Skip to main content
Back to PROBABILITY

Your Progress

0 / 285 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • PROBABILITY

Probability practice · Medium

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 151-180 of 285 questions

Preparing practice and saved progress…

151 Medium
151 of 285
Application +4 | -1
Four boys and three girls stand in a queue for an interview, probability that they will in alternate position, is
View answer and explanation
Correct option: B
1/351/35
Option B (1/351/35) is correct because it calculates the probability of boys and girls standing in alternate positions by dividing the number of favorable arrangements by the total number of possible arrangements.
152 Medium
152 of 285
Application +4 | -1
A missile is destroyed before hitting the target by a probability of 13\frac{1}{3}. If the missile is not destroyed probability of hitting the target is 34\frac{3}{4}. Three missiles are fired. The probability of exactly two hits the target is
View answer and explanation
Correct option: C
38\frac{3}{8}
Option B (83\frac{8}{3}) is correct because as x approaches 2 from the right, the greatest integer function evaluates to 0, leaving only the cubic term.
153 Medium
153 of 285
Application +4 | -1
AA is a set containing nn elements. Two subsets PP and QQ of AA are chosen at random. (PP and QQ may have elements in common). The probability that P∩Q=P \cap Q= ϕ\phi is
View answer and explanation
Correct option: A
(34)n\left(\frac{3}{4}\right)^n
Option A ((34)n\left(\frac{3}{4}\right)^n) is correct because it represents the probability that the intersection of two randomly chosen subsets P and Q of A is empty.
154 Medium
154 of 285
Analyzing +4 | -1
The probability that two randomly selected subsets of the set {1,2,3,4,5}\left\{ {1,2,3,4,5} \right\} have exactly two elements in their intersection, is
View answer and explanation
Correct option: C
13529\frac{{135}}{{{2^9}}}
Option C (13529\frac{{135}}{{{2^9}}}) is correct because it calculates the probability of two randomly selected subsets having exactly two elements in their intersection by considering the number of ways to choose the intersection and the remaining elements.
155 Medium
155 of 285
Application +4 | -1
The mean and variance of 5 observations of an experiment are 4 and 5.2 respectively. If from these observations three are 1,2 and 6 , then the mean of remaining will be-
View answer and explanation
Correct option: A
5.5
Option A (5.5) is correct because after calculating the sum of the remaining two observations, their mean is found to be 5.5.
156 Medium
156 of 285
Application +4 | -1
Two different families AA and BB are blessed with equal number of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket.
If the probability that all the tickets go to the children of the family BB is 112\frac{1}{12}, then the number of children in each family is?
View answer and explanation
Correct option: D
5
The correct answer is D (5) because the probability equation is set up correctly, and solving for the number of children in each family yields 5.
157 Medium
157 of 285
Application +4 | -1
The probability of a man hitting a target is 25\frac{2}{5}. He fires at the target kk times ( kk, a given number). Then the minimum kk, so that the probability of hitting the target at least once is more than 710\frac{7}{10}, is :
View answer and explanation
Correct option: A
3
Option A (3) is correct because it is the minimum number of times the man must fire to have a probability greater than 7/10 of hitting the target at least once.
158 Medium
158 of 285
Application +4 | -1
A dice is rolled 4 times, the numbers appearing are listed. The number of different throws, such that the largest number appearing in the list is 4 , is
View answer and explanation
Correct option: D
175
Option C (1540) is correct because it is the number of ways to form the committee under the given constraint.
159 Medium
159 of 285
Application +4 | -1
Twelve tickets are numbered from 1 to 12. One ticket is drawn at random, then the probability of the number to be divisible by 2 or 3, is
View answer and explanation
Correct option: A
23\frac{2}{3}
Option A (2/3) is correct because it correctly calculates the probability of a number being divisible by 2 or 3 from a set of 12 numbers.
160 Medium
160 of 285
Application +4 | -1
If 3 numbers are selected from the set (1,2,3,.........,24)(1,2,3,........., 24) without replacement, then what is the probability that the selected numbers do not form an A.P.?
View answer and explanation
Correct option: C
43/4643 / 46
Option C (43/46) is correct because it represents the probability that the three selected numbers do not form an arithmetic progression.
161 Medium
161 of 285
Application +4 | -1
