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JEE • MATHEMATICS • PROBABILITY

Probability practice · Medium

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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 61-90 of 285 questions
61 Medium
61 of 285
Analyzing +4 | -1
Let MM be the maximum value of the product of two positive integers when their sum is 66 . Let the sample space S={x∈Z:x(66−x)≥59M}S=\left\{x \in \mathbb{Z}: x(66-x) \geq \frac{5}{9} M\right\} and the event A={x∈S:xA=\{x \in S: x is a multiple of 3}\}. Then P(A)P(A) is equal to
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{1}{3}) is correct because we first find the maximum product M, then determine the sample space S and event A, and finally calculate the probability P(A).
62 Medium
62 of 285
Analyzing +4 | -1
The probability of getting a total of at least 6 in the simultaneously throw of three dice is
View answer and explanation
Correct option: D
103108\frac{{103}}{{108}}
The correct answer is D (103/108) because it represents the probability of the complement event (getting at least 6) calculated by subtracting the probability of the event not happening (getting less than 6) from 1.
63 Medium
63 of 285
Application +4 | -1
A and B throw a die alternatively. If A starts throwing then the probability that he throws 5 before B throws 3 or 6 is
View answer and explanation
Correct option: A
3/83/8
Option A (3/8) is correct because it represents the probability that A throws a 5 before B throws a 3 or 6, calculated using an infinite geometric series.
64 Medium
64 of 285
Application +4 | -1
Probability that Ajay will not go to office is 15\frac{1}{5} and probability that Ajay and Vijay will go to office is 27\frac{2}{7}, if their visits to office is independent of each other then find probability that Ajay will go to the office but Vijay will not go, is
View answer and explanation
Correct option: C
1835\frac{18}{35}
Option C (1835\frac{18}{35}) is correct because it is the probability of Ajay going to the office and Vijay not going, calculated using independent probabilities.
65 Medium
65 of 285
Application +4 | -1
Three people each flip two fair coins. The probability that exactly two of the people flipped one head and one tail, is
View answer and explanation
Correct option: B
3/83 / 8
Option B (3 / 8) is correct because it is the probability calculated using the binomial probability formula with n=3, r=2, and p=1/2.
66 Medium
66 of 285
Application +4 | -1
Of a total of 600 bolts, 20% are too large and 10% are too small. The remainder are considered to be suitable. If a bolt is selected at random, the probability that it will be suitable is
View answer and explanation
Correct option: B
710\frac{7}{{10}}
Option B (7/10) is correct because it calculates the probability of a bolt being suitable by dividing the number of suitable bolts by the total number of bolts.
67 Medium
67 of 285
Application +4 | -1
A sample of a 4 times is drawn at a random without replacement from a lot of 10 items containing 3 defectives. If xx denotes the number of defective items in the sample, then P(0<x<3)P(0 < x < 3) is equal to
View answer and explanation
Correct option: B
45\frac{4}{5}
Option B (4/5) is correct because it calculates the probability that the number of defective items in the sample is between 0 and 3 by considering the possible combinations.
68 Medium
68 of 285
Application +4 | -1
A five digit number if 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
15\frac{1}{5}
Option C (15\frac{1}{5}) is correct because it is the probability that the number is divisible by 4, calculated by dividing the number of favorable outcomes by the total number of outcomes.
69 Medium
69 of 285
Application +4 | -1
If two events AA and BB are such that P(Ac)=0.3,P(B)=0.4,P(A∩Bc)=0.5,P\left( {{A^c}} \right) = 0.3,P\left( B \right) = 0.4,P\left( {A \cap {B^c}} \right) = 0.5, then P(B/A∪Bc)=P\left( {B/A \cup {B^c}} \right) =
View answer and explanation
Correct option: B
0.25
Option B (0.25) is correct because it is calculated using the conditional probability formula and the given probabilities.
70 Medium
70 of 285
Application +4 | -1
A party is thrown for families having boys, girls, ladies and gentleman. When one member is randomly selected, odds against being a boy is 275\frac{27}{5}, odds in favour of being a girl is 725\frac{7}{25} and probability of being a lady is 1132\frac{11}{32}, then the odds in favour of the chosen person being a gentleman, is
View answer and explanation
Correct option: A
923\frac{9}{23}
Option A (923\frac{9}{23}) is correct because it calculates the probability of selecting a gentleman and then converts it to odds in favor.
71 Medium
71 of 285
Application +4 | -1
If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is
View answer and explanation
Correct option: A
1332\frac{13}{32}
Option A (1332\frac{13}{32}) is correct because it calculates the probability of drawing 2 white and 1 black ball from the three boxes.
