The probability distribution of a random variable $x$ is, $P(x) = \frac{k}{2^x}, x = 0, 1, 2, 3$. Then

Match List-I with List-II

List-I	List-II
(A) $k$	(I) $\frac{2}{15}$
(B) $P(x = 1)$	(II) $\frac{1}{5}$
(C) $P(1 < x < 3)$	(III) $\frac{8}{15}$
(D) $P(x \geq 2)$	(IV) $\frac{4}{15}$

Choose the correct answer from the options given below:

$P(x) = \dfrac{k}{2^x}, \quad x = 0, 1, 2, 3$

$x$	$P(x)$
0	$\frac{k}{2^0} = k$
1	$\frac{k}{2^1} = \frac{k}{2}$
2	$\frac{k}{2^2} = \frac{k}{4}$
3	$\frac{k}{2^3} = \frac{k}{8}$

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(A) Since the sum of all probabilities must equal 1:

$k + \dfrac{k}{2} + \dfrac{k}{4} + \dfrac{k}{8} = 1$

$\dfrac{8k + 4k + 2k + k}{8} = 1$

$\dfrac{15k}{8} = 1$

$k = \dfrac{8}{15} \longrightarrow$ (III)

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(B)

$P(x=1) = \dfrac{k}{2}$

$= \dfrac{8}{15 \times 2}$

$= \dfrac{4}{15} \longrightarrow$ (IV)

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(C) Since $x$ takes only integer values, $1 < x < 3$ means $x = 2$ only.

$P(1 < x < 3) = P(x = 2) = \dfrac{k}{4}$

$= \dfrac{8}{15 \times 4}$

$= \dfrac{2}{15} \longrightarrow$ (I)

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(D) $x \geq 2$ means $x = 2$ or $x = 3$.

$P(x \geq 2) = P(x=2) + P(x=3)$

$= \dfrac{k}{4} + \dfrac{k}{8}$

$= \dfrac{2k + k}{8}$

$= \dfrac{3k}{8}$

$= \dfrac{3 \times 8}{8 \times 15}$

$= \dfrac{1}{5} \longrightarrow$ (II)

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List-I	Match	List-II
(A) $k = \frac{8}{15}$	→	(III)
(B) $P(x=1) = \frac{4}{15}$	→	(IV)
(C) $P(1<x<3) = \frac{2}{15}$	→	(I)
(D) $P(x \geq 2) = \frac{1}{5}$	→	(II)

Therefore, the correct answer is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).