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JEE • MATHEMATICS • PROBABILITY

Probability practice · Mixed difficulty

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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 181-210 of 1130 questions

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181 Easy
181 of 1130
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.
182 Easy
182 of 1130
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.
183 Easy
183 of 1130
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.
184 Easy
184 of 1130
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.
185 Easy
185 of 1130
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.
186 Easy
186 of 1130
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.
187 Medium
187 of 1130
Application +4 | -1
Four persons are chosen at random from a group of 3 men, 2 women and 4 children. The chance that exactly 2 of them are children, is
View answer and explanation
Correct option: A
1021\frac{{10}}{{21}}
Option A (1021\frac{{10}}{{21}}) is correct because it calculates the probability of choosing exactly 2 children out of a group of 9 people.
188 Easy
188 of 1130
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.
189 Medium
189 of 1130
Application +4 | -1
In a college 25% boys and 10% girls offer Mathematics. There are 60% girls in the college. If a Mathematics student is chosen at random, then the probability that the student is a girl, will be
View answer and explanation
Correct option: B
38\frac{3}{8}
Option B (38\frac{3}{8}) is correct because it represents the probability that a randomly chosen Mathematics student is a girl, calculated using conditional probability.
190 Easy
190 of 1130
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.
191 Easy
191 of 1130
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.
192 Easy
192 of 1130
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.
193 Easy
193 of 1130
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.
194 Easy
194 of 1130
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.
195 Easy
195 of 1130
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.
196 Easy
196 of 1130
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.
197 Easy
197 of 1130
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).
198 Easy
198 of 1130
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.
199 Easy
199 of 1130
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.
200 Easy
200 of 1130
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.
201 Medium
201 of 1130
Application +4 | -1
A bag contains a mixed lot of red and blue balls. If two balls are drawn at random, the probability of drawing two red balls is five times the probability of drawing two blue balls and the probability of drawing one ball of each colour is six times the probability of drawing two blue balls. Then, the total number of red and blue balls in the bag is
View answer and explanation
Correct option: A
9
Option A (9) is correct because it is the total number of balls that satisfies the given probability conditions regarding red and blue balls drawn from the bag.
202 Easy
202 of 1130
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.
203 Easy
203 of 1130
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.
204 Medium
204 of 1130
Analyzing +4 | -1
Sum of two positive integers is 66 and μ\mu is the maximum value of their product S={x∈Z,x(66−x)≥5μ9},x≠0S=\left\{x \in \mathbb{Z}, x(66-x) \geq \frac{5 \mu}{9}\right\}, x \neq 0, then probability of AA when A={x∈S;x=3k,x∈N}A=\{x \in S ; x=3 k, x \in \mathbb{N}\} is:
View answer and explanation
Correct option: C
13\frac{1}{3}
Option C (1/3) is correct because after finding the range of x and the favorable set of values for event A, the probability is calculated as 1/3.
205 Easy
205 of 1130
Application +4 | -1
A fair coin is tossed 100 times. The probability of getting tails an odd number of times, is
View answer and explanation
Correct option: A
1/21 / 2
Option A (1/21 / 2) is correct because the probability of getting tails an odd number of times in 100 tosses of a fair coin is always 1/2.
206 Easy
206 of 1130
Application +4 | -1
Three faces of a fair dice are yellow, two faces red and one blue. The dice is tossed three times. The probability that the colours, yellow, red and blue appear in the first, second and third tosses respectively is
View answer and explanation
Correct option: C
136\frac{1}{36}
Option C (136\frac{1}{36}) is correct because it calculates the probability of getting yellow, red, and blue in that specific order by multiplying the individual probabilities.
207 Easy
207 of 1130
Application +4 | -1
Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has atleast one girl is
View answer and explanation
Correct option: D
47\frac{4}{7}
Option D (47\frac{4}{7}) is correct because it is the conditional probability of the eldest child being a girl given that the family has at least one girl.
208 Easy
208 of 1130
Application +4 | -1
Three identical dice are rolled. The probability that same number appears on them, is
View answer and explanation
Correct option: B
136\frac{1}{36}
Option C (754\frac{{7}}{{54}}) is correct because it calculates the probability that the sum of the numbers appearing on three dice is 9.
209 Easy
209 of 1130
Application +4 | -1
If AA and BB are two subsets of set X={1,2,3,4,5}X=\{1,2,3,4,5\}, then the probability that A∩BA \cap B contains just two elements is
View answer and explanation
Correct option: D
13529\frac{135}{2^9}
Option D (13529\frac{{135}}{{2^9}}) is correct because it calculates the probability of the intersection of A and B containing exactly two elements by considering the combinations and dividing by the total possible outcomes.
210 Easy
210 of 1130
Application +4 | -1
Let Ajay will not appear in JEE exam with probability p=27p=\frac{2}{7}, while both Ajay and Vijay will appear in the exam with probability q=15q=\frac{1}{5}. Then the probability, that Ajay will appear in the exam and Vijay will not appear is
View answer and explanation
Correct option: D
1835\frac{18}{35}
Option D (1835\frac{18}{35}) is correct because it is the probability that Ajay will appear in the exam and Vijay will not appear, calculated using the given probabilities.
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