2020 AMC 10B Problem 11

Fix Harold’s five books and count Betty’s choices

Betty must choose exactly two from Harold’s books and three from the other five

Assume that Harold has already picked his $5$ books. Of these five books, there are $\binom{5}{2}$ ways that Betty can have picked exactly two of the same books as Harold, and $\binom{5}{3}$ ways that Betty can choose her other three books from the $5$ books not on Harold’s list.

Thus there are $\binom{5}{2}\binom{5}{3}=100$ favorable choices for Betty out of $\binom{10}{5}=252$ equally likely choices. The probability is $$\frac{100}{252}=\frac{25}{63}.$$ Thus, D is the correct answer.