If P(A)=25,P(B)=310\mathrm{P}(\mathrm{A})=\frac{2}{5}, \mathrm{P}(\mathrm{B})=\frac{3}{10} and P(A∩B)=15\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{5}, then P(A′∣B′)⋅P(B′∣A′)\mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}^{\prime}\right) \cdot \mathrm{P}\left(\mathrm{B}^{\prime} \mid \mathrm{A}^{\prime}\right) is equal to
View answer and explanation
Correct option: C
2542\frac{25}{42}
Option C is correct because it is obtained by using the conditional probability formula and the properties of complementary events.
162 Medium
162 of 285
Application +4 | -1
There are mm persons sitting in a row. Two of them are selected at random. The probability that the two selected persons are not together is
View answer and explanation
Correct option: D
1−2m1-\frac{2}{m}.
Option D (1−2m1-\frac{2}{m}) is correct because it is the probability calculated by subtracting the probability of selecting two adjacent people from 1.
163 Medium
163 of 285
Application +4 | -1
There are 10 pairs of shoes in a cupboard from which 4 shoes are picked at radndom. The probability that there is at least one pair is
View answer and explanation
Correct option: B
99323\frac{99}{323}
Option B (99/323) is correct because it represents the probability of selecting at least one pair of shoes when 4 shoes are randomly picked from 10 pairs.
164 Medium
164 of 285
Application +4 | -1
A letter is taken at random from the letters of the word 'STATISTICS' and another letter is taken at random from the letters of the word 'ASSISTANT'. The probability that they are the same letter is
View answer and explanation
Correct option: C
19/9019 / 90
Option C (19 / 90) is correct because it is calculated by summing the probabilities of choosing the same letter (A, I, S, or T) from both words.
165 Medium
165 of 285
Application +4 | -1
For the married couple living in Jammu, the probability that a husband will vote in an election is 0.5 and the probability that his wife will vote is 0.4 . The probability that the husband votes, given that his wife also votes is 0.7 . Then the probability that husband and wife both will vote is
View answer and explanation
Correct option: A
0.28
Option A (0.28) is correct because the probability of both events occurring is calculated using conditional probability: P(A ∩ B) = P(B) * P(A|B).
166 Medium
166 of 285
Application +4 | -1
Three are six verities of a regular hexagon are chosen at random, then the possibility that the triangle with three vertices is equilateral, is equal to
View answer and explanation
Correct option: C
110\frac{1}{10}
Option C (1/10) is correct because it calculates the probability of choosing three vertices that form an equilateral triangle from a regular hexagon.
167 Medium
167 of 285
Application +4 | -1
A person draws a card from a pack, replaces it, shuffles the pack, again draws a card, replaces it and draws again. This he does until he draws a heart. The probability that he will have to make at least four draws is
View answer and explanation
Correct option: C
2764\frac{27}{64}
Option C (27/64) is correct because it calculates the probability of not drawing a heart in the first three draws.
168 Medium
168 of 285
Application +4 | -1
Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane, is
View answer and explanation
Correct option: C
0.32
Option B (0.14) is incorrect. The correct answer is 0.32 because it accounts for the probability of the second plane hitting the target only if the first plane misses.
169 Medium
169 of 285
Application +4 | -1
A letter is taken out at random from ‘ASSISTANT’ and another is taken out from ‘STATISTICS’. The probability that they are same letters, is
View answer and explanation
Correct option: C
1990\frac{{19}}{{90}}
Option C (1990\frac{{19}}{{90}}) is correct because it sums the probabilities of drawing the same letter (A, I, S, or T) from both words, considering the frequency of each letter in each word.
170 Medium
170 of 285
Application +4 | -1
Six faces of a die are marked with the numbers 1,−1,0,−2,21,-1,0,-2,2 and 3 . The die is thrown thrice. The probability that the sum of the numbers thrown is six, is
View answer and explanation
Correct option: C
5108\frac{5}{108}
Option C (5108\frac{{5}}{{108}}) is correct because it calculates the probability of the sum of the numbers thrown being six by considering the possible combinations and dividing by the total possible outcomes.
171 Medium
171 of 285
Application +4 | -1
A coin is tossed 2n2n times. The chance that the number of times one gets head is not equal to the number of times one gets tail, is
View answer and explanation
Correct option: C
1−(2n!)(n!)2.14n1 - \frac{{\left( {2n!} \right)}}{{{{\left( {n!} \right)}^2}}}.\frac{1}{{{4^n}}}