72 Medium
72 of 285
Analyzing +4 | -1
Let two fair six faced dice AA and BB be thrown simultaneously. If E1E_1 is the event that die A shows up four, E2E_2 is the event that BB shows up two and E3E_3 is the event that the sum of the numbers on both dice is odd, then which the following statements is not true?
View answer and explanation
Correct option: D
E1,E2E_1, E_2 and E3E_3 are independent
Option D (E1,E2E_1, E_2 and E3E_3 are independent) is the correct answer because while E1E_1, E2E_2, and E3E_3 are pairwise independent, they are not mutually independent.
73 Medium
73 of 285
Application +4 | -1
3%3 \% of the electric bulbs manufactured by a company are defective. Using Poisson distribution (approximation), the probability that a sample of 100 bulbs will contain exactly one defective, is

(Given antilog (.1742)=1.5)(.1742)=1.5)
View answer and explanation
Correct option: B
0.15
Option B (0.15) is correct because it approximates the probability of exactly one defective bulb using the Poisson distribution.
74 Medium
74 of 285
Application +4 | -1
Suppose that a die (with faces marked 1 to 6) is loaded in such a manner that for K=1,2,3,….,6K = 1,2,3, \ldots .,6 the probability of the face marked KK turning up when die is tossed is proportional to KK. The probability of the event that the outcome of a toss of the die will be an even number, is equal to
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A, 12\frac{1}{2} is correct because when the probability is proportional to K, the probability of an even number is simply the sum of probabilities of even numbers divided by the sum of probabilities of all numbers.
75 Medium
75 of 285
Application +4 | -1
An integer is chosen at random from the integers 1,2,3,…,501,2,3, \ldots, 50. The probability that the chosen integer is a multiple of atleast one of 4,6 and 7 is
View answer and explanation
Correct option: C
2150\frac{21}{50}
The correct answer is C (21/50) because it represents the probability calculated using the principle of inclusion-exclusion for multiples of 4, 6, and 7.
76 Medium
76 of 285
Analyzing +4 | -1
If XX is a binomial variate with parameters nn and pp, where 0<p<10 < p < 1 such that P(X=r)P(X=n−r)\frac{P(X=r)}{P(X=n-r)} is independent of nn and rr, then pp equals
View answer and explanation
Correct option: A
1/21 / 2
Option A (1/21/2) is correct because the condition P(X=r)P(X=n−r)\frac{P(X=r)}{P(X=n-r)} being independent of nn and rr implies that pp must be equal to 1/21/2.
77 Medium
77 of 285
Application +4 | -1
If X\mathrm{X} follows a binomial distribution with parameters n=100\mathrm{n}=100 and p=1/3\mathrm{p}=1 / 3, find r\mathrm{r} for which P(X=r)\mathrm{P}(\mathrm{X}=\mathrm{r}) is maximum
View answer and explanation
Correct option: C
33
Option C (33) is correct because the value of r for which P(X=r) is maximum is the mode of the binomial distribution, which is the integer part of (n+1)p = (100+1)(1/3) = 33.67, hence r=33.
78 Medium
78 of 285
Application +4 | -1
An urn contains 6 white and 9 black balls. Two successive draws of 4 balls are made without replacement. The probability that the first draw gives all white balls and second draw gives all black balls is
View answer and explanation
Correct option: D
3715\frac{3}{715}
Option D (3715\frac{3}{715}) is correct because the probability of drawing 4 white balls then 4 black balls without replacement is calculated as 6C415C4⋅9C411C4=3715\frac{{}^6C_4}{{}^{15}C_4} \cdot \frac{{}^9C_4}{{}^{11}C_4} = \frac{3}{715}.
79 Medium
79 of 285
Analyzing +4 | -1
For any two independent events E1{E_1} and E2,  P{(E1∪E2)∩(Eˉ1)∩(Eˉ2)}{E_2},\;P\left\{ {\left( {{E_1} \cup {E_2}} \right) \cap \left( {{{\bar E}_1}} \right) \cap \left( {{{\bar E}_2}} \right)} \right\}
View answer and explanation
Correct option: A
≤1/4\le 1/4
Option A (≤1/4\le 1/4) is correct because the expression simplifies to 0, which is less than or equal to 1/4.
80 Medium
80 of 285
Application +4 | -1
A bag contains 5 white and 3 black balls and 4 balls are successively drawn out and not replaced. The probability that they are alternately of different colours, is
View answer and explanation
Correct option: C
17\frac{1}{7}
Option C (1/7) is correct because it calculates the probability of alternately drawing white and black balls without replacement.
