Let X be a discrete random variable with probability distribution defined as :

$P(X=x)=\left\{\begin{matrix}kx& for & x=0, 1\\kx-\frac{1}{2} & for & x=2, 3\\0 & otherwise \end{matrix}\right.$

then k is equal to :

$\frac{1}{3}$

The correct answer is Option (2) → $\frac{1}{3}$

$P(X=x)=\left\{\begin{matrix}kx& for & x=0, 1\\kx-\frac{1}{2} & for & x=2, 3\\0 & otherwise \end{matrix}\right.$

Sum of all probabilities must be,

$k(0)+k(1)+k(2)-\frac{1}{2}+k(3)-\frac{1}{2}=1$

$k+2k+3k=2$

$6k=2$

$⇒k=\frac{2}{6}=\frac{1}{3}$