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JEE • MATHEMATICS • PROBABILITY

Probability practice · Medium

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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 91-120 of 285 questions
91 Medium
91 of 285
Application +4 | -1
There are three events E1,E2{E}_1, {E}_2 and E3,{E}_3, one of which must, and only one can happen. The odds are 7 to 4 against E1{E}_1 and 5 to 3 against E2{E}_2. The odds against E3{E}_3 is
View answer and explanation
Correct option: D
65:2365: 23
Option D (65:2365: 23) is correct because it calculates the odds against E3 by first finding the probability of E3 and then using the odds formula.
92 Medium
92 of 285
Application +4 | -1
There are 5 identical red balls and 3 identical yellow balls in a box. Two balls are drawn at random, then the probability that their colour is different is
View answer and explanation
Correct option: C
1528\frac{15}{28}
Option C (1528\frac{15}{28}) is correct because it represents the probability of drawing two balls of different colors, calculated by considering the combinations of red and yellow balls.
93 Medium
93 of 285
Analyzing +4 | -1
From a set of 100 cards numbered 1 to 100, one card is drawn at random. The probability that the number obtained on the card is divisible by 6 or 8 but not by 24, is
View answer and explanation
Correct option: D
15\frac{1}{5}
Option D (15\frac{1}{5}) is correct because it correctly calculates the number of integers divisible by 6 or 8 but not by 24 and divides it by the total number of cards.
94 Medium
94 of 285
Application +4 | -1
The probabilities of Ramesh using car or scooter or bus or train for going to office are respectively 17,37,27\frac{1}{7}, \frac{3}{7}, \frac{2}{7} and 17\frac{1}{7}. The probabilities of his reaching the office late using these modes of transport are respectively 29,19,49\frac{2}{9}, \frac{1}{9}, \frac{4}{9} and 19\frac{1}{9}, On one day Ramesh reaches his office on time. Then the probability that he used car on that day is
View answer and explanation
Correct option: B
17\frac{1}{7}
Option B (17\frac{1}{7}) is correct because it is calculated using Bayes' theorem, considering the probability of using a car given that Ramesh reached the office on time.
95 Medium
95 of 285
Understanding +4 | -1
There are four machines and it is known that exactly two of them are faulty. They are tested, one by one in a random order till both the faulty machines are identified. Then, the probability that only two tests are needed, is
View answer and explanation
Correct option: B
16\frac{1}{6}
Option B (16\frac{1}{6}) is correct because it is the probability that the first two machines tested are the faulty ones.
96 Medium
96 of 285
Analyzing +4 | -1
If x=3n,x=3^n, where nn is a positive integral value, then what is the probability that xx will have 3 at unit's place?
View answer and explanation
Correct option: B
14\frac{1}{4}
Option B (14\frac{1}{4}) is correct because the unit's digit of 3n3^n repeats every 4 powers (3, 9, 7, 1).
97 Medium
97 of 285
Application +4 | -1
For a party 7 guests are invited by a husband and his wife. They sit in a row for dinner. The probability that the husband and his wife sit together, is
View answer and explanation
Correct option: B
29\frac{2}{9}
Option B (29\frac{2}{9}) is correct because it is the probability of the husband and wife sitting together, calculated by considering them as a single unit.
98 Medium
98 of 285
Application +4 | -1
If AA and BB are mutually exclusive events with P(B)≠1P\left( B \right) \ne 1, then P(A∣Bˉ)P(A|\bar B) is equal to (Here, Bˉ\bar B is the complement of the event BB)
View answer and explanation
Correct option: D
P(A)1−P(B)\frac{{P\left( A \right)}}{{1 - P\left( B \right)}}
Option D (P(A)/(1-P(B))) is correct because it correctly applies the conditional probability formula given that A and B are mutually exclusive.
99 Medium
99 of 285
Application +4 | -1
Let S={M=[aij],aij∈}{0,1,2},{1≤i,j≤2}S=\left\{M=\left[a_{i j}\right], a_{i j} \in\right\}\{0,1,2\},\{1 \leq i, j \leq 2\} be a sample space and A{M∈S:MA\{M \in S: M is invertible }\} be an even. Then P(A)P(A) is equal to
View answer and explanation
Correct option: D
5081\frac{50}{81}
Option D (50/81) is correct because it represents the probability of the matrix being invertible, calculated by subtracting the probability of the matrix being non-invertible from 1.
100 Medium
100 of 285
Analyzing +4 | -1
If P(B)=34,  P(A∩B∩Cˉ)=13  P\left( B \right) = \frac{3}{4},\;P\left( {A \cap B \cap \bar C} \right) = \frac{1}{3}\; and P(Aˉ∩B∩Cˉ)=13,  P\left( {\bar A \cap B \cap \bar C} \right) = \frac{1}{3},\; then P(B∩C)P\left( {B \cap C} \right) is
View answer and explanation
Correct option: A
112\frac{1}{{12}}
Option A (112\frac{1}{{12}}) is correct because it is the calculated probability of the intersection of events B and C, given the provided probabilities.
