The random variable X has a probability distribution

\(P(X=x)=\begin{Bmatrix}5k, & x=2\\2k, & x=3\\3k, & x=1\\0, & otherwise\end{Bmatrix}\)

Then, the value of E(X) is:

1.9

The correct answer is option (3) → 1.9

Sum of probabilities must be equal to 1.

$⇒5k+2k+3k+0=1$

$⇒k=\frac{1}{10}$

Now,

$E(X)=∑x.P(X=x)$

$=\frac{5}{10}.2+\frac{2}{10}.3+\frac{3}{10}.1$

$=1+0.6+0.3$

$=1.9$