If the random variable X has the following probability distribution:

X	0	1	2	otherwise
P(X)	k	3k	5k	0

Match List-I with List-II

List-I	List-II
(A) k	(I) $\frac{13}{9}$
(B) E (X)	(II) $\frac{4}{9}$
(C) P (X ≤ 1)	(III) $\frac{8}{9}$
(D) P (1 ≤ X ≤ 2)	(IV) $\frac{1}{9}$

Choose the correct answer from the options given below:

1. (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
2. (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
3. (A) - (IV), (B) - (II), (C) - (I), (D) - (III)
4. (A) - (III), (B) - (II), (C) - (I), (D) - (IV)

The probability distribution has X taking values 0, 1, or 2, with probabilities k, 3k, and 5k respectively.

The sum of all probabilities must equal 1:

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

$k + 3k + 5k = 1$

$9k = 1$

$k = \dfrac{1}{9}$

(A) matches with (IV)

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The expected value is calculated as:

$E(X) = (0 \times k) + (1 \times 3k) + (2 \times 5k)$

$E(X) = 0 + 3k + 10k$

$E(X) = 13k$

$E(X) = 13 \times \dfrac{1}{9}$

$E(X) = \dfrac{13}{9}$

(B) matches with (I)

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For X ≤ 1, the random variable can be 0 or 1:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X \leq 1) = k + 3k$

$P(X \leq 1) = 4k$

$P(X \leq 1) = 4 \times \dfrac{1}{9}$

$P(X \leq 1) = \dfrac{4}{9}$

(C) matches with (II)

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For 1 ≤ X ≤ 2, the random variable can be 1 or 2:

$P(1 \leq X \leq 2) = P(X = 1) + P(X = 2)$

$P(1 \leq X \leq 2) = 3k + 5k$

$P(1 \leq X \leq 2) = 8k$

$P(1 \leq X \leq 2) = 8 \times \dfrac{1}{9}$

$P(1 \leq X \leq 2) = \dfrac{8}{9}$

(D) matches with (III)

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The correct matching is:

(A) → (IV): $k = \dfrac{1}{9}$

(B) → (I): $E(X) = \dfrac{13}{9}$

(C) → (II): $P(X \leq 1) = \dfrac{4}{9}$

(D) → (III): $P(1 \leq X \leq 2) = \dfrac{8}{9}$

Therefore, the answer is Option 2: (A) - (IV), (B) - (I), (C) - (II), (D) - (III)