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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 121-150 of 285 questions

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121 Medium
121 of 285
Analyzing +4 | -1
A number xx is chosen at random from the set {1,2,3,4,…\{1,2,3,4, \ldots, 100}100\}. Define the event: A=\mathrm{A}= the chosen number x\mathrm{x} satisfies (x−10)(x−50)(x−30)≥0\frac{(\mathrm{x}-10)(\mathrm{x}-50)}{(\mathrm{x}-30)} \geq 0
Then P (A) is:
View answer and explanation
Correct option: A
0.71
Option A (0.71) is correct because it represents the probability of x satisfying the given inequality within the specified set.
122 Medium
122 of 285
Analyzing +4 | -1
There are nn stations between two cities AA and BB. A train is to stop at three of these nn stations. What is the probability that no two of these three stations are consecutive?
View answer and explanation
Correct option: D
(n−3)(n−4)n(n−1)\frac{(n-3)(n-4)}{n(n-1)}
Option D ((n−3)(n−4)n(n−1)\frac{{(n-3)(n-4)}}{{n(n-1)}}) is correct because it represents the probability calculated by dividing the number of ways to choose 3 non-consecutive stations by the total number of ways to choose 3 stations.
123 Medium
123 of 285
Application +4 | -1
A class contains 20 boys and 20 girls of which half the boys and half the girls have cat eyes. If one student is selected from the class & if the probability that either the student is a boy or has cat eyes is ab\frac{{a}}{{b}}, then a2+b2=\sqrt{{a}^2+{b}^2}=
View answer and explanation
Correct option: A
5
Option A (5) is correct because it calculates the probability of either a boy or having cat eyes and then finds the square root of the sum of squares of the numerator and denominator.
124 Medium
124 of 285
Application +4 | -1
AA and BB are are two events. Odds against AA are 2 to 1 and odds in favour of A∪BA \cup B are 3 to1. If x≤P(B)≤yx \le P\left( B \right) \le y, then ordered pair (x,y)\left( {x,y} \right) is
View answer and explanation
Correct option: A
(512,34)\left( {\frac{5}{{12}},\frac{3}{4}} \right)
Option A ((512,34)\left( {\frac{5}{{12}},\frac{3}{4}} \right)) is correct because it provides the correct range for the probability of event B, derived from the given probabilities and the properties of probability.
125 Medium
125 of 285
Application +4 | -1
One hundred identical coins, each with probability, p, of showing up heads are tossed once. If 0<p<1,0 < p < 1, and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of p,p, is
View answer and explanation
Correct option: D
51101\frac{51}{101}
Option D (51101\frac{51}{101}) is correct because it's the value of p calculated using the given condition that the probability of 50 heads equals the probability of 51 heads.
126 Medium
126 of 285
Analyzing +4 | -1
A car is parked by an owner amongst 25 cars in a row, not at either end. On his return he finds that exactly 15 places are still occupied. The probability that both the neighboring places are empty is
View answer and explanation
Correct option: C
1592\frac{15}{92}
Option C (1592\frac{15}{92}) is correct because it accurately calculates the probability of the owner's neighboring spots being empty, given that 15 spots are occupied.
127 Medium
127 of 285
Application +4 | -1
A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is
View answer and explanation
Correct option: D
57\frac{5}{7}
The correct answer is D (57\frac{{5}}{{7}}) because it represents the conditional probability that the bag contains at least 5 black balls given that two balls drawn are black.
128 Medium
128 of 285
Application +4 | -1
An almirah stores 5 black and 4 white socks well mixed. A boy pull out 2 socks at random. The probability that 2 are of the same colour is
View answer and explanation
Correct option: A
04-Sep
Option A (4/9) is correct because it calculates the probability of drawing two socks of the same color by considering the combinations of drawing two black socks or two white socks.
129 Medium
129 of 285
Application +4 | -1
If the letters of the word ‘REGULATION’ be arranged at random, the probability that there will be exactly 4 letters between RR and EE is
View answer and explanation
Correct option: B
1/91/9
Option B (19\frac{1}{9}) is correct because it calculates the probability of having exactly 4 letters between R and E in the word REGULATION by considering the possible arrangements.
130 Medium
130 of 285
Analyzing +4 | -1
AA is a set containing nn 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 \cup Q contains just one element, is
View answer and explanation
Correct option: B
3n4n\frac{3 n}{4^n}
Option B (3n4n\frac{{3 n}}{{4^n}}) is correct because it represents the probability of P∪QP \cup Q containing exactly one element, considering all possible choices for sets P and Q.
131 Medium
131 of 285
Application +4 | -1
Suppose AA and BB are two mutually exclusive events such that P(A)=13(3p+1)P(A)=\frac{1}{3}(3 p+1) and P(B)=14(1−p)P(B)=\frac{1}{4}(1-p), then the set of possible values of pp lies in the interval:
View answer and explanation
Correct option: C
[−13,59]\left[-\frac{1}{3}, \frac{5}{9}\right]
Option D ([−79,49]\left[-\frac{7}{9}, \frac{4}{9}\right]) is incorrect. The correct answer is C ([−13,59]\left[-\frac{1}{3}, \frac{5}{9}\right]) because it considers the constraints on probabilities for mutually exclusive events.
