# negative hypergeometric distribution

An urn contains n red balls and m black balls. Balls are drawn sequentially from the urn one at a time without replacement. Let X denote the number of red balls removed before the first black ball is chosen. Find E[X].

According to the question, I got: $$P(X=k)={(m+n-k-1)!\over (m+n)!}\times{n!\over (n-k)!}\times m$$ $$E[X]=\sum XP(X=k)$$ $$=\sum_{k=0}^nk\times {(m+n-k-1)!\over (m+n)!}\times{n!\over (n-k)!}\times m$$ So my question is how to simplified this term above?
This is so far I can get:
$$E[X]=m\sum_{k=0}^n{\begin{pmatrix}n\\k\end{pmatrix}\over\ \begin{pmatrix} m+n \\ k+1 \end{pmatrix}}\times{k\over k+1}$$ Any way I can simplified this?

• – StubbornAtom Oct 2 '18 at 20:03