# Closed form for Chu–Vandermonde identity-like summations

I have been trying to analyze a randomized algorithm I cooked up and have found the need to compute some moments of a random variable modeling an aspect of the algorithm but I am unsure if there is a nice closed form. Suppose the random variable in question is $$X$$ and it can take a value $$i \in \lbrace 0, 1, \cdots, k \rbrace$$ with probability $$\text{Pr}\lbrace X = i \rbrace = \binom{n}{k}^{-1}\binom{M}{i}\binom{n-M}{k-i}$$ where $$n \geq M \geq 1$$ and $$n \geq k \geq 1$$. Note that I assume that $$\binom{a}{b} = 0$$ if $$a < b$$ since there are $$0$$ ways to choose size $$b$$ sets out of a size $$a$$ set when this inequality holds. It is clear by the Chu-Vandermonde identity that indeed these values are probabilities because $$\sum_{i=0}^{k} \text{Pr}\lbrace X = i \rbrace = \binom{n}{k}^{-1} \sum_{i=0}^k \binom{M}{i}\binom{n-M}{k-i} = 1$$ The first moment of $$X$$, namely its expectation, can be computed analytically as \begin{align} \mathbb{E}(X) &= \sum_{i=0}^k i \text{Pr}\lbrace X = i \rbrace \\ &= \binom{n}{k}^{-1} \sum_{i=1}^k i \binom{M}{i}\binom{n-M}{k-i} \\ &= M \binom{n}{k}^{-1} \sum_{i=1}^k \binom{M-1}{i-1}\binom{(n-1)-(M-1)}{(k-1)-(i-1)} \\ &= M \binom{n}{k}^{-1} \binom{n-1}{k-1} \\ &= \frac{Mk}{n} \end{align}

I also need to compute $$\mathbb{E}(X^2)$$ so I can do a variance calculation but I am not sure how to go about it. One can find a loose upper bound of $$Mk^2/n$$ but at least empirically, it seems for the valid ranges of $$k$$, the upper bound can at least be made to be $$Mk^{3/2}/n$$. If I could find an exact expression to this moment, I would feel a lot better of where I might be able to go with this algorithm.

So does anyone recognize this probability distribution and have insight into the desired moment I want to compute?