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JEE • MATHEMATICS • PROBABILITY

Probability practice · Easy

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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 121-150 of 822 questions

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121 Easy
121 of 822
Application +4 | -1
A man and his wife appear for an interview for two posts. The probability of the man’s selection is 1/5 and that of his wife’s selection is 1/7. The probability that at least one of them is selected, is
View answer and explanation
Correct option: D
1135\frac{{11}}{{35}}
Option D (1135\frac{{11}}{{35}}) is correct because it calculates the probability of at least one selection using the formula for the union of two independent events.
122 Easy
122 of 822
Application +4 | -1
A die is tossed twice. The probability of having a number greater than 4 on each toss is
View answer and explanation
Correct option: B
19\frac{1}{9}
Option B (19\frac{1}{9}) is correct because it represents the probability of getting a number greater than 4 on each of the two tosses of a die.
123 Easy
123 of 822
Application +4 | -1
A bag A contains 2 white and 3 red balls, and a bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and it is found to be red. Let pp be the probability that it was drawn from the bag B then find the value of [1p]\left[\frac{1}{p}\right] where [.] denotes G.I.F.
View answer and explanation
Correct option: A
2
Option A (2) is correct because the probability that the red ball was drawn from bag B is 25/52, and the greatest integer function of its reciprocal is 2.
124 Easy
124 of 822
Application +4 | -1
Three persons, A, B and C\mathrm{C}, fire at a target in turn, starting with A\mathrm{A}. Their probability of hitting the target are 0.4,0.30.4,0.3 and 0.2 respectively. The probability of two hits is
View answer and explanation
Correct option: B
0.188
Option B (0.188) is correct because it is the sum of probabilities of exactly two hits by A, B, and C in all possible combinations.
125 Easy
125 of 822
Analyzing +4 | -1
If nn integers taken at random are multiplied together, then the probability that the last digit of the product is 1, 3, 7 or 9, is
View answer and explanation
Correct option: A
2n5n\frac{{{2^n}}}{{{5^n}}}
Option A, 2n5n\frac{{{2^n}}}{{{5^n}}}, is correct because it represents the probability of the last digit being 1, 3, 7, or 9 after multiplying n random integers.
126 Easy
126 of 822
Application +4 | -1
The probability of getting the sum more than 7 when a pair of dice is tossed is
View answer and explanation
Correct option: D
512\frac{5}{12}
Option D (512\frac{5}{12}) is correct because it represents the probability of getting a sum greater than 7 when tossing a pair of dice, calculated by counting favorable outcomes and dividing by the total possible outcomes.
127 Easy
127 of 822
Application +4 | -1
′X′'X' speaks truth in 60% and ′Y′'Y' in 50% of the cases. The probability that they contradict each other narrating the same incident is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (1/2) is correct because it correctly calculates the probability of the two people contradicting each other by considering the cases where one speaks the truth and the other lies.
128 Easy
128 of 822
Application +4 | -1
Two coins and a die are tossed. The probability that both coins fall heads and the die shows a 3 or 6, is
View answer and explanation
Correct option: B
1/121/12
Option B (112\frac{1}{12}) is correct because it multiplies the probabilities of each independent event occurring.
129 Easy
129 of 822
Application +4 | -1
The probability that in a family of 5 members, exactly 2 members have birthday on sunday, is
View answer and explanation
Correct option: D
10×6375\frac{{10 \times {6^3}}}{{{7^5}}}
Option D is correct because it applies the binomial probability formula to calculate the probability of exactly 2 members having a birthday on Sunday.
130 Easy
130 of 822
Application +4 | -1
Let a die is loaded in such a way that prime number faces are twice as likely to occur as a non-prime number faces. Then, the probability that an odd number will be show up when the die is tossed, is
View answer and explanation
Correct option: D
59\frac{5}{9}
Option D is correct because the probability is calculated by assigning weights to prime and non-prime numbers and then finding the probability of an odd number appearing.
