2020 AMC 10B Problem 11

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.