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Q1:

2026: 31 May Shift 2

Probability

Easy

Common

A die is rolled. If the outcome is an odd number, then the probability that the number is prime, is:

Answer options
Option 3
Correct Answer
Explanation for 2026: 31 May Shift 2 MAT question 1

Q2:

2026: 31 May Shift 2

Probability

Medium

Core

If AA and BB are two events such that P(A‾∪B‾)=23P(\overline{A} \cup \overline{B}) = \dfrac{2}{3} and P(A∪B)=59P(A \cup B) = \dfrac{5}{9}, then which of the following are TRUE?

A. P(A∩B)=13P(A \cap B) = \dfrac{1}{3}

B. P(A∪B)+P(A∩B)=79P(A \cup B) + P(A \cap B) = \dfrac{7}{9}

C. P(A)+P(B)=29P(A) + P(B) = \dfrac{2}{9}

D. P(A‾)+P(B‾)=109P(\overline{A}) + P(\overline{B}) = \dfrac{10}{9}

Choose the correct answer from the options given below:

Answer options
Option 1
Correct Answer
Explanation for 2026: 31 May Shift 2 MAT question 2

Q3:

2026: 31 May Shift 2

Probability

Easy

Core

In a factory of fans, machine A, B and C manufacture respectively 20%, 40% and 40% of total fans. Out of their output 2, 3 and 5 percent are defective fans respectively. A fan is drawn at random from the products then the probability that the drawn fan is defective is:

Answer options
Option 2
Correct Answer
Explanation for 2026: 31 May Shift 2 MAT question 3

Q4:

2026: 31 May Shift 2

Probability

Easy

Core

Match the LIST-I with LIST-II

LIST-ILIST-II
A.If AA and BB are mutually exclusive events then P(A∪B)=P(A \cup B) =I.P(A∩B)P(B),P(B)≠0\dfrac{P(A \cap B)}{P(B)}, P(B) \neq 0
B.If AA and BB are independent events then P(A∩B)=P(A \cap B) =II.P(A∩B)P(A),P(A)≠0\dfrac{P(A \cap B)}{P(A)}, P(A) \neq 0
C.If AA and BB 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 AA and BB 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
Option 1
Correct Answer
Explanation for 2026: 31 May Shift 2 MAT question 4

Q5:

2026: 31 May Shift 2

Probability

Hard

Core

A letter is known to have come either from LONDON or CLIFTON. On the envelop just two consecutive letter 'ON' are visible. The probability that the letter has come from LONDON is:

Answer options
Option 2
Correct Answer
Explanation for 2026: 31 May Shift 2 MAT question 5

Q6:

2026: 31 May Shift 2

Probability

Medium

Applied

If the mean and variance of a binomial distribution are 4 and 3 respectively, then the probability of getting exactly 5 successes in the distribution is:

Answer options
Option 3
Correct Answer
Explanation for 2026: 31 May Shift 2 MAT question 6

Q7:

2026: 31 May Shift 2

Probability

Medium

Applied

A random variable X has the following probability distribution:

X01234567
P(X)0k2k2k3kk2k^22k22k^27k2+k7k^2+k

Then, which of the following are correct ?

A. k=0.1k = 0.1

B. P(X<3)=0.3P(X<3) = 0.3

C. P(0<X<3)=0.8P(0<X<3) = 0.8

D. P(X>6)=0.17P(X>6) = 0.17

Choose the correct answer from the options given below:

Answer options
Option 1
Correct Answer
Explanation for 2026: 31 May Shift 2 MAT question 7

Q8:

2026: 31 May Shift 2

Probability

Medium

Applied

Let X denote the number of hours a person watches television during a randomly selected day. The probability that X can take the value xix_i, has the following form, where kk is some unknown constant.

P(X=xi)={0.5if xi=0kxiif xi=1k(4−xi)if xi=2 or 30otherwiseP(X=x_i) = \begin{cases} 0.5 & \text{if } x_i=0 \\ kx_i & \text{if } x_i=1 \\ k(4-x_i) & \text{if } x_i=2 \text{ or } 3 \\ 0 & \text{otherwise} \end{cases}

The probability that the person watches at most 2 hours of television on a selected day is :

Answer options
Option 2
Correct Answer
Explanation for 2026: 31 May Shift 2 MAT question 8

Q9:

2026: 31 May Shift 2

Probability

Medium

Applied

The probability distribution table of a random variable X is given below.

