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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 121-150 of 1130 questions
121 Medium
121 of 1130
Application +4 | -1
The chance of an event happening is the square of the chance of a second event but the odds against the first are the cube of the odds against the second. The chances of the events are
View answer and explanation
Correct option: A
19,13\frac{1}{9}, \frac{1}{3}
Option A (19,13\frac{1}{9}, \frac{1}{3}) is correct because it satisfies both given conditions: the chance of the first event is the square of the second, and the odds against the first are the cube of the odds against the second.
122 Medium
122 of 1130
Application +4 | -1
The probability that in a random arrangement of the letters of the word ‘UNIVERSITY’, the two I′ sI'\,\mathrm{s} do not come together is
View answer and explanation
Correct option: A
04-May
Option A (04-May) is correct because the probability is calculated by finding the total arrangements and subtracting arrangements where the two I's are together, then dividing by the total arrangements.
123 Easy
123 of 1130
Understanding +4 | -1
A die is thrown. If it shows a six, we draw a ball from a bag consisting 2 black balls and 6 white balls. If it does not show a six, then we toss a coin. Then, the sample space of this experiment consists of
View answer and explanation
Correct option: B
18 points
Option B (18 points) is correct because the sample space consists of 8 outcomes when a six is rolled and 10 outcomes when a six is not rolled, totaling 18.
124 Easy
124 of 1130
Application +4 | -1
Two non-negative integers are chosen at random. The probability that the sum of the square is divisible by 5 is
View answer and explanation
Correct option: B
925\frac{9}{25}
Option B (925\frac{9}{25}) is correct because it represents the probability that the sum of the squares of two randomly chosen non-negative integers is divisible by 5.
125 Easy
125 of 1130
Application +4 | -1
An unbiased die is rolled until two consecutive trails result in even numbered faces. The probability that exactly six trials are required to get two consecutive even numbered faces, is
View answer and explanation
Correct option: A
5(12)65\left(\frac{1}{2}\right)^6
Option A (5(1/2)6) is correct because it calculates the probability of getting two consecutive even numbers on the 6th trial, with 5 possible sequences of odd and even numbers leading to this outcome.
126 Medium
126 of 1130
Application +4 | -1
Two numbers xx and yy are chosen at random from the set {1,2,3,…,30}\{1,2,3, \ldots, 30\}. The probability that x2−y2x^2-y^2 is divisible by 3 is
View answer and explanation
Correct option: D
None of these
The correct probability is 4787\frac{47}{87}, which is not among the options. x2−y2x^2 - y^2 is divisible by 3 if and only if both xx and yy are divisible by 3, or neither are divisible by 3.
127 Easy
127 of 1130
Application +4 | -1
A person draws a card from a pack of playing cards, replaces it and shuffles the pack. He continues doing this until he draws a spade. The chance that he fail the first two times is
View answer and explanation
Correct option: D
916\frac{9}{{16}}
Option D (916\frac{9}{{16}}) is correct because it calculates the probability of failing to draw a spade in the first two attempts with replacement.
128 Easy
128 of 1130
Application +4 | -1
Three dice are thrown. The probability that the same number will appear on each of them, is
View answer and explanation
Correct option: C
1/361/36
Option C (1/36) is correct because it is the probability of getting the same number on all three dice, calculated by dividing the number of favorable outcomes by the total number of outcomes.
129 Easy
129 of 1130
Application +4 | -1
A coin is tossed 4 times. The probability that at least one head turns up, is
View answer and explanation
Correct option: D
1516\frac{{15}}{{16}}
Option A (1/41/4) is correct because the probability is calculated by considering the number of ways to choose 3 white balls out of 4 white balls, divided by the total number of ways to choose 3 balls out of 8 balls, considering identical balls.
130 Easy
130 of 1130
Application +4 | -1
Probability is 0.45 that a dealer will sell at least 20 television sets during a day, and the probability is 0.74 that he will sell less than 24 televisions. The probability that he will sell 20,21,2220,21,22, or 23 televisions during the day, is
View answer and explanation
Correct option: A
0.19
Option A (0.19) is correct because it represents the probability of the dealer selling 20, 21, 22, or 23 televisions, calculated using the addition rule of probability.
131 Easy
131 of 1130
Understanding +4 | -1
If CC and DD are two events such that C⊂DC \subset D and P(D)≠0P(D) \neq 0, then the correct statement among the following is:
View answer and explanation
Correct option: D
P(C/D)=P(D)P(C)P(C / D)=\frac{P(D)}{P(C)}
The correct answer is not listed in the options. The correct answer should be P(C/D)=P(C)P(D)P(C / D)=\frac{P(C)}{P(D)} since C⊂DC \subset D.