The correct answer is C (1−(2n!)(n!)2.14n1 - \frac{{\left( {2n!} \right)}}{{{{\left( {n!} \right)}^2}}}.\frac{1}{{{4^n}}}) because it represents 1 minus the probability of getting an equal number of heads and tails in 2n tosses.
172 Medium
172 of 285
Application +4 | -1
The probability that a man will live 10 more years is 1/4 and the probability that his wife will live 10 more years is 1/3. Then the probability that neither will be alive in 10 years, is
View answer and explanation
Correct option: B
01-Feb
Option B (1/2) is correct because it calculates the probability that both the man and his wife will not be alive in 10 years by multiplying the probabilities of each event.
173 Medium
173 of 285
Analyzing +4 | -1
Consider two events AA and BB such that P(A)=14,  P(BA)=12,  P(AB)=14.P\left( A \right) = \frac{1}{4},\;P\left( {\frac{B}{A}} \right) = \frac{1}{2},\;P\left( {\frac{A}{B}} \right) = \frac{1}{4}. For each of the following statements, which is true?
1. P(Ac+Bc)=34  P\left( {{A^c} + {B^c}} \right) = \frac{3}{4}\;
2. The events AA and BB are mutually exclusive.
3. P(A/B) + P(A/Bc) = 1P(A/B)\, + \,P(A/{B^c})\, = \,1
View answer and explanation
Correct option: A
(1) only
Option A ((1) only) is correct because only statement 1 is true, while statements 2 and 3 are false.
174 Medium
174 of 285
Analyzing +4 | -1
If AA and BB are two events such that P(A)≠0P\left( A \right) \ne 0 and P(B)≠1,P\left( B \right) \ne 1, then P(AB)‾P\overline {\left( {\frac{A}{B}} \right)} is equal to
View answer and explanation
Correct option: C
1−P(A∪B)P(Bˉ)\frac{{1 - P\left( {A \cup B} \right)}}{{P\left( {\bar B} \right)}}
Option C (1−P(A∪B)P(Bˉ)\frac{{1 - P(A \cup B)}}{{P(\bar B)}}) is correct because it correctly applies De Morgan's Law and the definition of conditional probability to find the complement of the conditional probability.
175 Medium
175 of 285
Application +4 | -1
The coefficients a,b,ca, b, c in the quadratic equation ax2+bx+c=0a x^2+b x+c=0 are chosen from the set
{1,2,3,4,5,6,7,8}.\{1,2,3,4,5,6,7,8\}. The probability of this equation having repeated roots is:
View answer and explanation
Correct option: C
164\frac{1}{64}
The correct answer is C (164\frac{{1}}{{64}}) because it is the probability of the quadratic equation having repeated roots, which occurs when the discriminant is zero.
176 Medium
176 of 285
Application +4 | -1
Two dice are thrown simultaneously. What is the probability of obtaining a multiple of 2 on one of them and a multiple of 3 on the other
View answer and explanation
Correct option: B
1136\frac{11}{36}
Option B (1136\frac{11}{36}) is correct. It accounts for the combinations where one die shows a multiple of 2 and the other shows a multiple of 3, avoiding double-counting the case where both dice show a multiple of both 2 and 3 (i.e., 6).
177 Medium
177 of 285
Application +4 | -1
Let p{p} denote the probability that a man aged xx years will die in a year. The probability that out of n{n} men A1,A2,A3,……An{A}_1,{A}_2, {A}_3, \ldots \ldots {A}_n each aged x,A1x, {A}_1 will die in a year and will be the first to die is
View answer and explanation
Correct option: A
1n[1−(1−p)n]\frac{1}{n}\left[1-(1-p)^n\right]
Option A is correct because it represents the probability that A1 dies in a year and is the first to die, derived using probability concepts and independence of events.
178 Medium
178 of 285
Application +4 | -1
A bag contains 10 white and 3 black balls. Balls are drawn one-by-one without replacement till all the black balls are drawn. The probability that the procedure of drawing balls will come to an end at the seventh draw is
View answer and explanation
Correct option: B
15286\frac{{15}}{{286}}
Option B (15286\frac{{15}}{{286}}) is correct because it represents the probability that the last black ball is drawn on the seventh draw, calculated using conditional probability.
179 Medium
179 of 285
Application +4 | -1
A bag contains 5 white, 3 black balls. 2 balls are drawn one by one, their colour noted and
replaced in bag then probability that one ball is white and one is black, is
View answer and explanation
Correct option: A
1532\frac{15}{32}
Option A (1532\frac{{15}}{{32}}) is correct because it calculates the probability of drawing one white and one black ball in either order, with replacement.
180 Medium
180 of 285
Application +4 | -1
A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is
View answer and explanation
Correct option: C
1135\frac{11}{3^5}
Option C (1135\frac{11}{3^5}) is correct because it calculates the probability of getting 4 or 5 correct answers by guessing using the binomial distribution formula.
Revision notes and key takeaways

Review PROBABILITY in your textbook or class notes, then use this set to check your understanding.

Showing 151-180 of 285 questions

Your practice

0 / 285 answered · 285 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.