If a random variable X has the following probability distribution:, X, 0, 1, 2, 3, P(X), K, K/2, K/4, K/8, then, Match List-I with List-II, List-I, List-II, (A) The value of K is, (I) 2/15, (B) P(0 2) is, (IV) 4/15, Choose the correct answer from the options given below: Option 1: (A) - (I), (B) - (II), (C) - (III), (D) - (IV) Option 2: (A) - (II), (B) - (III), (C) - (I), (D) - (IV) Option 3: (A) - (III), (B) - (IV), (C) - (I), (D) - (II) Option 4: (A) - (IV), (B) - (I), (C) - (III), (D) - (II)

Answer: (A) - (III), (B) - (IV), (C) - (I), (D) - (II). Solution: Given probability distribution: - P(X=0) = K - P(X=1) = K/2 - P(X=2) = K/4 - P(X=3) = K/8 All probabilities must sum to 1: K + K/2 + K/4 + K/8 = 1 Converting to common denominator 8: 8K/8 + 4K/8 + 2K/8 + K/8 = 1 15K/8 = 1 K = 8/15 Therefore, (A) matches with (III) For P(0 For P(1 For P(X > 2): The inequality X > 2 includes only X = 3: P(X > 2) = P(X=3) = K/8 = (8/15)/8 = 8/15 x 1/8 = 1/15 Therefore, (D) matches with (II) Final matching: - (A) → (III): K = 8/15 - (B) → (IV): P(0 2) = 1/15

Algebra

Probability

Medium

common

If a random variable X has the following probability distribution:

X 0 1 2 3
P(X) K K/2 K/4 K/8

then,

Match List-I with List-II

List-I List-II
(A) The value of K is (I) 2/15
(B) P(0 < X < 2) is (II) 1/15
(C) P(1 < X < 3) is (III) 8/15
(D) P(X > 2) is (IV) 4/15

Choose the correct answer from the options given below:

X	0	1	2	3
P(X)	K	K/2	K/4	K/8

List-I	List-II
(A) The value of K is	(I) 2/15
(B) P(0 < X < 2) is	(II) 1/15
(C) P(1 < X < 3) is	(III) 8/15
(D) P(X > 2) is	(IV) 4/15

(A) - (I), (B) - (II), (C) - (III), (D) - (IV)

(A) - (II), (B) - (III), (C) - (I), (D) - (IV)

(A) - (III), (B) - (IV), (C) - (I), (D) - (II)

(A) - (IV), (B) - (I), (C) - (III), (D) - (II)

✅ Correct Option: 3

Given probability distribution:

$P(X=0) = K$
$P(X=1) = K/2$
$P(X=2) = K/4$
$P(X=3) = K/8$

All probabilities must sum to 1:

$K + \frac{K}{2} + \frac{K}{4} + \frac{K}{8} = 1$

Converting to common denominator 8:

$\frac{8K}{8} + \frac{4K}{8} + \frac{2K}{8} + \frac{K}{8} = 1$

$\frac{15K}{8} = 1$

$K = \frac{8}{15}$

Therefore, (A) matches with (III)

For $P(0 < X < 2)$ :

The inequality $0 < X < 2$ includes only $X = 1$ :

$P(0 < X < 2) = P(X=1)$

$= \frac{K}{2}$

$= \frac{8/15}{2}$

$= \frac{8}{15} \times \frac{1}{2}$

$= \frac{4}{15}$

Therefore, (B) matches with (IV)

For $P(1 < X < 3)$ :

The inequality $1 < X < 3$ includes only $X = 2$ :

$P(1 < X < 3) = P(X=2)$

$= \frac{K}{4}$

$= \frac{8/15}{4}$

$= \frac{8}{15} \times \frac{1}{4}$

$= \frac{2}{15}$

Therefore, (C) matches with (I)

For $P(X > 2)$ :

The inequality $X > 2$ includes only $X = 3$ :

$P(X > 2) = P(X=3)$

$= \frac{K}{8}$

$= \frac{8/15}{8}$

$= \frac{8}{15} \times \frac{1}{8}$

$= \frac{1}{15}$

Therefore, (D) matches with (II)

Final matching:

(A) → (III): $K = \frac{8}{15}$
(B) → (IV): $P(0 < X < 2) = \frac{4}{15}$
(C) → (I): $P(1 < X < 3) = \frac{2}{15}$
(D) → (II): $P(X > 2) = \frac{1}{15}$

2026: 31 May Shift 2

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

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

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

2026: 19 May Shift 1

Two cards are drawn successively with replacement from a well shuffled pack of 52 playing cards. Which of the following is correct probability distribution of number of Queens(X) as $\binom{X}{P(X)}$ ?

2023: 23 May Shift 3

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 $\frac{3}{10}, \frac{1}{5}, \frac{1}{10}$ and $\frac{2}{5}$ . The probabilities that he will be late are $\frac{1}{4}, \frac{1}{3}$ and $\frac{1}{12}$ , 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:

2025: 19 May Shift 1

Match List-I with List-II

Let A and B be any two events

List-I	List-II
(A) $P(A')$	(I) $\frac{P(A \cap B)}{P(A)}; P(A) \neq 0$
(B) $P(\phi)$	(II) $\frac{P(A \cap B)}{P(B)}; P(B) \neq 0$
(C) $P(A\vert B)$	(III) $1 - P(A)$
(D) $P(B\vert A)$	(IV) 0

Choose the correct answer from the options given below:

2026: 22 May Shift 1

Bag I contains 2 red and 3 blue balls and bag II contains $\alpha$ red and 5 blue balls. One ball is drawn at random from one of the bags and is found to be blue. If the probability that it was drawn from bag II is $\frac{25}{52}$ , then $\alpha$ is equal to:

2023: 22 May Shift 3

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 :

If a random variable X has the following probability distribution: X0123P(X)KK/2K/4K/8 then, Match List-I with List-II List-IList-II(A) The value of K is(I) 2/15(B) P(0 < X < 2) is(II) 1/15(C) P(1 < X < 3) is(III) 8/15(D) P(X > 2) is(IV) 4/15 Choose the correct answer from the options given below:

Solution

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