then the probability that the sum of any two consecutive numbers must not be equal to 7 is
View answer and explanation
Correct option: C
415\frac{4}{15}
Option C is correct because the probability is calculated by finding the number of arrangements where no two consecutive numbers sum to 7 and dividing by the total number of arrangements.
1097 Easy
1097 of 1130
Application +4 | -1
In a survery, it was found that 21 persons liked product AA, 26 liked product BB and 29 liked product CC. If 14 persons liked products AA and B,B, 12 liked products CC and A,A, 13 persons liked products BB and CC and 8 liked all the three products then
(i) Find the number of persons who liked the product CC only
(ii) The number of persons who like the products A{A} and B{B} but not C{C}
View answer and explanation
Correct option: A
12; 6
The correct answer is A (12; 6) because it correctly calculates the number of people who liked product C only and the number of people who liked products A and B but not C using set theory principles.
1098 Medium
1098 of 1130
Application +4 | -1
Two distinct numbers are chosen at random from the first 15 natural numbers. The probability that the sum will be divisible by 3 is
View answer and explanation
Correct option: B
3191\frac{31}{91}
Option B (3191\frac{{31}}{{91}}) is correct because it accurately calculates the probability of the sum of two distinct numbers chosen from the first 15 natural numbers being divisible by 3.
1099 Medium
1099 of 1130
Application +4 | -1
A box contains 9 tickets numbered 1 to 9 inclusive. If 3 tickets are drawn from the box without replacement. The probability that they are alternatively either {odd, even, odd} of {even, odd, even} is
View answer and explanation
Correct option: D
518\frac{5}{{18}}
Option D (518\frac{5}{{18}}) is correct because it calculates the probability of drawing tickets in alternating odd-even or even-odd sequences without replacement.
1100 Easy
1100 of 1130
Remembering +4 | -1
In a poisson distribution mean is 16, then standard deviation is
View answer and explanation
Correct option: D
4
Option D (4) is correct because in a Poisson distribution, the variance is equal to the mean, and the standard deviation is the square root of the variance. Since the mean is 16, the standard deviation is 16=4\sqrt{16} = 4.
1101 Medium
1101 of 1130
Application +4 | -1
A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is
View answer and explanation
Correct option: B
25\frac{2}{5}
Option B (25\frac{2}{5}) is correct because it is the probability of drawing a red ball on the second draw, calculated using the law of total probability considering the two possible outcomes of the first draw (red or black).
1102 Easy
1102 of 1130
Understanding +4 | -1
A three digit number is selected at random from the set of all three-digit numbers. The probability that the number selected has all the three digits same, is
View answer and explanation
Correct option: D
1/1001 / 100
Option D (1/1001 / 100) is correct because there are 900 three-digit numbers, and only 9 of them have all three digits the same (111, 222, ..., 999). The probability is therefore 9/900 = 1/100.
1103 Easy
1103 of 1130
Application +4 | -1
Bag P\mathrm{P} contains 6 red and 4 blue balls and bag Q\mathrm{Q} contains 5 red and 6 blue balls. A ball is transferred from bag P to bag Q\mathrm{Q} and then a ball is drawn from bag QQ. What is the probability that the ball drawn is blue?
View answer and explanation
Correct option: B
815\frac{8}{15}
Option B (815\frac{8}{15}) is correct because it is the probability of drawing a blue ball from bag Q after transferring a ball from bag P.
1104 Medium
1104 of 1130
Understanding +4 | -1
Let A,BA, B and CC be three events such that P(C)=0P(C)=0.

Statement-1: P(A∩B∩C)=0P(A \cap B \cap C)=0

Statement-2: P(A∪B∪C)=P(A∪B)P(A \cup B \cup C)=P(A \cup B)
View answer and explanation
Correct option: B
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
Option B (STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1) is correct because both statements are true, but the second statement doesn't explain the first.
1105 Easy
1105 of 1130
Application +4 | -1
A number is selected at random from the first twenty-five natural numbers. If it is a composite number, then it is divided by 5 . But if it is not a composite number, then it is divided by 2 . The probability that there will be no remainder in the division, is
View answer and explanation
Correct option: C
0.20.2
Option D is correct because the probability is 0.2, calculated by finding the number of composite numbers divisible by 5 and non-composite numbers divisible by 2, then dividing by the total number of selections.
1106 Easy
1106 of 1130
Application +4 | -1
It is given that the events AA and BB are such that P(A)=14,P(A/B)=12P(A)=\frac{1}{4}, P(A / B)=\frac{1}{2} and P(B/A)=23P(B / A)=\frac{2}{3}. Then, P(B)P(B) is
View answer and explanation
Correct option: D
13\frac{1}{3}
Option D (13\frac{1}{3}) is correct because it uses the conditional probability formulas to find P(B).
1107 Easy
1107 of 1130
Application +4 | -1
If the numbers appeared on the two throws of a fair six faced die are α\alpha and β\beta, then the probability that x2+αx+β>0x^2+\alpha x+\beta>0, for all x∈Rx \in R, is :
View answer and explanation
Correct option: A
1736\frac{17}{36}
Option A (17/36) is correct because it represents the probability calculated by finding favorable outcomes where the discriminant is negative.
1108 Easy
1108 of 1130
Application +4 | -1
A pair of dice is thrown 5 times. For each throw, a total of 5 is considered a success. If the probability of at least 4 successes is k311\frac{k}{3^{11}}, then kk is equal to
View answer and explanation
Correct option: D
123
Option D (123) is correct because it is the numerator of the probability of at least 4 successes, calculated using binomial probability.
1109 Easy
1109 of 1130
Analyzing +4 | -1
Let AA and BB be two events such that P(A∪B‾)=16,P(A∩B)=14P(\overline{A \cup B})=\frac{1}{6}, P(A \cap B)=\frac{1}{4} and P(Aˉ)=14P(\bar{A})=\frac{1}{4}, where Aˉ\bar{A} stands for complement of event AA. Then events AA and BB are
View answer and explanation
Correct option: B
independent but not equally likely
Option B (independent but not equally likely) is correct because the events A and B are independent, but their probabilities are not equal.
1110 Medium
1110 of 1130
Application +4 | -1
Two numbers are randomly selected from
the first 100 natural numbers. The probability that the
product of the numbers is divisible by 7 is
View answer and explanation
Correct option: C
48594950\frac{4859}{4950}
Option C (48594950\frac{4859}{4950}) is correct because it calculates the probability that at least one of the two numbers is divisible by 7.
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