Ms. Carr asks her students to read any $5$ of the $10$ books on a reading list. Harold randomly selects $5$ books from this list, and Betty does the same. What is the probability that there are exactly $2$ books that they both select?
$\textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{14}{45} \qquad\textbf{(D)}\ \frac{25}{63} \qquad\textbf{(E)}\ \frac{1}{2}$
We don't care about which books Harold selects. We just care that Betty picks $2$ books from Harold's list and $3$ that aren't on Harold's list.
The total amount of combinations of books that Betty can select is $\binom{10}{5}=252$. 
There are $\binom{5}{2}=10$ ways for Betty to choose $2$ of the books that are on Harold's list.
From the remaining $5$ books that aren't on Harold's list, there are $\binom{5}{3}=10$ ways to choose $3$ of them.
$\frac{10\cdot10}{252}=\boxed{\textbf{(D) }\frac{25}{63}}$ ~quacker88