132 Medium
132 of 285
Analyzing +4 | -1
An unbiased cubical die is thrown nn times. If xi=x_i= outcome on the ith i^{\text {th }} throw and X=gcd⁡(x1,x2,…..,xn)X=\operatorname{gcd}\left(x_1, x_2, \ldots . ., x_n\right), then E(X)E(X) is equal to
View answer and explanation
Correct option: C
16n(6n+3n+2n+1+8)\frac{1}{6^n}\left(6^n+3^n+2^{n+1}+8\right)
Option D (16n(6n+3n+2n+2+8)\frac{1}{6^n}\left(6^n+3^n+2^{n+2}+8\right)) is correct because it represents the expected value of the greatest common divisor of n throws of a die.
133 Medium
133 of 285
Application +4 | -1
Nine balls of the same size and color, numbered 1,2,…..,91,2, \ldots . ., 9, were put into a packet. Now A draws a ball from the packet, noted that it is of number a, and puts back it. Then B also draws a ball from the packet and noted that it is of number bb. Then the probability for the inequality a−2b+a-2 b+ 10>010>0 to hold is
View answer and explanation
Correct option: D
6181\frac{61}{81}
Option D (61/81) is correct because it represents the probability calculated by counting favorable outcomes where a - 2b + 10 > 0 and dividing by the total possible outcomes.
134 Medium
134 of 285
Application +4 | -1
The probability that in a year of the 22nd century chosen at random there will be 53 Sundays, is
View answer and explanation
Correct option: D
528\frac{5}{{28}}
Option D (5/28) is correct because it calculates the probability of having 53 Sundays in a randomly chosen year of the 22nd century considering both leap and non-leap years.
135 Medium
135 of 285
Application +4 | -1
A random variate XX takes the values 0, 1, 2, 3 and its mean is 1.3. If P(X=3)=2P(X=1)P\left( {X = 3} \right) = 2P\left( {X = 1} \right) and P(X=2)=0.3P\left( {X = 2} \right) = 0.3, then P(X=0)P\left( {X = 0} \right) is equal to
View answer and explanation
Correct option: D
0.4
Option D (0.4) is correct because it satisfies the given mean and probability conditions for the random variable X.
136 Medium
136 of 285
Application +4 | -1
If the letters of the word ‘REGULATIONS’ be arranged at random, the probability that there will be exactly 4 letters between RR and EE is
View answer and explanation
Correct option: A
655\frac{6}{{55}}
Option A is correct because it represents the probability of having exactly 4 letters between R and E in a random arrangement of the word 'REGULATIONS', calculated by considering the arrangements of the letters with the given constraint.
137 Medium
137 of 285
Application +4 | -1
There are 4 white and 3 black balls in a box. In another box there are 3 white and 4 black balls. An unbiased dice is rolled. If it shows a number less than or equal to 3 then a ball is drawn from the first box but if it shows a number more than 3 then a ball is drawn from the second box. If the ball drawn is black then the probability that the ball was drawn from the first box is
View answer and explanation
Correct option: D
3/73 / 7
Option D (3/73 / 7) is correct because it is calculated using Bayes' theorem, considering the probabilities of drawing from each box and the probability of drawing a black ball.
138 Medium
138 of 285
Application +4 | -1
If P(A∪B)=0.8P\left( {A \cup B} \right) = 0.8 and P(A∩B)=0.3,P\left( {A \cap B} \right) = 0.3, then P(A′)+P(B′)P\left( {A'} \right) + P\left( {B'} \right) equals to
View answer and explanation
Correct option: D
0.9
Option D (0.9) is correct because it is calculated using the properties of probability and De Morgan's laws.
139 Medium
139 of 285
Application +4 | -1
A person throws two dice, one the common cube and the other regular tetrahedron with numbers 1,2,3,41,2,3,4 on its faces, the number on the lowest face being taken in the case of a tetrahedron. The chance that the sum of numbers thrown is not less than 5 is
View answer and explanation
Correct option: B
34\frac{3}{4}
Option B (34\frac{{3}}{{4}}) is correct because the probability that the sum is not less than 5 is 18/24, which simplifies to 3/4.
140 Medium
140 of 285
Application +4 | -1
If AA and BB are events such that P(Aˉ∪Bˉ)=34,P(Aˉ∩Bˉ)=14P(\bar{A} \cup \bar{B})=\frac{3}{4}, \quad P(\bar{A} \cap \bar{B})=\frac{1}{4} and P(Aˉ)=23P(\bar{A})=\frac{2}{3}, then P(Aˉ∩B)P(\bar{A} \cap B) is
View answer and explanation
Correct option: A
512\frac{5}{12}
Option A (5/12) is correct because using De Morgan's laws and probability rules, we can find P(A ∪ B) and then calculate P(¬A ∩ B).
