Problem source

I have a bag containing $M$ tokens, $m$ of which are red. I remove
$n$ tokens from the bag at random without replacement. Let 
\[
X_{i}=\begin{cases}
1 & \mbox{ if the ith token I remove is red;}\\
0 & \mbox{ otherwise.}
\end{cases}
\]
Let $X$ be the total number of red tokens I remove. 
\begin{questionparts}
\item Explain briefly why $X=X_{1}+X_{2}+\cdots+X_{n}.$
\item Find the expectation $\mathrm{E(}X_{i}).$
\item Show that $\mathrm{E}(X)=mn/M$. 
\item Find $\mathrm{P}(X=k)$ for $k=0,1,2,\ldots,n$. 
\item Deduce that 
\[
\sum_{k=1}^{n}k\binom{m}{k}\binom{M-m}{n-k}=m\binom{M-1}{n-1}.
\]
\end{questionparts}

Solution source

\begin{questionparts}
\item The left hand side counts the number of red tokens we have taken. The right hand side counts the number of red tokens we have taken at each point, across all points. Therefore these must be the same.

\item $\E[X_i] = \mathbb{P}(i\text{th token is red}) = \frac{m}{M}$ (since there is nothing special about the $i$th token.
\item Therefore $\E[X] = \E[X_1 + \cdots + X_n] = n\E[X_i] = \frac{nm}{M}$
\item $\mathbb{P}(X=k) = \binom{m}{k}\binom{M-m}{n-k}/\binom{M}{n}$ since this is the number of ways we can choose $k$ of the $m$ red objects, $n-k$ of the $M-m$ non-red objects divided by the total number of ways we can choose our $n$ tokens.

\item $\,$ \begin{align*}
&& \frac{mn}{M} &= \E[X] \\
&&&= \sum_{k=1}^n k \mathbb{P}(X=k) \\
&&&= \sum_{k=1}^n k \binom{m}{k}\binom{M-m}{n-k}/\binom{M}{n} \\
\Rightarrow &&  \sum_{k=1}^n k \binom{m}{k}\binom{M-m}{n-k} &= m \frac{n}{M} \binom{M}{n} = m \binom{M-1}{n-1}
\end{align*}
\end{questionparts}

This question is a nice example of how to find the mean of the hypergeometric distribution