The question is:
An urn contains $20$ chips, $10$ marked $H$ and $10$ marked $C$. Randomly select chips, one at a time, without replacement. After $10$ chips have been selected, let $X$ be the half of the difference between the numbers of $H$ and $C$ chips that have been selected. For example, if there $7$ $H$-chips and $3$ $C$-chips, then $X = \frac{7-3}2 = 2$. If there are $2$ $H$-chips and $8$ $C$-chaps, then $X = \frac{2 - 8}2 = -3$. Find the probability of $X = 2$.
The given answer is: $.0779$
I'm not sure where to go with this one. For $X = 2, H-C = 4$, and $H + C = 10$. So, $H$ is $7$ and $C$ is $3$.
From there I'm not sure what to do. Any help?
$$H+C=10$$ $$H-C=4$$
Hence$(H,C)=(7,3)$.
That is out of $10$ $H$'s, pick $7$ of them and out of $10$ $C$'s, pick $3$ of them. Without restriction it is pick any $10$ chips from $20$ chips.
Hence the answer is $$\frac{\binom{10}{7}\binom{10}{3}}{\binom{20}{10}}= \frac{\binom{10}{3}^2}{\binom{20}{10}}$$