141 Medium
141 of 285
Understanding +4 | -1
A,B{A}, {B} are two events in a random experiment such that 0<p(A)<10 < p(A) < 1 and P(B)≠1P(B) \neq 1
Assertion (A): P(ABc)=P(A)−P(A∩B)1−P(B)P\left(\frac{A}{B^c}\right)=\frac{P(A)-P(A \cap B)}{1-P(B)}
Reason (R): P(AB)=P(A∩B)P(B)P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}
View answer and explanation
Correct option: A
Both A and R are true and R is the correct explanation of A
Option A (Both A and R are true and R is the correct explanation of A) is correct because both the assertion and the reason are true, and the reason correctly explains the assertion.
142 Medium
142 of 285
Analyzing +4 | -1
Let XX and YY are two events such that P(X∪Y)=P(X∩Y)P(X \cup Y)=P(X \cap Y).
Statement 1: P(X∩Y′)=P(X′∩Y)=0P\left(X \cap Y^{\prime}\right)=P\left(X^{\prime} \cap Y\right)=0
Statement 2: P(X)+P(Y)=2P(X∩Y)P(X)+P(Y)=2 P(X \cap Y)
View answer and explanation
Correct option: B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1 .
Option B (Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1) is correct because both statements are true, but Statement 2 doesn't directly explain Statement 1.
143 Medium
143 of 285
Application +4 | -1
A coin is biased so that a head is twice as likely to
occur as a tail. If the coin is tossed 3 times, the
probability of getting two tails and one head is
View answer and explanation
Correct option: A
29\frac{2}{9}
Option A (2/9) is correct because it is the probability of getting two tails and one head when the coin is tossed 3 times.
144 Medium
144 of 285
Application +4 | -1
Three six-faced dice are thrown together. The probability that the sum of the numbers appearing on the dice is k(3≤k≤8)k(3 \leq k \leq 8), is
View answer and explanation
Correct option: C
(k−1)(k−2)432\frac{(k-1)(k-2)}{432}
Option C ((k−1)(k−2)432\frac{{(k-1)(k-2)}}{{432}}) is correct because it represents the probability calculated by finding the coefficient of xkx^k in the expansion of (x+x2+...+x6)3(x+x^2+...+x^6)^3 and dividing by the total possible outcomes.
145 Medium
145 of 285
Application +4 | -1
A determinant of the second order is made with the elements 0 and 1. If mn\frac{{m}}{{n}} be the probability that the determinant made is non-negative, where m{m} and n{n} are relative primes, then the value of n−m{n}-{m} is
View answer and explanation
Correct option: A
3
Option A (3) is correct because it represents the value of n-m, where m/n is the probability that the determinant is non-negative.
146 Medium
146 of 285
Application +4 | -1
Out of 400 people, 160 are smokers and non-vegetarian, 100 are vegetarian smokers, 140 are non-smokers and vegetarian. If a person is smoker and non-vegetarian, then the chances of his death is 35%35 \%, if he is vegetarian and smoker then chances of his death is 15%15 \%. If he is non-smoker and vegetarian then chances of his death is 10%10 \%. If a persons is found to be dead, then the probability that he is smoker and non-vegetarian, is
View answer and explanation
Correct option: A
5685\frac{56}{85}
Option A (5685\frac{56}{85}) is correct because it's the probability calculated using Bayes' theorem, considering the probabilities of death given each lifestyle choice.
147 Medium
147 of 285
Application +4 | -1
If the integers mm and nn are chosen at random between 1 and 100, then the probability that a number of the form 7m+7n{7^m} + {7^n} is divisible by 5, equals
View answer and explanation
Correct option: A
14\frac{1}{4}
Option A (1/4) is correct because the probability that 7m+7n{7^m} + {7^n} is divisible by 5 depends on the last digits of 7m{7^m} and 7n{7^n}, which cycle through 7, 9, 3, 1.
148 Medium
148 of 285
Analyzing +4 | -1
Two dice are thrown independently. Let AA be the event that the number appeared on the 1st 1^{\text {st }} die is less than the number appeared on the 2nd 2^{\text {nd }} die, BB be the event that the number appeared on the 1st 1^{\text {st }} die is even and that one the second die is odd, and CC be the event that the number appeared on the 1st 1^{\text {st }} die is odd and that on the 2nd 2^{\text {nd }} is even. Then
View answer and explanation
Correct option: C
The number of favourable cases of the event (A∪B)∩C(A \cup B) \cap C is 6
Option C is correct. The number of favourable cases of the event (A∪B)∩C(A \cup B) \cap C is 6.
149 Medium
149 of 285
Application +4 | -1
Three randomly chosen non negative integers x,yx, y and zz are found to satisfy eq. x+y+z=10x+y+z=10 and the probability that zz is even is p/qp / q (p,qp, q coprime) then p+q={p}+{q}= ____
View answer and explanation
Correct option: A
17
Option A (17) is correct because the probability that z is even is 6/11, and thus p+q = 6+11 = 17.
150 Medium
150 of 285
Application +4 | -1
The probability that a certain kind of component will survive a given shock test is 34\frac{3}{4}. The probability that exactly 2 of the next 4 components tested survive is
View answer and explanation
Correct option: D
27/128
Option D (27/128) is correct because it is the probability calculated using the binomial probability formula for exactly 2 successes out of 4 trials.
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