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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 691-720 of 1130 questions

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691 Easy
691 of 1130
Application +4 | -1
If two events AA and BB such that P(A′)=0.3,P(B)=0.5P\left(A^{\prime}\right)=0.3, P(B)=0.5 and P(A∩B)=0.3P(A \cap B)=0.3, then P(B∣A∪B′)P\left(B \mid A \cup \mathrm{B}^{\prime}\right) is
View answer and explanation
Correct option: A
0.6
Option A (0.6) is correct because it represents the conditional probability of B given A∪B′A \cup B', calculated using the formula for conditional probability and the given probabilities.
692 Medium
692 of 1130
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.
693 Easy
693 of 1130
Application +4 | -1
A student appears for tests I, II and III. The student is successful if he passes either in tests I and II or tests I and IV. The probabilities of the student passing in tests I, II, III are p,q\mathrm{p}, \mathrm{q} and 12\frac{1}{2} respectively. The probability that the student is successful is 12\frac{1}{2} then the relation between p\mathrm{p} and qq is given by
View answer and explanation
Correct option: A
pq+p=1p q+p=1
Option A (pq+p=1p q+p=1) is correct because it represents the probability of the student being successful, which is given as 12\frac{1}{2}.
694 Medium
694 of 1130
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.
695 Medium
695 of 1130
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.
696 Easy
696 of 1130
Application +4 | -1
Five coins whose faces are marked 2, 3 are tossed. The chance of obtaining a total of 12 is
View answer and explanation
Correct option: D
May-16
The correct answer is May-16, which represents 5/16. This is calculated using binomial probability, considering the probability of getting a 2 or 3 on each coin toss.
697 Easy
697 of 1130
Application +4 | -1
A box contains 100 tickets numbered 1, 2, …, 100. Two tickets are choosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The probability that the minimum number is 5 is
View answer and explanation
Correct option: D
01-Sep
The correct answer is 1/9. The probability is calculated using conditional probability, considering the event that the maximum number is not more than 10.
698 Hard
698 of 1130
Analyzing +4 | -1
A ship is fitted with three engines E1,E2E_1, E_2 and E3E_3. The engines function independently of each other with respective probabilities 12,14\frac{1}{2}, \frac{1}{4} and 14\frac{1}{4}. For the ship to be operational at least two of its engines must function. Let XX denote the event that ship is operational and let X1,X2X_1, X_2 and X3X_3 denote respectively the events that the engines E1,E2E_1, E_2 and E3E_3 are functioning. Which of the following is not true?
View answer and explanation
Correct option: C
P(X/X2)=516P\left(X / X_2\right)=\frac{5}{16}
Option C (P(X/X2)=516P(X / X_2)=\frac{{5}}{{16}}) is not true because the correct conditional probability P(X/X2)P(X/X_2) is 58\frac{{5}}{{8}}, not 516\frac{{5}}{{16}}.
699 Easy
699 of 1130
Application +4 | -1
A bag contains 5 brown and 4 white socks. A man pulls out two socks. The probability that these are of the same colour, is
View answer and explanation
Correct option: D
48/108
Option D (48/108) is correct because it calculates the probability of drawing two socks of the same color by considering both brown and white socks.
700 Easy
700 of 1130
Application +4 | -1
A man is known to speak truth 3 out of 4
times. He takes out a card at random from a well shuffled
pack of 52 playing cards, and reports it is a king. The
probability that its actually a king is
View answer and explanation
Correct option: D
15\frac{1}{5}
Option D (15\frac{1}{5}) is correct because it uses Bayes' theorem to calculate the probability that the card is actually a king, given that the man reports it is a king, considering his truth-telling probability.
701 Easy
701 of 1130
Analyzing +4 | -1
A book contains 1000 pages. A page is chosen at random. The probability that the sum of the digits of the marked number on the page is equal to 9 , is
View answer and explanation
Correct option: B
11200\frac{11}{200}
Option B (11200\frac{11}{200}) is correct because it represents the ratio of favorable outcomes (sum of digits equals 9) to the total possible outcomes (1000 pages).
702 Easy
702 of 1130
Application +4 | -1
There are 4 envelopes corresponding to 4 letters. If the letters are placed in the envelopes at random. The probability that all the letters are not placed in the right envelopes, is
View answer and explanation
Correct option: B
2324\frac{23}{24}
Option B (23/24) is correct because it calculates the probability that not all letters are placed in the right envelopes by subtracting the probability that all letters are in the correct envelopes from 1.
703 Easy
703 of 1130
Understanding +4 | -1
If E‾\overline{{E}} and F‾\overline{{F}} are the complementary events of events E{E} and F{F} respectively and if 0<P(F)<10 < {P}({F}) < 1, then
View answer and explanation
Correct option: A
P(E∣F)+P(E‾∣F)=1{P}({E} \mid {F})+{P}(\overline{{E}} \mid {F})=1
Option A (P(E∣F)+P(E‾∣F)=1{P}({E} \mid {F})+{P}(\overline{{E}} \mid {F})=1) is correct because given event F, either E occurs or its complement E‾\overline{E} occurs, and the sum of their conditional probabilities is 1.
