<s>[INST] Finding mathematical expectation Suppose that a distribution of a random variable $X$ is given by,
$P(X=-1)=\frac{1}{6}=P(X=4)$ and
$P(X=0)=\frac{1}{3}=P(X=2)$. 
Then, Find the value of $E\left(\frac{X}{X+2}\right)$.
(A) $\frac{1}{36}$
(B) $\frac{1}{18}$
(C) $\frac{1}{9}$
(D) $\frac{1}{6}$.
I am finding $E(X)=\frac{7}{6}$ and $E(X+2)=\frac{19}{6}$. 
 [/INST] $E(X)$ and $E(X+2)$ will not help. You want to find the value of $\frac{X}{X+2}$ in the cases $X=-1$, $X=4$, $X=0$, and $X=2$, multiply by the relevant probability, and add up. 
It will not be difficult. For example, when $X=-1$, we have $\frac{X}{X+2}=-1$, so one term of your sum will be $(-1)(\frac{1}{6})$.
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