2021 AMC Fall 10A Problem 9

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

The actual face values are irrelevant: the question is about parity, so collapse the six faces into two outcomes, "even" and "odd," with probabilities $\tfrac34$ and $\tfrac14$. After that it is a standard two-independent-trials computation.

Complement: the sum is odd exactly when the parities differ, which happens with probability $2 \cdot \tfrac34 \cdot \tfrac14 = \tfrac{3}{8}$. So the sum is even with probability $1 - \tfrac38 = \tfrac58$. Note that $\tfrac38$ is choice (A), placed there for anyone who stops at the complement.

• Reading 'even is 3 times as likely as odd' as P(even) = 3/4 per face or as P(even) = 3/5, rather than 3/4 for the whole even class.
• Computing the probability that the sum is odd ($\tfrac38$) and forgetting to subtract from $1$.