If the random variable X has the following distribution:

Match List-I with List-II:

Choose the correct answer from the options given below:

(A) - (IV), (B) - (III), (C) - (II), (D) - (I)

The correct answer is Option (2) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)

$\sum P(X)=1$

so $k+2k+3k+0=1$

$6k=1⇒k=\frac{1}{6}$

(A) k → (IV) $\frac{1}{6}$

(B) $P(X<2)=k+2k=3k=\frac{1}{2}$ (III)

(C) $\sum(X)=0×k+2k+6k+0$

$=8k=\frac{4}{3}$ (II)

(D) $P(X∈[1,2])=2k+3k=\frac{5}{6}$ (I)