2020 AMC 10B Problem 11

1. It does not matter which books Harold picks; every set of five looks the same from Betty's point of view. So imagine Harold's five books are already chosen. The list is now split into "Harold's $5$" and "the other $5$".
2. All of Betty's choices. She picks $5$ books from $10$:
$$\binom{10}{5} = 252 \text{ equally likely choices}.$$
3. Good choices. Exactly $2$ books in common means Betty takes $2$ of Harold's books and her other $3$ from the books he did not pick:
- $2$ of Harold's $5$: $\binom{5}{2} = 10$ ways,
- $3$ of the other $5$: $\binom{5}{3} = 10$ ways.

Together: $10 \times 10 = 100$ good choices.
4. Probability.
$$\frac{100}{252} = \frac{25}{63}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{25}{63}}$.

When two people choose at random and on their own, you can fix the first person's choice and count only the second person's options, since every first choice gives the same answer. Then "exactly $2$ in common" becomes a simple product: pick some from one group and the rest from the other.

Whatever Harold picks, the list splits into his $5$ books and the other $5$. Betty's selection is a uniformly random $5$-subset, $\binom{10}{5} = 252$ possibilities in all.

For exactly $2$ books in common, Betty must take $2$ of Harold's $5$ and $3$ of the remaining $5$:
$$\binom{5}{2}\binom{5}{3} = 10 \cdot 10 = 100.$$

The probability is $\dfrac{100}{252} = \dfrac{25}{63}$.

The answer is $\boxed{\textbf{(D)}\ \frac{25}{63}}$.