704 Easy
704 of 1130
Application +4 | -1
A bag contain 5 black balls, 4 white balls and 3 red balls. If a ball is selected randomly, the probability that it is a black or red ball, is
View answer and explanation
Correct option: D
02-Mar
Option D (2/3) is correct because it calculates the probability of selecting a black or red ball by summing the individual probabilities and dividing by the total.
705 Easy
705 of 1130
Application +4 | -1
Six faces of an unbiased die are numbered with 2, 3, 5, 7, 11 and 13. If two such dice are thrown, then the probability that the sum on the uppermost faces of the dice is an odd number, is
View answer and explanation
Correct option: A
518\frac{5}{{18}}
Option A (5/18) is correct because the sum is odd only when one die shows an odd number and the other shows an even number.
706 Easy
706 of 1130
Application +4 | -1
A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six, is
View answer and explanation
Correct option: A
3/83 / 8
Option A (3/83 / 8) is correct because it uses Bayes' theorem to find the probability that it was actually a six, given the man reported it was a six.
707 Easy
707 of 1130
Application +4 | -1
If events are independent and P(A)=13,  P(B)=13,  P(C)=14,  P\left( A \right) = \frac{1}{3},\;P\left( B \right) = \frac{1}{3},\;P\left( C \right) = \frac{1}{4},\; then P(A′∩B′∩C′)\mathrm{P}\left( {A' \cap B' \cap C'} \right) is equal to
View answer and explanation
Correct option: C
13\frac{1}{3}
Option C (1/3) is correct because the probability of independent events is calculated by multiplying the probabilities of the complements of each event.
708 Easy
708 of 1130
Analyzing +4 | -1
One Indian and four American men and their wives are to be seated randomly around a circular table. The conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife, is
View answer and explanation
Correct option: C
25\frac{2}{5}
Option C (25\frac{{2}}{{5}}) is correct because it represents the conditional probability of the Indian man sitting next to his wife given that each American man is seated next to his wife around the circular table.
709 Easy
709 of 1130
Analyzing +4 | -1
If two squares are chosen at random on a chess board, the probability that they have a side in common is
View answer and explanation
Correct option: B
Jan-18
The correct answer is Jan-18 (118\frac{1}{18}) because it calculates the probability of two randomly chosen squares on a chessboard sharing a side.
710 Easy
710 of 1130
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 more correct answers by guessing using the binomial distribution formula.
711 Medium
711 of 1130
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.
712 Easy
712 of 1130
Application +4 | -1
AA and BB toss a coin alternatively till one of them gets a head and wins the game. If AA begins the game, the probability BB wins the game is
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{{1}}{{3}}) is correct because it is calculated by summing the probabilities of B winning on their first, second, third, etc. turns, which forms a geometric series.
713 Easy
713 of 1130
Application +4 | -1
If the probability for AA to fail in an examination is 0.2 and that for BB is 0.3, then the probability that either AA or BB fails, is
View answer and explanation
Correct option: B
0.44
Option B (0.44) is correct because it calculates the probability of either A or B failing by considering the probabilities of A failing, B failing, and both failing.
714 Medium
714 of 1130
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.
715 Easy
715 of 1130
Application +4 | -1
Two numbers a and b are chosen at random from the set of first 30 natural numbers. The probability that a2−b2{a^2} - {b^2} is divisible by 3 is
View answer and explanation
Correct option: D
4787\frac{{47}}{{87}}
Option D (4787\frac{{47}}{{87}}) is correct because it calculates the probability that a2−b2{a^2} - {b^2} is divisible by 3 when a and b are chosen from the first 30 natural numbers.
716 Medium
716 of 1130
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.
717 Easy
717 of 1130
Application +4 | -1
The records of a hospital show that 10% of the cases of a certain disease are fatal. If 6 patients are suffering from the disease, then the probability that only three will die, is
View answer and explanation
Correct option: B
1458×10−51458 \times {10^{ - 5}}
Option A (8748×10−58748 \times {10^{ - 5}}) is incorrect. The correct answer is B (1458×10−51458 \times {10^{ - 5}}) because it represents the probability of exactly three patients dying out of six, calculated using the binomial probability formula.
718 Medium
718 of 1130
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).
719 Medium
719 of 1130
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.
720 Easy
720 of 1130
Application +4 | -1
A fair coin is tossed a fixed number of times. If the probability of getting 4 heads equals the probability of getting 7 heads, then the probability of getting 2 heads is
View answer and explanation
Correct option: A
552048\frac{{55}}{{2048}}
Option A (552048\frac{{55}}{{2048}}) is correct because it is the probability of getting 2 heads when the number of tosses is 11, derived from the given condition P(4 heads) = P(7 heads).
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