The probability distribution of a discrete random variable is given by :

$P(X-x)=\left\{\begin{matrix} 3kx & for & x=1, 2,3 \\2k(x+3) & for & x=4, 5, 6\\0 & & otherwise \end{matrix}\right.$

where k is a constant.

The mean of X is :

$\frac{10}{3}$

The correct answer is Option (4) → $\frac{10}{3}$

Sum of probabilities,

$3k(1)^2+3k(2)^2+3k(3)^2+k(4+3)+k(5+3)+k(6+3)=1$

$3k+12k+27k+7k+8k+9k=1$

$k=\frac{1}{66}$

$E(X)=\sum\limits_xx.P(X=x)$

$=1×3k(1)+2.3k(4)+3.3k(9)+4.k(7)+5.k(8)+6.k(9)$

$=\frac{115}{33}$