A random variable y has the following probability distribution, y, 1, 2, 3, 4, 5, P(y), 2k, 3k, k, 4k, 5k, Match List-I with List-II, List-I, List-II, (A) P(y > 2), (I) 2/5, (B) k, (II) 2/3, (C) P(y ≤ 3), (III) 8/15, (D) P(2 ≤ y ≤ 4), (IV) 1/15, Choose the correct answer from the options given below: Option 1: (A) - (III), (B) - (IV), (C) - (I), (D) - (II) Option 2: (A) - (II), (B) - (I), (C) - (IV), (D) - (III) Option 3: (A) - (II), (B) - (IV), (C) - (I), (D) - (III) Option 4: (A) - (IV), (B) - (III), (C) - (II), (D) - (I)

Answer: (A) - (II), (B) - (IV), (C) - (I), (D) - (III). Solution: The probability distribution is given as:, y, 1, 2, 3, 4, 5, P(y), 2k, 3k, k, 4k, 5k, For any probability distribution, all probabilities must sum to 1: 2k + 3k + k + 4k + 5k = 1 15k = 1 k = 1/15 This matches (B) with (IV). For P(y > 2), the values of y are 3, 4, and 5: P(y > 2) = P(y=3) + P(y=4) + P(y=5) P(y > 2) = k + 4k + 5k P(y > 2) = 10k P(y > 2) = 10 x 1/15 P(y > 2) = 10/15 = 2/3 This matches (A) with (II). For P(y ≤ 3), the values of y are 1, 2, and 3: P(y ≤ 3) = P(y=1) + P(y=2) + P(y=3) P(y ≤ 3) = 2k + 3k + k P(y ≤ 3) = 6k P(y ≤ 3) = 6 x 1/15 P(y ≤ 3) = 6/15 = 2/5 This matches (C) with (I). For P(2 ≤ y ≤ 4), the values of y are 2, 3, and 4: P(2 ≤ y ≤ 4) = P(y=2) + P(y=3) + P(y=4) P(2 ≤ y ≤ 4) = 3k + k + 4k P(2 ≤ y ≤ 4) = 8k P(2 ≤ y ≤ 4) = 8 x 1/15 P(2 ≤ y ≤ 4) = 8/15 This matches (D) with (III). The correct matching is: (A) → (II) (B) → (IV) (C) → (I) (D) → (III)

Algebra

Probability

Medium

applied

A random variable y has the following probability distribution

y 1 2 3 4 5
P(y) 2k 3k k 4k 5k

Match List-I with List-II

List-I List-II
(A) $P(y > 2)$ (I) $2/5$
(B) k (II) $2/3$
(C) $P(y \leq 3)$ (III) $8/15$
(D) $P(2 \leq y \leq 4)$ (IV) $1/15$

Choose the correct answer from the options given below:

List-I	List-II
(A) $P(y > 2)$	(I) $2/5$
(B) k	(II) $2/3$
(C) $P(y \leq 3)$	(III) $8/15$
(D) $P(2 \leq y \leq 4)$	(IV) $1/15$

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

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

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

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

✅ Correct Option: 3

The probability distribution is given as:

y	1	2	3	4	5
P(y)	2k	3k	k	4k	5k

For any probability distribution, all probabilities must sum to 1:

$2k + 3k + k + 4k + 5k = 1$

$15k = 1$

$k = \dfrac{1}{15}$

This matches (B) with (IV).

For $P(y > 2)$ , the values of $y$ are 3, 4, and 5:

$P(y > 2) = P(y=3) + P(y=4) + P(y=5)$

$P(y > 2) = k + 4k + 5k$

$P(y > 2) = 10k$

$P(y > 2) = 10 \times \dfrac{1}{15}$

$P(y > 2) = \dfrac{10}{15} = \dfrac{2}{3}$

This matches (A) with (II).

For $P(y \leq 3)$ , the values of $y$ are 1, 2, and 3:

$P(y \leq 3) = P(y=1) + P(y=2) + P(y=3)$

$P(y \leq 3) = 2k + 3k + k$

$P(y \leq 3) = 6k$

$P(y \leq 3) = 6 \times \dfrac{1}{15}$

$P(y \leq 3) = \dfrac{6}{15} = \dfrac{2}{5}$

This matches (C) with (I).

For $P(2 \leq y \leq 4)$ , the values of $y$ are 2, 3, and 4:

$P(2 \leq y \leq 4) = P(y=2) + P(y=3) + P(y=4)$

$P(2 \leq y \leq 4) = 3k + k + 4k$

$P(2 \leq y \leq 4) = 8k$

$P(2 \leq y \leq 4) = 8 \times \dfrac{1}{15}$

$P(2 \leq y \leq 4) = \dfrac{8}{15}$

This matches (D) with (III).

The correct matching is:

(A) → (II)

(B) → (IV)

(D) → (III)

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