The probability distribution of a random variable X is

X	0	1	2	3	4
P(X)	0.2	k	k	2k	k

Match List-I with List-II

List-I	List-II
(A) value of k	(I) $\frac{16}{25}$
(B) $P(x \geq 2)$	(II) $\frac{9}{25}$
(C) $P(x = 3)$	(III) $\frac{4}{25}$
(D) $P(x < 2)$	(IV) $\frac{8}{25}$

Choose the correct answer from the options given below:

The probability distribution table has unknown values represented by $k$. The sum of all probabilities must equal 1.

$P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1$

$0.2 + k + k + 2k + k = 1$

$0.2 + 5k = 1$

$5k = 0.8$

$k = 0.16$

$k = \dfrac{4}{25}$

Therefore, (A) value of $k = \dfrac{4}{25}$ = (III)

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For $P(x \geq 2)$, include $X = 2, 3,$ and $4$:

$P(x \geq 2) = P(X=2) + P(X=3) + P(X=4)$

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

$P(x \geq 2) = 4k$

$P(x \geq 2) = 4 \times \dfrac{4}{25}$

$P(x \geq 2) = \dfrac{16}{25}$

Therefore, (B) $P(x \geq 2) = \dfrac{16}{25}$ = (I)

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From the table, $P(X=3) = 2k$:

$P(x = 3) = 2k$

$P(x = 3) = 2 \times \dfrac{4}{25}$

$P(x = 3) = \dfrac{8}{25}$

Therefore, (C) $P(x = 3) = \dfrac{8}{25}$ = (IV)

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For $P(x < 2)$, include $X = 0$ and $X = 1$:

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

$P(x < 2) = 0.2 + k$

$P(x < 2) = \dfrac{5}{25} + \dfrac{4}{25}$

$P(x < 2) = \dfrac{9}{25}$

Therefore, (D) $P(x < 2) = \dfrac{9}{25}$ = (II)

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Final matching:

(A) value of $k$ → (III) $\dfrac{4}{25}$

(B) $P(x \geq 2)$ → (I) $\dfrac{16}{25}$

(C) $P(x = 3)$ → (IV) $\dfrac{8}{25}$

(D) $P(x < 2)$ → (II) $\dfrac{9}{25}$