XiX_iP(Xi)P(X_i)
01/8
1a
2b
31/8

If E(X)=32E(X) = \dfrac{3}{2}, then the values of a and b are:

Answer options
Option 1
Correct Answer
Explanation for 2026: 31 May Shift 2 MAT question 9

Q10:

2026: 30 May Shift 2

Probability

Easy

Common

Probability of solving a specific problem independently by A and B are 12\frac{1}{2} and 13\frac{1}{3} respectively, then which of the following are True?

[where P(X) denotes the probability of an event X]

(A) P(Problem is solved) = 23\frac{2}{3}

(B) P(Exactly one of them solves the problem) = 12\frac{1}{2}

(C) P(Both of them solve the problem) = 56\frac{5}{6}

(D) P(Atmost one solves the problem) = 56\frac{5}{6}

Choose the correct answer from the options given below:

Answer options

Q11:

2026: 30 May Shift 2

Probability

Medium

Core

If A and B are two independent events such that P(A) = 1/3, P(A∪B)=3/5P(A\cup B) = 3/5 and P(B) = p, then which of the following are true?

(A) The value of p is 25\frac{2}{5}

(B) P(A∩B)=215P(A\cap B) = \frac{2}{15}

(C) P(B/A)=13P(B/A) = \frac{1}{3}

(D) P(A/B)=35P(A/B) = \frac{3}{5}

Choose the correct answer from the options given below:

Answer options

Q13:

2026: 30 May Shift 2

Probability

Medium

Core

A man is known to speak truth 4 out of 5 times. A coin is tossed. He reports that a head appears. The probability that actually there was a head, is

Answer options

Q14:

2026: 30 May Shift 2

Probability

Easy

Core

If two cards are drawn at random without replacement from a pack of 52 playing cards, then the probability that both the cards are red, is

Answer options

Q15:

2026: 30 May Shift 2

Probability

Hard

Applied

A random variable X has the following probability distribution:

X0123456
P(X)z2z2z3zz2z^22z22z^27z2+z7z^2+z

Which of the following are TRUE?

(A) P(X < 3) = 0.8

(B) P(X > 4) = 0.2

(C) The value of z is 0.1

(D) P(3≤X<5)=0.31P(3\le X<5) = 0.31

Choose the correct answer from the options given below:

Answer options

Q16:

2026: 30 May Shift 1

Probability

Medium

Common

A box has 5 blue and 4 red balls. One ball is drawn at random and not replaced. Its colour is also not noted, then another ball is drawn at random. The probability of second ball being red is

Answer options

Q17:

2026: 30 May Shift 1

Probability

Easy

Common

If E and F are two events such that P(E) = 0.8, P(F) = 0.7 and P(E ∩ F) = 0.6 then P(Eˉ/Fˉ)P(\bar{E}/\bar{F}) is:

Answer options

Q18:

2026: 30 May Shift 1

Probability

Easy

Core

If A and B are two independent events such that P(A) = 0.5, P(B) = 0.3 then

Match the LIST-I with LIST-II

LIST-ILIST-II
A. P(A/B)P(A/B)I. 0.35
B. P(B/A)P(B/A)II. 0.5
C. P(A∩B)P(A\cap B)III. 0.3
D. P(Aˉ∩Bˉ)P(\bar{A}\cap\bar{B})IV. 0.15

Choose the correct answer from the options given below:

Answer options

Q19:

2026: 30 May Shift 1

Probability

Medium

Core

A bag contains 17 tickets numbered from 1 to 17. A ticket is drawn, and then another ticket is drawn without replacing the first one. Then the probability that both the tickets show at least one even number, is

Answer options

Q20:

2026: 30 May Shift 1

Probability

Easy

Core

Probabilities of solving a specific mathematical problem independently by three students are 12,13\frac{1}{2},\frac{1}{3} and 14\frac{1}{4}. If all the three students try to solve the problem independently, the probability that the problem is solved, is:

Answer options

Q21:

2026: 30 May Shift 1

Probability

Hard

Core

A company manufactured a product through 4 units. A, B, C and D. The probability that a product is manufactured are 310,15,110\frac{3}{10},\frac{1}{5},\frac{1}{10} and 25\frac{2}{5} by units A, B, C and D respectively. The probability that it will be defective are 14,13\frac{1}{4},\frac{1}{3} and 112\frac{1}{12} if it is produced by A, B, C respectively and there is no defective from unit D. A product is chosen, it is found to be defective.

