Proving Expected Value of a Random Variable Let $r$ and $b$ be positive integers and define $\alpha = \frac{r}{r+b}$. 
A bowl contains $r$ red balls and $b$ blue balls; thus, $\alpha$ is the 
fraction of the balls that are red. Consider the following experiment:   
Choose one ball uniformly at random. 


*

*If the chosen ball is red, then put it back, together with 
        an additional red ball.

*If the chosen ball is blue, then put it back, together with 
        an additional blue ball. 
Define the random variable $X$ to be the fraction of the balls that are
red, after this experiment. Prove that $E(X) = \alpha$. 
this what I have so far:
E(X)=$\sum\limits_{w\in S}X(w)∙Pr(w)$ 
Let X be the fraction of the balls that are red, after this experiment
$X = \alpha =\frac{r}{r+b}$
$\Pr(w)=0*a+\frac{1}{r+b}*a+\frac{2}{r+b}*a+....\frac{r+b}{r+b}*a$
=$\frac{r}{(r+b)^2}(1+2+3+...+r+b)$
=$\frac{r}{(r+b)^2}*\frac{r+b(r+b+1)}{2}$        after expending I get:
=$\frac{r^2+rb}{2}$
I don't know if what I have done so far is correct or how I should go from here.
Any ideas?
 A: We calculate. Let $n=r+b$. After we have done the experiment, we have $n+1$ balls. The probability that $r+1$ of them are red is $\frac{r}{n}$,  and the probability $r$ of them are red is $\frac{b}{n}$. 
So if $Y$ is the number of red after the experiment, then
$$E(Y)=(r+1)\cdot\frac{r}{n}+r\cdot \frac{b}{n}.$$
Simplify. We get $E(Y)=\frac{r^2+r+rb}{n}$. But $r^2+rb=rn$, so $E(Y)=\frac{(n+1)r}{n}$.
Since $X=\frac{Y}{n+1}$, we have $E(X)=\frac{E(Y)}{n+1}=\frac{r}{n}$.
A: To streamline André’s answer
You have $\mathsf E(X) = \sum\limits_{\omega\in S} \Pr(\omega)X(\omega)$
However, the outcomes for the experiment are simply either: $R$: we add a red ball else $B$: we add a blue ball.  So, $S=\{R, B\}$
Now the probability of adding a ball is equal to the proportion of balls before the increment.  Then we measure $X$, the proportion of red balls after the increment.
So we want to find:   $\mathsf E(X)
 = \Pr(R)\;X(R) + \Pr(B)\; X(B)
\\[1ex] = \dfrac{r}{r+b}\;\dfrac{r+1}{r+b+1}+ \dfrac{b}{r+b}\;\dfrac{r}{r+b+1}
\\[1ex] = \dfrac{r}{r+b}\;\dfrac{r+b+1}{r+b+1}
\\[2ex] = \alpha$
