CUET Mathematics: AlgebraProbability. Free, no login required.

Q1:

2024: 16 May Shift 1

Probability

Medium

Applied

There are 66 cards numbered 11 to 66, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Then P(X>3)\mathrm{P}(\mathrm{X}>3) is :

Answer options
Option 1
Correct Answer
Explanation for 2024: 16 May Shift 1 MAT question 1

Q2:

2024: 16 May Shift 1

Probability

Medium

Applied

A coin is tossed K times. If the probability of getting 33 heads is equal to the probability of getting 77 heads, then the probability of getting 88 tails is :

Answer options
Option 3
Correct Answer
Explanation for 2024: 16 May Shift 1 MAT question 2

Q3:

2023: 23 May Shift 3

Probability

Easy

Common

If a fair coin is tossed 10 times, then the probability of obtaining at least one head is :

Answer options
Option 3
Correct Answer
Explanation for 2023: 23 May Shift 3 MAT question 3

Q4:

2023: 23 May Shift 3

Probability

Hard

Common

Two cards are drawn simultaneously from a well shuffled pack of 52 cards. Then variance of the number of kings is

Answer options
Option 4
Correct Answer
Explanation for 2023: 23 May Shift 3 MAT question 4

Q5:

2023: 23 May Shift 3

Probability

Hard

Core

A doctor is to visit a patient. It is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively 310,15,110\frac{3}{10}, \frac{1}{5}, \frac{1}{10} and 25\frac{2}{5}. The probabilities that he will be late are 14,13\frac{1}{4}, \frac{1}{3} and 112\frac{1}{12}, if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he arrives late.

The probability that he comes by bus is:

Answer options
Option 1
Correct Answer
Explanation for 2023: 23 May Shift 3 MAT question 5

Q6:

2023: 23 May Shift 3

Probability

Medium

Core

If a fair coin is tossed 10 times the probability of atleast 6 heads is:

Answer options
Option 4
Correct Answer
Explanation for 2023: 23 May Shift 3 MAT question 6

Q7:

2023: 22 May Shift 3

Probability

Easy

Common

The probability that a student is not a swimmer is 15\frac{1}{5}. Then the probability that out of five students, four are swimmers is :

Answer options
Option 1
Correct Answer
Explanation for 2023: 22 May Shift 3 MAT question 7

Q9:

2023: 22 May Shift 3

Probability

Easy

Core

The black and red die are rolled. The conditional probability of obtaining a sum greater than 9 given that the black die resulted in a 5 is :

Answer options
Option 1
Correct Answer
Explanation for 2023: 22 May Shift 3 MAT question 9

Q10:

2023: 22 May Shift 3

Probability

Medium

Core

Urn I contains 6 red balls and 4 black balls and Urn II contains 4 red balls and 6 black balls. One ball is drawn at random from Urn I and placed in Urn II. If one ball is drawn at random from Urn II, then the probability that it is a red ball is :

Answer options

Q11:

2023: 30 May Shift 3

Probability

Easy

Common

If the probability distribution of a random variable X is as given below :

X-10123
P(X)K15\frac{1}{5}2K310\frac{3}{10}K

Then the value of K is :

Answer options

Q13:

2023: 30 May Shift 3

Probability

Easy

Core

Match List - I with List - II.

List - IList - II
(A) If A and B are mutually exclusive events, then P(A∪B)=P(A \cup B) =(I) P(A∩B)P(B),P(B)≠0\frac{P(A \cap B)}{P(B)}, P(B) \neq 0
(B) If A and B are independent events, then P(A∩B)=P(A \cap B) =(II) P(A∩B)P(A),P(A)≠0\frac{P(A \cap B)}{P(A)}, P(A) \neq 0
(C) If A and B are two events of a sample space of an experiment, then P(A/B)=P(A/B) =(III) P(A)⋅P(B)P(A) \cdot P(B)
(D) If A and B are two events of a sample space of an experiment, then P(B/A)=P(B/A) =(IV) P(A)+P(B)P(A) + P(B)

Choose the correct answer from the options given below :

Answer options

Q15:

2023: 15 June Shift 2

Probability

Easy

Common

The random variable X has a probability distribution P(X) of the following form, where k is some number.

P(X=x)={k,if x=02k,if x=13k,if x=20,otherwiseP(X=x) = \begin{cases} k, & \text{if } x=0 \\ 2k, & \text{if } x=1 \\ 3k, & \text{if } x=2 \\ 0, & \text{otherwise} \end{cases}

Then P(x≤2)P(x \leq 2) is :

Answer options

Q17:

2023: 15 June Shift 2

Probability

Easy

Core

If in a binomial distribution n=4n = 4, P(X=0)=1681P(X=0) = \frac{16}{81}, then P(X=4)P(X=4) equals :

Answer options

Q18:

2023: 15 June Shift 2

Probability

Medium

Core

A and B throw a die alternatively till one of them gets a number more than 4 and wins the game. Then the probability of winning the game by B, if A starts first :

Answer options

Q19:

