If a random variable $X$ has the following probability distribution:

X	0	1	2	3	4
P(X)	k	2k	3k	k²	6k²

, then

Match List-I with List-II

List-I	List-II
(A) k	(I) 3/7
(B) $P(X < 2)$	(II) 6/49
(C) $P(X > 3)$	(III) 1/7
(D) $P(2 \leq X \leq 3)$	(IV) 22/49

Choose the correct answer from the options given below:

For any probability distribution, the sum of all probabilities must equal 1.

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

$k + 2k + 3k + k^2 + 6k^2 = 1$

$6k + 7k^2 = 1$

$7k^2 + 6k - 1 = 0$

$k = \dfrac{-6 \pm \sqrt{36 - 4(7)(-1)}}{2 \times 7}$

$k = \dfrac{-6 \pm \sqrt{64}}{14}$

$k = \dfrac{-6 \pm 8}{14}$

This gives two values:

$k = \dfrac{-6 + 8}{14} = \dfrac{2}{14} = \dfrac{1}{7}$ (valid)

$k = \dfrac{-6 - 8}{14} = \dfrac{-14}{14} = -1$ (rejected, probability cannot be negative)

Therefore $k = \dfrac{1}{7}$

(A) matches with (III)

---

$P(X < 2)$ means $X$ can be 0 or 1.

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

$P(X < 2) = k + 2k$

$P(X < 2) = 3k$

$P(X < 2) = 3 \times \dfrac{1}{7}$

$P(X < 2) = \dfrac{3}{7}$

(B) matches with (I)

---

$P(X > 3)$ means $X$ can only be 4.

$P(X > 3) = P(X=4)$

$P(X > 3) = 6k^2$

$P(X > 3) = 6 \times \left(\dfrac{1}{7}\right)^2$

$P(X > 3) = 6 \times \dfrac{1}{49}$

$P(X > 3) = \dfrac{6}{49}$

(C) matches with (II)

---

$P(2 \leq X \leq 3)$ means $X$ can be 2 or 3.

$P(2 \leq X \leq 3) = P(X=2) + P(X=3)$

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

$P(2 \leq X \leq 3) = 3\left(\dfrac{1}{7}\right) + \left(\dfrac{1}{7}\right)^2$

$P(2 \leq X \leq 3) = \dfrac{3}{7} + \dfrac{1}{49}$

$P(2 \leq X \leq 3) = \dfrac{21}{49} + \dfrac{1}{49}$

$P(2 \leq X \leq 3) = \dfrac{22}{49}$

(D) matches with (IV)

---

The correct matching is:

(A) k → (III) $\dfrac{1}{7}$

(B) $P(X < 2)$ → (I) $\dfrac{3}{7}$

(C) $P(X > 3)$ → (II) $\dfrac{6}{49}$

(D) $P(2 \leq X \leq 3)$ → (IV) $\dfrac{22}{49}$