131 Easy
131 of 822
Analyzing +4 | -1
Two non-negative integers xx and yy are chosen at random with replacement. The probability that x2+y2x^2+y^2 is divisible by 10, is
View answer and explanation
Correct option: A
950\frac{9}{50}
Option A (950\frac{{9}}{{50}}) is correct because it represents the probability calculated by dividing the number of favorable outcomes (pairs whose squares sum to a multiple of 10) by the total number of possible outcomes.
132 Easy
132 of 822
Analyzing +4 | -1
A dice is rolled three times. The probability of getting a larger number than the previous number each time is
View answer and explanation
Correct option: B
554\frac{5}{{54}}
Option B (5/54) is correct because it represents the probability of getting a larger number than the previous number each time when rolling a dice three times.
133 Easy
133 of 822
Application +4 | -1
Four persons are selected from a group of 4 men, 2 women and 3 children. The probability that exactly two of them are men is
View answer and explanation
Correct option: D
1021\frac{10}{21}
Option D (10/21) is correct because it calculates the probability of selecting exactly two men from a group of 4 men, 2 women, and 3 children.
134 Easy
134 of 822
Application +4 | -1
Girl students constitute 10%10 \% of I year and 5%5 \% of II year at Roorkee University. During summer holidays 70%70 \% of the I year and 30%30 \% of II year students are given a project. The girls take turns on duty in canteen. The chance that I year girl student is on duty in a randomly selected day is
View answer and explanation
Correct option: B
1417\frac{14}{17}
Option B (1417\frac{14}{17}) is correct because it represents the conditional probability of a randomly selected girl on duty being from the I year.
135 Easy
135 of 822
Application +4 | -1
If P(A)=P(B)=xP\left( A \right) = P\left( B \right) = x and P(A∩B)=P(A′∩B′)=13P\left( {A \cap B} \right) = P\left( {A' \cap B'} \right) = \frac{1}{3}, then xx is equal to
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because it satisfies the probability conditions given in the problem, derived from the properties of union and intersection of events.
136 Easy
136 of 822
Application +4 | -1
A five digit 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
1/51/5
Option C (1/51/5) is correct because the probability is calculated by dividing the number of favorable outcomes (divisible by 4) by the total number of possible outcomes.
137 Easy
137 of 822
Application +4 | -1
Let AA and BB be independent events such that P(A)=13P(A)=\frac{1}{3} and P(B)=16P(B)=\frac{1}{6}. Then,
View answer and explanation
Correct option: B
P(A/Bˉ)=13P(A / \bar{B})=\frac{1}{3}
Option B (P(A/Bˉ)=13P(A / \bar{B})=\frac{1}{3}) is correct because for independent events, the conditional probability is equal to the probability of the event itself.
138 Easy
138 of 822
Application +4 | -1
An integer is chosen at random from first two hundred digits. Then, the probability that the integer chosen is divisible by 6 or 8, is
View answer and explanation
Correct option: A
14\frac{1}{4}
Option A (1/4) is correct because it is calculated by finding the probability of divisibility by 6 or 8 using the inclusion-exclusion principle.
139 Easy
139 of 822
Application +4 | -1
Consider 5 independent Bernulli's trials each with probability of success pp. If the probability of at least one failure is greater than or equal to 3132\frac{31}{32}, then pp lies in the interval:
View answer and explanation
Correct option: C
[0,12]\left[0, \frac{1}{2}\right]
The correct answer is [0,12]\left[0, \frac{1}{2}\right] because it represents the interval in which p lies, given the probability condition.
140 Easy
140 of 822
Application +4 | -1
If (1+3p)/3,(1−p)/4(1+3 p) / 3,(1-p) / 4 and (1−2p)/2(1-2 p) / 2 are the probabilities of three mutually exclusive events, then the set of all values of pp is
View answer and explanation
Correct option: A
1/3≤p≤1/21 / 3 \leq p \leq 1 / 2
Option A (1/3≤p≤1/21/3 \leq p \leq 1/2) is correct because it satisfies the conditions for probabilities of mutually exclusive events, ensuring each probability is non-negative and their sum is less than or equal to 1.
