2021 AMC Fall 10A Problem 9

Only whether each roll is even or odd matters, so think of each roll as just "even" or "odd".

1. Chances for one roll. Say the chance of odd is some amount $q$. Even is $3$ times as likely, so its chance is $3q$. These two cover everything, so $q + 3q = 1$, which gives $q = \frac14$.
- Chance of even: $\frac34$.
- Chance of odd: $\frac14$.
2. When is the sum even? Look at examples: $2 + 4 = 6$ (even + even is even), $3 + 5 = 8$ (odd + odd is even), $2 + 3 = 5$ (even + odd is odd). So the sum is even exactly when both rolls are even or both are odd.
3. Both even. The rolls do not affect each other, so multiply: $\frac34 \times \frac34 = \frac{9}{16}$.
4. Both odd. $\frac14 \times \frac14 = \frac{1}{16}$.
5. Add the two cases. $\frac{9}{16} + \frac{1}{16} = \frac{10}{16} = \frac58$.

The answer is $\boxed{\textbf{(E)}\ \frac{5}{8}}$.

The actual numbers on the die do not matter, only even or odd. Once you shrink six faces down to two outcomes with chances $\frac34$ and $\frac14$, the problem is two independent rolls: multiply for "and", add for separate cases.

Find the chance the sum is odd, then subtract from $1$. The sum is odd when one roll is even and the other odd. That can happen as even-then-odd, $\frac34 \times \frac14 = \frac{3}{16}$, or odd-then-even, also $\frac{3}{16}$. Together that is $\frac{6}{16} = \frac38$. So the sum is even with chance $1 - \frac38 = \frac58$. Be careful: $\frac38$ is choice (A), waiting for anyone who stops one step early.

Only parity matters. On one roll, let the probability of an odd number be $q$; then the probability of an even number is $3q$, and $q + 3q = 1$ gives $q = \tfrac14$. So each roll is even with probability $\tfrac34$ and odd with probability $\tfrac14$.

The sum of two rolls is even exactly when the rolls have the same parity. The two rolls are independent, so
$$P(\text{both even}) + P(\text{both odd}) = \left(\tfrac34\right)^2 + \left(\tfrac14\right)^2 = \tfrac{9}{16} + \tfrac{1}{16} = \tfrac{10}{16} = \tfrac{5}{8}.$$

The answer is $\boxed{\textbf{(E)}\ \frac{5}{8}}$.