81 Medium
81 of 285
Application +4 | -1
Two integers rr and ss are drawn one at a time without replacement from the set 1,2,⋯ ,n1,2, \cdots, n. If pk=P(r≤k∣s≤k)p_k=P(r \leq k \mid s \leq k), then 4p74 p_7 is if n=25n=25.
View answer and explanation
Correct option: A
1
Option A (1) is correct because given the condition p_k = P(r <= k | s <= k), and n=25, k=7, then 4p_7 = 1.
82 Medium
82 of 285
Application +4 | -1
Bag A contains 4 green and 3 red balls and Bag B\mathrm{B} contains 4 red and 3 green balls. One bag is taken at random and a ball is noted it is green. The probability that it comes from Bag  B\mathrm{Bag} \,\, \mathrm{B} is -
View answer and explanation
Correct option: C
37\frac{3}{7}
Option C (37\frac{{3}}{{7}}) is correct because the probability is calculated using Bayes' theorem, considering the probabilities of drawing a green ball from each bag.
83 Medium
83 of 285
Application +4 | -1
Let 0.8 the probability that a man aged 90 years will die in a year. If pp is the probability that out of 4 men A1,A2,A3A_1, A_2, A_3 and A4A_4 each aged 90,A190, A_1 will die in a year and will be the first to die, find 10000p−249010000 p-2490 is
View answer and explanation
Correct option: D
6
Option D (6) is correct because it is calculated by finding the probability that A1 dies first and then subtracting 2490 from 10000 times that probability.
84 Medium
84 of 285
Application +4 | -1
A bag contains 8 red and 7 black balls. Two balls are drawn at random. The probability that both the balls are of the same colour, is
View answer and explanation
Correct option: C
715\frac{7}{{15}}
Option C (715\frac{7}{{15}}) is correct because it represents the probability of drawing two balls of the same color, calculated by summing the probabilities of drawing two red balls and drawing two black balls.
85 Medium
85 of 285
Application +4 | -1
Two dice are rolled. If the probability of getting different numbers on the these dice is pq\frac{p}{q} where p,qp, q are relatively prime then q−pq-p is
View answer and explanation
Correct option: C
1
Option B (2-4a) is correct because applying L'Hopital's rule to the limit yields the result 2-4a.
86 Medium
86 of 285
Application +4 | -1
One percent of the population suffers from a certain disease. There is blood test for this disease, and it is 99%99 \% accurate, in other words, the probability that it gives the correct answer is 0.99 , regardless of whether the person is sick or healthy. A person takes the blood test, and the result says that he has the disease. The probability that he actually has the disease, is
View answer and explanation
Correct option: C
50%50 \%
Option C (50 %) is correct because the probability is calculated using Bayes' theorem, considering the prior probability of having the disease and the accuracy of the test.
87 Medium
87 of 285
Application +4 | -1
A bag contains 50 tickets numbered 1, 2, 3, …, 50 of which five are drawn at random and arranged in ascending order of magnitude (x1<x2<x3<x4<x5).\left( {{x_1} < {x_2} < {x_3} < {x_4} < {x_5}} \right). The probability that x3=30,{x_3} = 30, is
View answer and explanation
Correct option: C
  20C2×  29C2  50C5\frac{{{\;^{20}}{C_2} \times {\;^{29}}{C_2}}}{{{\;^{50}}{C_5}}}
Option C (  20C2×  29C2  50C5\frac{{{\;^{20}}{C_2} \times {\;^{29}}{C_2}}}{{\;^{50}}{C_5}}) is correct because it represents the probability of having x3 = 30, with two numbers less than 30 and two numbers greater than 30.
88 Medium
88 of 285
Application +4 | -1
Three letters are written to there different persons and addresses on the three envelopes are also written. Without looking at the addresses, the letters are kept in these envelopes. The probability that all the letters are not placed into their right envelopes is
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (1/3) is correct because it calculates the probability that all letters are not placed into their right envelopes using derangements.
89 Medium
89 of 285
Understanding +4 | -1
If AA and BB are independent events and P(C)=0,P\left( C \right) = 0, then
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because if P(C) = 0, then A and C, B and C, and A, B, and C are all independent events.
90 Medium
90 of 285
Analyzing +4 | -1
10 mangoes are to be distributed among 5 persons. The probability that at least one of them will receive none, is
View answer and explanation
Correct option: D
125143\frac{125}{143}
Option D (125143\frac{125}{143}) is correct because it calculates the probability that at least one person receives no mangoes by subtracting the probability that each person receives at least one mango from 1.
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