The random variable X has a probability distribution \(P(X=x)=\begin{Bmatrix}5k, & x=1\\2k, & x=1\\3k, & x=2\\0, & otherwise\end{Bmatrix}\) Then, the value of E(X) is:

0.8 0.4 1 0.5

1

x 0 1 2 E(x) 5k 2k 3k Since x is random variable its sum of probabilities is equal to 1 $\sum_0^4E(x)=1$ $E(x=0)+E(x=1)+E(x=2)=1$ $5k+2k+3k=1$ $10k = 1$ $k=\frac{1}{10}$