The question is:
$X$ is a random variable, and $f(x) = (x-1)/2$ for $1 \le x \le 3$
Find $\Bbb E (X^2)$
Here's my solution: \begin{align} \Bbb P(1)&= 0/2= 0 \\ \Bbb P(2)&= 1/2 \\ \Bbb P(3)&= 2/2= 1 \\ \end{align} \begin{align} \Bbb E(X^2) = & 1^2 \Bbb P(1) + 2^2 \Bbb P(2) + 3^2 \Bbb P(3) \\ = & 1 \cdot 0 + 4 \cdot 1/2 + 9 \cdot 1 \\ = & 11 \end{align} This is my solution, but it is wrong. I need help in understanding where my mistake is.
Thanks for your help!