Match the LIST-I with LIST-II

LIST-ILIST-II
A. The Probability that the defective product is from AI. 4/9
B. The Probability that the defective product is from BII. 0
C. Probability of defective productIII. 1/2
D. Probability that the defective product is from DIV. 3/20

Choose the correct answer from the options given below:

Answer options

Q22:

2026: 29 May Shift 2

Probability

Medium

Common

If P(B)=25,P(A∪B)=35,P(A∩B)=310P(B) = \frac{2}{5}, P(A \cup B) = \frac{3}{5}, P(A \cap B) = \frac{3}{10} then P(A/B)+2P(B/A)P(A/B) + 2P(B/A) equals

Answer options

Q23:

2026: 29 May Shift 2

Probability

Medium

Core

If A, B and C are three independent events such that P(A)=P(B)=P(C)=x, then P(at least two of A, B, C occur) =

Answer options

Q24:

2026: 29 May Shift 2

Probability

Easy

Core

If A and B are two events such that A≠∅A \neq \emptyset, B ≠∅\neq \emptyset, then which of following is/are TRUE:

A. P(B/A)=P(A).P(B)P(B/A)=P(A).P(B)

B. P(B/A)=P(A∩B)P(A)P(B/A)=\frac{P(A\cap B)}{P(A)}

C. P(A′/B′)=1−P(A∪B)1−P(B)P(A'/B')=\frac{1-P(A\cup B)}{1-P(B)}

D. P(B/A).P(A/B)=2P(B/A).P(A/B)=2

Choose the correct answer from the options given below:

Answer options

Q25:

2026: 29 May Shift 2

Probability

Medium

Core

An insurance company insured 200 scooter drivers, 400 car drivers and 600 Truck drivers. The probability of an accident involving scooter driver, car driver and truck driver are 0.01, 0.03 and 0.15 respectively. One of the insured person meets an accident, then the probability that he is a truck driver is:-

Answer options

Q26:

2026: 29 May Shift 2

Probability

Easy

Core

A box contains 3 red and 2 black balls and another box contains 2 red and 4 black balls. One box chosen at random. From the selected box one ball is drawn. The probability of getting a red ball is

Answer options

Q27:

2026: 29 May Shift 2

Probability

Hard

Applied

Ten cards are numbered from 1 to 10. A and B are playing a game by drawing a card, alternately, with replacement, till one gets a prime number then the probability of B's winning if A started the game is

Answer options

Q28:

2026: 29 May Shift 1

Probability

Medium

Common

If A and B are event such that P(A)=12P(A)=\dfrac{1}{2}, P(B)=pP(B)=p and P(A∪B)=35P(A \cup B)=\dfrac{3}{5} then which of following statements is/are TRUE?

(A) Events A and B are independent if p=15p=\dfrac{1}{5}

(B) Events A and B are independent if p∈(0,1)∼{15}p \in (0,1) \sim \left\{\dfrac{1}{5}\right\}

(C) Events A and B are independent then P(A∩B)=110P(A \cap B)=\dfrac{1}{10}

(D) Event A and B are independent then P(A∩B)=910P(A \cap B)=\dfrac{9}{10}

Choose the correct answer from the options given below:

Answer options

Q29:

2026: 29 May Shift 1

Probability

Easy

Core

Bag A contains 3 red and 5 white balls and bag B contains 4 red and 4 white balls. A bag is selected at random, and a ball is drawn. Then the probability of getting a red ball is:

Answer options

Q30:

2026: 29 May Shift 1

Probability

Medium

Core

Let A and B be two independent events such that P(A)=kP(A)=k, P(B)=2kP(B)=2k and probability of occurrence of exactly one of them is 59\dfrac{5}{9}. Then the value(s) of kk is/are:

Answer options