Expected value in an urn problem In an urn there are $n$ red balls and $m$ blue balls. I extract them without replacement. Let $X$=time of first blue. What is $E(X)$?
I found PMF of $X$ and it is, if $k > n+1$
$$P(X=k) =\frac{n(n-1) \dots (n-k+1)m}{(m+n)(m+n-1) \dots (n+m-k+1)}$$
How could I evaluate $E(X)$?
Edit: I'm looking for an explicit form of $E(X)$ 
 A: For clarity, let there be $b$ blue balls and $r$ red ones.
By symmetry the gaps between the blue balls should be of equal length.  Thus the answer is $$\boxed{E[r,b]=\frac r{b+1
}+1}$$
Sanity check:  suppose there are $1$ red balls and $2$ blue ones.  Then the answer is $1$ with probability $\frac 23$ and $2$ with probability $\frac 13$ so $E=\frac 43$ and of course $\frac {1}{1+2}+1=\frac 43$.  More generally, with $1$ red and $b$ blue then the answer is $$E=1\times \frac b{b+1}+2\times (1-\frac b{b+1})=\frac {b+2}{b+1}=\frac 1{b+1}+1$$
A: Assume there are $r$ red and $b$ blue balls. We define the random variables as
$X_1$ : number of red balls preceding the first blue ball.
$X_i$ : number of red balls drawn following the appearance of the $(i-1)$th blue but before the appearance of the $i$th blue ball,$\quad i=2,3,...$
$X_{b+1}:$ number of red balls drawn following the appearance of the $b$th blue.
So, $X_1+X_2+...+X_{b+1}=r$; note that the $(X_{i})_{i=1,2,...,b+1}$'s are identically distributed.
Thus from the linearity of expectation, $(b+1)\mathbb{E}(X_1)=r\Rightarrow \mathbb{E}(X_1)=\dfrac{r}{b+1}$
Now, let $Y:$ number of draws needed to get the first blue ball.
Then, $Y=X_1+1\Rightarrow \mathbb{E}(Y)=\mathbb{E}(X_1)+1$, which is the required answer.

We can also calculate $\mathbb{E}(X_1)$ directly from the pmf 
$$\mathbb{P}(X_1=k)=\frac{b}{r+b}.\frac{(r)_k}{(r+b-1)_k}\quad,k=0,1,...,r$$
