A random variable X has the following probability distribution:

Then:

(A) $k=\frac{1}{6}$

(B) $P(X<2)=\frac{1}{2}$

(C) $E(X)=\frac{3}{4}$

(D) $P(1<X<2)=\frac{5}{6}$

Choose the correct answer from the options given below:

(A) and (B) only

The correct answer is Option (1) → (A) and (B) only

Since sum of all probabilities must be 1.

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

$⇒6k=1$

$⇒k=\frac{1}{6}$ → (A)

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

$=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}$ → (B)