2022: 7 Aug Shift 2

Probability

Easy

Core

Let X be a discrete random variable and probability distribution X is

X−1-101
P(X)12\frac{1}{2}15\frac{1}{5}310\frac{3}{10}

Then E(X) is equal to :

Answer options
Option 2
Correct Answer
Explanation for 2022: 7 Aug Shift 2 MAT question 19

Q20:

2022: 7 Aug Shift 2

Probability

Medium

Core

In Binomial distribution with parameters n=12n = 12 and p=13p = \frac{1}{3}, value of E(X2)+E(X)E(X^2) + E(X) is :

Answer options
Option 4
Correct Answer
Explanation for 2022: 7 Aug Shift 2 MAT question 20

Q21:

2022: 7 Aug Shift 2

Probability

Medium

Core

Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. The probability that it was drawn from Bag II is :

Answer options
Option 1
Correct Answer
Explanation for 2022: 7 Aug Shift 2 MAT question 21

Q22:

2022: 7 Aug Shift 2

Probability

Easy

Core

In Binomial distribution with parameters n=100n = 100 and p, Variance of distribution is maximum when p is equal to :

Answer options
Option 3
Correct Answer
Explanation for 2022: 7 Aug Shift 2 MAT question 22

Q23:

2022: 7 Aug Shift 2

Probability

Easy

Core

A shopkeeper sells three types of flower seeds A1,A2,A3A_1, A_2, A_3. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases.

A shopkeeper sells three types of flower seeds A1,A2,A3A_1, A_2, A_3. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases.

The probability of a randomly chosen seed to germinate is :

Answer options
Option 4
Correct Answer
Explanation for 2022: 7 Aug Shift 2 MAT question 23

Q24:

2022: 7 Aug Shift 2

Probability

Easy

Core

A shopkeeper sells three types of flower seeds A1,A2,A3A_1, A_2, A_3. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases.

A shopkeeper sells three types of flower seeds A1,A2,A3A_1, A_2, A_3. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases.

The probability that seed will not germinate, given that the seed is of type A3A_3.

Answer options
Option 3
Correct Answer
Explanation for 2022: 7 Aug Shift 2 MAT question 24

Q25:

2022: 7 Aug Shift 2

Probability

Medium

Core

A shopkeeper sells three types of flower seeds A1,A2,A3A_1, A_2, A_3. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases.

A shopkeeper sells three types of flower seeds A1,A2,A3A_1, A_2, A_3. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases.

The probability that seed is of type A2A_2 given that seed germinate.

Answer options
Option 2
Correct Answer
Explanation for 2022: 7 Aug Shift 2 MAT question 25

Q26:

2022: 7 Aug Shift 2

Probability

Medium

Core

A shopkeeper sells three types of flower seeds A1,A2,A3A_1, A_2, A_3. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases.

A shopkeeper sells three types of flower seeds A1,A2,A3A_1, A_2, A_3. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases.

The probability that seed is not of type A1A_1, given that seed germinates.

Answer options
Option 2
Correct Answer
Explanation for 2022: 7 Aug Shift 2 MAT question 26

Q27:

2022: 7 Aug Shift 2

Probability

Medium

Core

A shopkeeper sells three types of flower seeds A1,A2,A3A_1, A_2, A_3. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases.

A shopkeeper sells three types of flower seeds A1,A2,A3A_1, A_2, A_3. They are sold as a mixture where the proportions are 4 : 4 : 2 respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively. Calculate the probability in the following cases.

The probability that seed is of type A1A_1 given that seed doesn't germinate.

Answer options
Option 4
Correct Answer
Explanation for 2022: 7 Aug Shift 2 MAT question 27

Q28:

2022: 17 Aug Shift 2

Probability

Easy

Core

The random variable XX has a probability distribution

P(X=x)={5k,x=02k,x=13k,x=20,otherwise.P(X=x) = \begin{cases} 5k, & x=0 \\ 2k, & x=1 \\ 3k, & x=2 \\ 0, & \text{otherwise.} \end{cases}

Then, the value of E(X)E(X) is :

Answer options

Q29:

2022: 17 Aug Shift 2

Probability

Medium

Core

Match List - I with List - II.

List - IList - II
(A) Two events E and F will be independent if P(E′F′)P(E'F') is equal to(I) 1−P(E/F)1 - P(E/F)
(B) If P(F)≠0P(F) \neq 0, then P(E′/F)P(E'/F) is equal to(II) P(E)=P(F)P(E) = P(F)
(C) If E and F are independent events, then(III) P(E∩F′)=P(E)⋅P(F′)P(E \cap F') = P(E) \cdot P(F')
(D) If P(E∩F)≠0P(E \cap F) \neq 0 and P(E/F)=P(F/E)P(E/F) = P(F/E), then(IV) [1−P(E)][1−P(F)][1-P(E)][1-P(F)]

Choose the correct answer from the options given below :

Answer options

Q30:

2022: 17 Aug Shift 2

Probability

Easy

Core

In a hospital, there are 300 patients, out of which 120 are female. It is known that out of 120 females, 10% of the patients are below 40 years of age. What is the probability that a patient chosen randomly is below 40 yrs of age given that the chosen patient is a female.

Answer options