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JEE • MATHEMATICS • PROBABILITY

Probability practice · Easy

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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 822 questions
91 Easy
91 of 822
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.
92 Easy
92 of 822
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.
93 Easy
93 of 822
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.
94 Easy
94 of 822
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.
95 Easy
95 of 822
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.
96 Easy
96 of 822
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.
97 Easy
97 of 822
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.
98 Easy
98 of 822
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.
99 Easy
99 of 822
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.
100 Easy
100 of 822
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.
101 Easy
101 of 822
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.
102 Easy
102 of 822
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.
103 Easy
103 of 822
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.
104 Easy
104 of 822
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.
105 Easy
105 of 822
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.
106 Easy
106 of 822
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.
107 Easy
107 of 822
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.
108 Easy
108 of 822
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.
109 Easy
109 of 822
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).
110 Easy
110 of 822
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.
111 Easy
111 of 822
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.
112 Easy
112 of 822
Analyzing +4 | -1
One Indian and four American men and their wifes are to be seated randomly around a circular table. Then 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 (2/5) is correct because it represents the conditional probability that the Indian man is seated next to his wife, given that each American man is seated next to his wife.
113 Easy
113 of 822
Application +4 | -1
An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is
View answer and explanation
Correct option: D
2/72 / 7
Option D (2/7) is correct because it calculates the probability of drawing three balls of different colors from the urn.
114 Easy
114 of 822
Application +4 | -1
If the range of a random variable XX is {0,1,2,3,4,…} with P(X=k)=(k+1)a3kP\left( {X = k} \right) = \frac{{\left( {k + 1} \right)a}}{{{3^k}}} for k≥0,k \ge 0, then aa is equal to
View answer and explanation
Correct option: B
49\frac{4}{9}
Option B (49\frac{4}{9}) is correct because it is the value of 'a' that makes the sum of probabilities for all possible values of X equal to 1.
115 Easy
115 of 822
Understanding +4 | -1
Probability of all 3 digit numbers having all the digits same is
View answer and explanation
Correct option: A
1100\frac{1}{{100}}
Option A (1100\frac{1}{{100}}) is correct because there are 9 favorable cases (111, 222, ..., 999) out of 900 total 3-digit numbers.
116 Easy
116 of 822
Analyzing +4 | -1
For two events AA and BB, if P(A)=P(AB)=14P\left( A \right) = P\left( {\frac{A}{B}} \right) = \frac{1}{4} and P(BA)=12,P\left( {\frac{B}{A}} \right) = \frac{1}{2}, then
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because given the probabilities, events A and B are independent, and the conditional probabilities P(A′B)P(\frac{{A'}}{B}) and P(B′A)P(\frac{{B'}}{A}) can be derived.
117 Easy
117 of 822
Application +4 | -1
A random variable X takes values −1,0,1,2-1,0,1,2 with probabilities 1+3p4,1−p4,1+2p4,1−4p4\frac{1+3 p}{4}, \frac{1-p}{4}, \frac{1+2 p}{4}, \frac{1-4 p}{4} respectively, where pp varies over RR. Then the minimum and maximum values of the mean of XX are respectively
View answer and explanation
Correct option: D
−116-\frac{1}{16} and 54\frac{5}{4}
Option D (−116-\frac{1}{16} and 54\frac{5}{4}) is correct because the minimum and maximum values of the mean are found by considering the range of possible values for p.
118 Easy
118 of 822
Application +4 | -1
Two integers xx and yy are chosen with replacement out of the set {0,1,2,3,…,10}.\{0,1,2,3, \ldots, 10\}. Then the probability that ∣x−y∣>5,|x-y|>5, is
View answer and explanation
Correct option: B
30121\frac{30}{121}
Option B (30121\frac{30}{121}) is correct because it is the probability that |x-y|>5, calculated by finding the number of favorable pairs and dividing by the total number of pairs.
119 Easy
119 of 822
Application +4 | -1
A die is tossed thrice. A success is getting 1 or 6 on a toss. The mean and the variance of number of successes
View answer and explanation
Correct option: A
μ=1,σ2=2/3\mu=1, \sigma^2=2 / 3
Option A (μ=1,σ2=2/3\mu=1, \sigma^2=2 / 3) is correct because the mean and variance are calculated using the formulas for binomial distribution: mean = np and variance = npq.
120 Easy
120 of 822
Application +4 | -1
A coin is tossed three times. The probability of getting head and tail alternatively, is
View answer and explanation
Correct option: C
14\frac{1}{4}
Option C (213\frac{2}{{13}}) is correct because the mean of the number of aces is calculated using the probabilities of selecting and not selecting aces in two draws with replacement.
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