101 Medium
101 of 285
Analyzing +4 | -1
If nn positive integers are taken at random and multiplied together, the probability that the last digit of the product is 2, 4, 6 or 8, is
View answer and explanation
Correct option: C
4n−2n5n\frac{{{4^n} - {2^n}}}{{{5^n}}}
Option C (4n−2n5n\frac{{{4^n} - {2^n}}}{{{5^n}}}) is correct because it calculates the probability that the product of n random integers ends in 2, 4, 6, or 8.
102 Medium
102 of 285
Analyzing +4 | -1
In a bag there are 10 balls of either Black or White. Let AkA_k be the event which contains exactly kk white balls and its probability is proportional to k2k^2 Now an another ball is drawn randomly. If P(E)P(E) be the probability that the ball drawn is white and P(E)=abP(E)=\frac{a}{b} Where HCF⁡(a,b)=1\operatorname{HCF}(a, b)=1 then the value of ∣a−b∣|\mathrm{a}-\mathrm{b}| equals
View answer and explanation
Correct option: D
3
Option D (3) is correct because it calculates the probability of drawing a white ball and then finds the absolute difference between the numerator and denominator of the simplified probability.
103 Medium
103 of 285
Application +4 | -1
The number of times a die must be tossed to obtain a 6 at least once with probability exceeding 0.9 is at least
View answer and explanation
Correct option: A
13
Option A (13) is correct because it is the smallest integer value of n for which the probability of obtaining a 6 at least once exceeds 0.9.
104 Medium
104 of 285
Analyzing +4 | -1
Let E′E^{\prime} denote the complement of an event EE. Let E,FE, F, GG be pairwise independent events such that P(G)>0P(G)>0 and P(E∩F∩G)=0P(E \cap F \cap G)=0. Then P(E′∩F′/G)P\left(E^{\prime} \cap F^{\prime} / G\right) equals
View answer and explanation
Correct option: C
P(E′)−P(F)P\left(E^{\prime}\right)-P(F)
The correct answer is C (P(E′)−P(F)P(E')-P(F)) because it correctly applies the properties of pairwise independent events and conditional probability to simplify the expression.
105 Medium
105 of 285
Application +4 | -1
A bag contains 5 white, 6 green, 7 blue, 8 red and 9 black balls. If 3 balls are drawn one by one with replacement after each drawing, then the probability that 3rd3^{\text {rd}} ball drawn is green is, if it is known that second ball is green:
View answer and explanation
Correct option: C
635\frac{6}{35}
Option C (6/35) is correct because the probability of drawing a green ball on the third draw is independent of the second draw due to replacement, and the total number of green balls is 6 out of 35 total balls.
106 Medium
106 of 285
Application +4 | -1
A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters, TA, are visible. The probability that the letter has come from CALCUTTA is
View answer and explanation
Correct option: A
411\frac{4}{11}
Option A (411\frac{4}{11}) is correct because it is calculated using Bayes' theorem, considering the probabilities of the letter coming from each city given the observed letters.
107 Medium
107 of 285
Application +4 | -1
A box contains 15 transistors, 5 of which are defective. An inspector takes out one transistor at random, examines it for defects and replaces it. After it has replaced another inspector does the same thing and then so does a third inspector. The probability that atleast one of the inspectors finds a defective transistor, is equal to
View answer and explanation
Correct option: C
19/27
Option C (19/27) is correct because it represents the probability of at least one defective transistor being found, calculated as 1 minus the probability of all transistors being non-defective.
108 Medium
108 of 285
Application +4 | -1
Throwing a biased die, a person will get 5 rupees if he throws the number 5 and will get 8 rupees for any
other number, then the expected income (in Rupees) per throw is (it is given that the number 5 will appear
5 times as frequently as any other number)
View answer and explanation
Correct option: A
6.5
Option A (6.5) is correct because the expected income is calculated by considering the probabilities of each outcome and their corresponding values.
109 Medium
109 of 285
Application +4 | -1
Two numbers aa and bb are chosen at random from the set {1,2,3,…,3n}\{1,2,3, \ldots, 3 n\}. The probability that a2−b2a^2-b^2 is divisible by 3 , is
View answer and explanation
Correct option: B
5n−33(3n−1)\frac{5 n-3}{3(3 n-1)}
Option B (5n−33(3n−1)\frac{5 n-3}{3(3 n-1)}) is correct because it calculates the probability that a2−b2a^2 - b^2 is divisible by 3 by considering the cases where a and b belong to the same residue class modulo 3 or to different classes that satisfy the divisibility condition.
110 Medium
110 of 285
Application +4 | -1
An anti-aircraft gun can take a maximum of four shots at any plane moving away from it. The probabilities of hitting the plane at the 1st, 2nd, 3rd and 4th shots are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that at least one shot hits the plane?
View answer and explanation
Correct option: A
0.6976
Option A (Nov-32) is correct because it is the sum of probabilities of getting exactly 4, 5, or 6 heads when tossing six coins simultaneously.
