Got this question:
Let X be Hypergeometric with parameters w, b, n, find E${x \choose 2}$ by thinking, without any complicated calculations.
So I guess there must be some story proof here, but I can not think of any. Could someone help on this?
Thanks.
I can't match your parameter names with the parameter names in Hypergeometric Distribution wikipedia page, so I'll have to use their parameters.
Specifically, we can use $\binom{x}{2}\cdot \binom{K}{x} = \binom{K}{2}\cdot \binom{K-2}{x-2}$ to simplify the sum in the expected value as
$$\begin{align*} \text{E}\binom{X}{2} &= \sum_{k=2}^n \binom{k}{2} \cdot \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}} \\ &= \frac{\binom{K}{2}}{\binom{N}{n}}\sum_{k=2}^{n} \binom{K-2}{k-2}\binom{(N-2)-(K-2)}{(n-2)-(k-2)} \\ &= \frac{\binom{K}{2}\binom{N-2}{n-2}}{\binom{N}{n}}\sum_{k=0}^{n-2} \frac{\binom{K-2}{k}\binom{(N-2)-(K-2)}{(n-2)-k}}{\binom{N-2}{n-2}} \\ &= \frac{\binom{K}{2}\binom{N-2}{n-2}}{\binom{N}{n}} \end{align*}$$
Here, the last equation follows because the last sum is the sum of probabilities of a hypergeometric random variables with parameters $N-2, K-2, n-2.$