132 Medium
132 of 1130
Analyzing +4 | -1
Let 0<P(A)<1,0<P(B)<10 < P\left( A \right) < 1,0 < P\left( B \right) < 1 and P(A∪B)=P(A)+P(B)−P(A)  P(B).P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( A \right)\;P\left( B \right). Then,
View answer and explanation
Correct option: C
P((A∪B)c)=P(Ac)P(Bc)P\left( {{{\left( {A \cup B} \right)}^c}} \right) = P\left( {{A^c}} \right)P\left( {{B^c}} \right)
Option C (P((A∪B)c)=P(Ac)P(Bc)P({{\left( {A \cup B} \right)}^c}) = P({A^c})P({B^c})) is correct because it correctly identifies that A and B are independent events, leading to the given relationship for their complements.
133 Easy
133 of 1130
Application +4 | -1
From a bag containing 30 tickets numbered 1,2,3,…30,1,2,3, \ldots 30, three tickets are selected at random without replacement. Suppose these tickets are arranged as x1<x2<x3x_1 < x_2 < x_3. The probability that x2=15x_2=15 and x3>20x_3>20, is
View answer and explanation
Correct option: A
1/291 / 29
Option A (1/29) is correct because it represents the probability of selecting three tickets such that x2 = 15 and x3 > 20, calculated by dividing the number of favorable outcomes by the total number of possible outcomes.
134 Easy
134 of 1130
Application +4 | -1
If A\mathrm{A} and B\mathrm{B} be two events such that P(A)=38,P(B)=58P(A)=\frac{3}{8}, P(B)=\frac{5}{8} and P(A∪B)=34P(A \cup B)=\frac{3}{4}, then P(A/B)×P(A′/B)P(A / B) \times P\left(A^{\prime} / B\right) is equal to
View answer and explanation
Correct option: D
625\frac{6}{25}
Option D (625\frac{6}{25}) is correct because it is the product of P(A/B) and P(A'/B) calculated using conditional probability and the given probabilities.
135 Easy
135 of 1130
Application +4 | -1
A six faced die is so biased that it is twice likely to show an even number as an odd number when thrown. If the die is thrown twice the probability that sum of the numbers is even is
View answer and explanation
Correct option: D
59\frac{5}{9}
Option D (59\frac{5}{9}) is correct because it is the probability of getting an even sum when a biased die is thrown twice, considering the die is twice as likely to show an even number as an odd number.
136 Easy
136 of 1130
Analyzing +4 | -1
Odds in favour of an event AA are 2 to 1 and odds in favour of A∪BA \cup B are 3 to 1 . Consistent with this information the smallest and largest values for the probability of event BB are given by
View answer and explanation
Correct option: C
112≤P(B)≤34\frac{1}{12} \leq P(B) \leq \frac{3}{4}
Option C (112≤P(B)≤34\frac{{1}}{{12}} \leq P(B) \leq \frac{{3}}{{4}}) is correct because it represents the range of possible values for the probability of event B, derived from the given odds and the properties of probability.
137 Medium
137 of 1130
Application +4 | -1
The probability that at least one of the events AA and BB occurs is 0.7 and they occur simultaneously with probability 0.2. Then, P(Aˉ)+P(Bˉ)=P\left( {\bar A} \right) + P\left( {\bar B} \right) =
View answer and explanation
Correct option: C
1.1
Option C (1.1) is correct because it is derived from the properties of probability and complementary events.
138 Easy
138 of 1130
Application +4 | -1
A person draws a card from a pack of 52 playing cards, replaces it and shuffles the pack. He continues doing this until he draws a spade, the chance that he will fail in the first two draws is
View answer and explanation
Correct option: B
916\frac{9}{16}
Option B (9/16) is correct because the probability of failing in the first two draws is the product of the probabilities of failing in each individual draw, since the draws are independent.
139 Easy
139 of 1130
Application +4 | -1
A coin is biased so that the head is 3 times as likely to occur as tail. This coin is tossed until a head or three tails occur. If XX denotes the number of tosses of the coin, then the mean of XX is
View answer and explanation
Correct option: C
2116\frac{21}{16}
Option C (21/16) is correct because it is the expected value (mean) of the number of tosses calculated using probabilities of getting a head or three tails.
140 Easy
140 of 1130
Analyzing +4 | -1
A number xx is chosen from the set A={33n:n∈N}.A=\left\{33^n: n \in N\right\}. The probability that xx has 3 in units place, is
View answer and explanation
Correct option: C
14\frac{1}{4}
Option C (14\frac{1}{4}) is correct because the units digit of powers of 3 cycle through 3, 9, 7, and 1, resulting in a probability of 1/4 for a 3 in the units place.