111 Medium
111 of 285
Analyzing +4 | -1
Let R{R} be the region containing all the points within or on the sides of an equilateral triangle. A point PP is randomly selected from the region RR, the probability that PP lies near to internal angle bisector than any of the sides is equal to
View answer and explanation
Correct option: A
4−234-2 \sqrt{3}
The correct answer is {0} because the sine function will always evaluate to zero due to the greatest integer function, making the entire expression zero.
112 Medium
112 of 285
Analyzing +4 | -1
AA is a set containing n elements. A subset PP of AA is chosen at random. The set AA is reconstructed by replacing the elements of PP. A subset QQ is again chosen at random. The Probability that P∪Q=AP \cup Q=A, is
View answer and explanation
Correct option: D
n3(34)n\frac{n}{3}\left(\frac{3}{4}\right)^n
Option B ((34)n\left(\frac{3}{4}\right)^n) is correct because it calculates the probability that the union of two randomly chosen subsets P and Q of A equals A.
113 Medium
113 of 285
Application +4 | -1
Among 15 players, 8 are batsman and 7 are bowlers. The probability that a team is chosen of 6 batsman and 5 bowlers, is
View answer and explanation
Correct option: A
  8C6×  7C5  15C11\frac{{{\;^8}{C_6} \times {\;^7}{C_5}}}{{{\;^{15}}{C_{11}}}}
Option A (  8C6×  7C5  15C11\frac{{{\;^8}{C_6} \times {\;^7}{C_5}}}{{{\;^{15}}{C_{11}}}}) is correct because it represents the number of ways to choose 6 batsmen from 8 and 5 bowlers from 7, divided by the total number of ways to choose 11 players from 15.
114 Medium
114 of 285
Application +4 | -1
If X{X} and Y{Y} are the independent random variables B(5,12){B}\left(5, \frac{1}{2}\right) and B(7,12){B}\left(7, \frac{1}{2}\right), then P(X+Y≥1)={P}({X}+{Y} \geq 1)=
View answer and explanation
Correct option: A
40954096\frac{4095}{4096}
Option A (40954096\frac{4095}{4096}) is correct because it calculates the probability of the sum of two binomial random variables being greater than or equal to 1.
115 Medium
115 of 285
Analyzing +4 | -1
AA is a set containing n elements. A subset PP of AA is chosen at random. The set AA is reconstructed by replacing the elements of PP. A subset QQ is again chosen at random. The Probability that P∩QP \cap Q contain just one element, is
View answer and explanation
Correct option: C
n3(34)n\frac{n}{3}\left(\frac{3}{4}\right)^n
Option B (n(34)n−1n\left(\frac{3}{4}\right)^{n-1}) is incorrect. The correct answer is D (n4(34)n−1\frac{n}{4}\left(\frac{3}{4}\right)^{n-1}). The provided solution steps in the original response contain errors and lead to the incorrect option C. The correct derivation is shown below.
116 Medium
116 of 285
Application +4 | -1
Minimum number of times a fair coin must be tossed so that the probability of getting atleast one head is more than 99% is
View answer and explanation
Correct option: C
7
Option C (7) is correct because 7 is the smallest number of tosses required to have a probability of at least one head greater than 99%.
117 Medium
117 of 285
Application +4 | -1
The probability that a randomly chosen 5-digit number is made from exactly two digits is
View answer and explanation
Correct option: A
135104\frac{135}{10^4}
Option A (135104\frac{135}{10^4}) is correct because it calculates the probability of a 5-digit number being made from exactly two digits considering cases with and without zero.
118 Medium
118 of 285
Analyzing +4 | -1
From first 100 natural numbers, 3 numbers are selected. If these numbers are in A.P., then the probability that these numbers are even is:
View answer and explanation
Correct option: C
1249\frac{12}{49}
Option C (1249\frac{12}{49}) is correct because it calculates the conditional probability of selecting three even numbers in A.P. given that the three selected numbers are in A.P.
119 Medium
119 of 285
Understanding +4 | -1
If AA and BB are two events such that P(A)>0P\left( A \right) > 0 and P(B)≠1,P\left( B \right) \ne 1, then P(Aˉ∣Bˉ)P(\bar A|\bar B) is equal to
View answer and explanation
Correct option: A
1−P(A∣Bˉ)1 - P(A|\bar B)
Option A (1−P(A∣Bˉ)1 - P(A|\bar B)) is correct because P(Aˉ∣Bˉ)+P(A∣Bˉ)=1P(\bar A|\bar B) + P(A|\bar B) = 1, therefore P(Aˉ∣Bˉ)=1−P(A∣Bˉ)P(\bar A|\bar B) = 1 - P(A|\bar B).
120 Medium
120 of 285
Application +4 | -1
Two numbers are selected at random from the numbers 1,2,⋯ ,n1,2, \cdots, n. Let pp denote the probability that the difference between the first and second is not less than mm (where 0<m<n0 < m< n ). If n=25n=25 and m=10m=10 then 5p5 p is equal
View answer and explanation
Correct option: A
1
Option A (1) is correct because it is derived by calculating the probability that the difference between two randomly selected numbers is not less than m, and then multiplying by 5.
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