2005 AMC 10B Problem 9

Only parity matters: a sum is odd exactly when one die is odd and the other is even.

First die ($1,1,2,2,3,3$): four odd faces, two even, so $P(\text{odd}) = \frac{2}{3}$ and $P(\text{even}) = \frac{1}{3}$.

Second die ($4,4,5,5,6,6$): two odd faces, four even, so $P(\text{odd}) = \frac{1}{3}$ and $P(\text{even}) = \frac{2}{3}$.

The two dice are independent, so
$$P(\text{odd sum}) = P(\text{odd, even}) + P(\text{even, odd}) = \frac{2}{3}\cdot\frac{2}{3} + \frac{1}{3}\cdot\frac{1}{3} = \frac{4}{9} + \frac{1}{9} = \frac{5}{9}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{5}{9}}$.

Questions about whether a sum is odd or even depend only on the parity of each summand, which collapses each die to a two-outcome experiment. Read the actual face lists to get those two probabilities; unusual dice are designed so that the naive $\frac{1}{2}$ fails.

Complement: the sum is even when both dice are odd or both are even, with probability $\frac{2}{3}\cdot\frac{1}{3} + \frac{1}{3}\cdot\frac{2}{3} = \frac{4}{9}$, so the odd-sum probability is $1 - \frac{4}{9} = \frac{5}{9}$.

• Assuming each die is odd half the time and answering 1/2; the faces are not balanced between odd and even.
• Counting only the case "first odd, second even" ($\frac{4}{9}$, choice (B)) and forgetting the reverse case.