Match List-I with List-II

If the random variable x has the following distribution:

x	0	1	2	otherwise
P(x)	k	k	2k	0

List-I	List-II
(A) k	(I) $\frac{3}{4}$
(B) P(x ≥ 2)	(II) $\frac{1}{4}$
(C) P(x ≤ 2)	(III) $\frac{1}{2}$
(D) P(0 < x ≤ 2)	(IV) 1

Choose the correct answer from the options given below:

The probability distribution shows:

• When $x = 0$, probability $= k$
• When $x = 1$, probability $= k$
• When $x = 2$, probability $= 2k$
• For any other value of $x$, probability $= 0$

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In any probability distribution, all probabilities must add up to 1.

$P(x=0) + P(x=1) + P(x=2) = k + k + 2k$

$= 4k$

Since total probability $= 1$:

$4k = 1$

$k = \frac{1}{4}$

Therefore, (A) $k$ matches with (II) $\frac{1}{4}$

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For $P(x \geq 2)$, $x$ can be $2, 3, 4, 5...$

Since the probability is $0$ for values other than $0, 1, 2$:

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

$= 2k$

$= 2 \times \frac{1}{4}$

$= \frac{1}{2}$

Therefore, (B) $P(x \geq 2)$ matches with (III) $\frac{1}{2}$

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For $P(x \leq 2)$, $x$ can be $0, 1,$ or $2$

$P(x \leq 2) = P(x=0) + P(x=1) + P(x=2)$

$= k + k + 2k$

$= 4k$

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

$= 1$

This equals $1$ because the distribution only has values at $x = 0, 1, 2$, so $P(x \leq 2)$ covers all possible outcomes.

Therefore, (C) $P(x \leq 2)$ matches with (IV) $1$

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For $P(0 < x \leq 2)$, $x$ is greater than $0$ and less than or equal to $2$

So $x$ can be $1$ or $2$ (not $0$)

$P(0 < x \leq 2) = P(x=1) + P(x=2)$

$= k + 2k$

$= 3k$

$= 3 \times \frac{1}{4}$

$= \frac{3}{4}$

Therefore, (D) $P(0 < x \leq 2)$ matches with (I) $\frac{3}{4}$

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Final Matching:

(A) $k$ → (II) $\frac{1}{4}$

(B) $P(x \geq 2)$ → (III) $\frac{1}{2}$

(C) $P(x \leq 2)$ → (IV) $1$

(D) $P(0 < x \leq 2)$ → (I) $\frac{3}{4}$

The correct answer is: (A) - (II), (B) - (III), (C) - (IV), (D) - (I)