141 Easy
141 of 822
Application +4 | -1
If P(A∩B)=12,P(Aˉ∩Bˉ)=13,P(A)=p,P(B)=2  p,P\left( {A \cap B} \right) = \frac{1}{2},P\left( {\bar A \cap \bar B} \right) = \frac{1}{3},P\left( A \right) = p,P\left( B \right) = 2\;p, then the value of pp is given by
View answer and explanation
Correct option: B
Jul-18
The correct answer is Jul-18, which translates to p = 7/18. This is found by using the formula for P(A∪B) and the given probabilities to solve for p.
142 Easy
142 of 822
Analyzing +4 | -1
In a bag there are 20 cards 10 names AA and another 10 names B. Cards are drawn randomly one by one with replacement, the probability that second AA comes before third BB is
View answer and explanation
Correct option: B
1116\frac{11}{16}
Option B (1116\frac{11}{16}) is correct because it calculates the probability of the second A occurring before the third B by considering all possible sequences of A and B.
143 Easy
143 of 822
Application +4 | -1
A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested, then probability that both are dead is
View answer and explanation
Correct option: D
328\frac{3}{28}
Option D (328\frac{3}{28}) is correct because it is the probability of selecting two dead batteries without replacement.
144 Easy
144 of 822
Understanding +4 | -1
Twenty five coins are tossed simultaneously. The probability that the fifth coin will fall with head upward, is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because the probability of a fair coin landing heads is always 1/2, regardless of other coin tosses.
145 Easy
145 of 822
Analyzing +4 | -1
If P(A∩B)=13,  P(A∪B)=56,  P\left( {A \cap B} \right) = \frac{1}{3},\;P\left( {A \cup B} \right) = \frac{5}{6},\; and P(A)=12P\left( A \right) = \frac{1}{2} then which one of the following is correct?
View answer and explanation
Correct option: A
AA and BB are independent events
Option A (A and B are independent events) is correct because the given probabilities satisfy the condition for independence: P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B).
146 Easy
146 of 822
Application +4 | -1
Ram and Shyam throw with one dice for a prize of Rs. 88 which is to be won by the player who throws 1 first. If Ram starts, then mathematical expectation for Shyam is
View answer and explanation
Correct option: B
Rs. 40
Option B (Rs. 40) is correct because it accurately calculates Shyam's mathematical expectation based on the probability of him winning the prize.
147 Easy
147 of 822
Application +4 | -1
The probability that Krishna will be alive 10 year hence is 715\frac{7}{15} and Hari will be alive is 710\frac{7}{10}. The probability that both Krishna and Hari will be dead 10 years hence is
View answer and explanation
Correct option: B
24150\frac{24}{150}
Option B (24150\frac{24}{150}) is correct because it calculates the probability of both Krishna and Hari being dead by multiplying their individual probabilities of being dead.
148 Easy
148 of 822
Application +4 | -1
Suppose AA is a set containing nn elements. Two subsets PP and QQ of AA are chosen at random. If the probability that PP is a subset of QQ is 243/1024243 / 1024, then nn is equal to
View answer and explanation
Correct option: C
5
Option C (5) is correct because it represents the number of elements in set A, given the probability that P is a subset of Q.
149 Easy
149 of 822
Application +4 | -1
If three distinct numbers are chosen randomly from the first 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3 is
View answer and explanation
Correct option: D
41155\frac{4}{1155}
Option D (41155\frac{4}{1155}) is correct because it is the probability of selecting three numbers divisible by both 2 and 3, calculated by dividing favorable combinations by total combinations.
150 Easy
150 of 822
Application +4 | -1
A and B are two independent events. The probability that both A\mathrm{A} and B\mathrm{B} occur is 1/61 / 6 and the probability that at least one of them occurs is 2/32 / 3. Find 8P(A)+9P(B)8 \mathrm{P}(\mathrm{A})+9 \mathrm{P}(\mathrm{B}) if P(A)>\mathrm{P}(\mathrm{A})> P(B)\mathrm{P}(\mathrm{B}).
View answer and explanation
Correct option: A
7
Option A (7) is correct because by solving the equations derived from the given probabilities, we find P(A) = 1/2 and P(B) = 1/3, leading to 8P(A) + 9P(B) = 7.
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