141 Easy
141 of 1130
Analyzing +4 | -1
If two events AA and BB are such that P(Ac)=0.3,  P(B)=0.4P\left( {{A^c}} \right) = 0.3,\;P\left( B \right) = 0.4 and P(A∩Bc)=0.5,P\left( {A \cap {B^c}} \right) = 0.5, then  P[B(A∪Bc)]\;P\left[ {\frac{B}{{\left( {A \cup {B^c}} \right)}}} \right] is equal to
View answer and explanation
Correct option: C
14\frac{1}{4}
Option C (14\frac{1}{4}) is correct because it represents the conditional probability of event B occurring given that event A or the complement of event B has occurred, calculated using the given probabilities.
142 Easy
142 of 1130
Application +4 | -1
A bag contains 5 red, 3 white and 2 black balls. If a ball is picked at random, the probability that it is red is
View answer and explanation
Correct option: B
1/21/2
Option B (1/21/2) is correct because the probability of picking a red ball is the number of red balls divided by the total number of balls.
143 Easy
143 of 1130
Application +4 | -1
In a multiple choice question there are four alternative answers, of which one or more are correct. A candidate will get marks in the question only if he ticks all the correct answers. The candidate decides to tick answers at random. If he is allowed upto three chanes to answer the question, the probability that he will get marks in the question, is
View answer and explanation
Correct option: B
15\frac{1}{5}
Option B (15\frac{1}{5}) is correct because it represents the probability of selecting the correct answer within three attempts, considering all possible combinations.
144 Medium
144 of 1130
Application +4 | -1
If the probability density function of a random variable XX is f(x)=x2f\left( x \right) = \frac{x}{2} in 0≤x≤2,0 \le x \le 2, then P(X>15X>1)P\left( {\frac{{X > 15}}{{X > 1}}} \right) is equal to
View answer and explanation
Correct option: C
712\frac{7}{{12}}
Option C (712\frac{7}{{12}}) is correct because it calculates the conditional probability P(X > 1.5 | X > 1) by finding the individual probabilities P(X > 1.5) and P(X > 1) using integration of the probability density function.
145 Easy
145 of 1130
Application +4 | -1
Kiran is deciding whether to visit Russia or Ukrain for the holidays. She makes her decision by rolling a regular 6-sided die. If she gets 1 or 2 she goes to Ukrain, if she rolls a 3,4 or 5 she goes to Russia. If she rolls a 6 she rolls again, the probability that she goes to Ukrain is pp, then 5p5 p is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the probability that Kiran goes to Ukrain is 2/5, and therefore 5p = 5*(2/5) = 2.
146 Easy
146 of 1130
Application +4 | -1
Out of 40 consecutive integers, two are chosen at random, the probability that their sum is odd is
View answer and explanation
Correct option: B
20/3920/39
Option B (2039\frac{20}{39}) is correct because it calculates the probability of the sum being odd by considering the number of ways to choose one even and one odd number.
147 Easy
147 of 1130
Understanding +4 | -1
If A   and  BA\;\,\mathrm{and}\,\,B are arbitrary events, then
View answer and explanation
Correct option: B
P(A∪B)≤P(A)+P(B)P\left( {A \cup B} \right) \le P\left( A \right) + P\left( B \right)
Option B (P(A∪B)≤P(A)+P(B)P(A \cup B) \le P(A) + P(B)) is correct because P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B), and since P(A∩B)≥0P(A \cap B) \ge 0, it follows that P(A∪B)≤P(A)+P(B)P(A \cup B) \le P(A) + P(B).
148 Easy
148 of 1130
Application +4 | -1
In a binomial distribution the mean is 15 and variance is 10. Then parameter nn is
View answer and explanation
Correct option: C
45
Option C (45) is correct because it is the value of n that satisfies the given mean and variance for a binomial distribution.
149 Medium
149 of 1130
Application +4 | -1
If aa is an integer lying in [−5,30][-5,30], then the probability that the graph of y=x2+2(a+4)xy=x^2+2(a+4) x −5a+64-5 a+64 is strictly above the xx-axis is
View answer and explanation
Correct option: C
2/92 / 9
Option C (2/9) is correct because the probability that the graph is strictly above the x-axis is 8/36 = 2/9, derived from the condition that the discriminant is negative.
150 Easy
150 of 1130
Application +4 | -1
If a die is thrown and a card is selected at random from a deck of 52 playing cards, then the probability of getting an even number on the die and a spade card is
View answer and explanation
Correct option: C
18\frac{1}{8}
Option C (18\frac{1}{8}) is correct because it is the product of the probability of getting an even number on the die and the probability of getting a spade card.
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