Let X be a distance random variable whose probability distribution is defined as :

$P(X=x)= \left\{\begin{matrix}0.5 & ,if & x=0\\k(x+1) & ,if & x=1\, or \, 2\\k(6-x) & ,if & x=3\, or\, 4\\0 & , & otherwise\end{matrix}\right.$

The, value of k is :

$\frac{2}{21}$

The correct answer is Option (2) → $\frac{2}{21}$

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

$0.5k+k(1+1)+k(2+1)+k(6-3)+k(6-4)=1$

$0.5k+2k+3k+3k+2k=1$

$10.5k=1$

$k=\frac{1}{10.5}=\frac{10}{105}=\frac{2}{21}$