2005 AMC 10B Problem 9

Whether a sum is odd depends only on whether each number is odd or even: odd + even is odd, but odd + odd and even + even are both even.

1. First die ($1, 1, 2, 2, 3, 3$): four odd faces ($1, 1, 3, 3$) and two even faces. So it is odd with probability $\frac{4}{6} = \frac{2}{3}$ and even with probability $\frac{1}{3}$.
2. Second die ($4, 4, 5, 5, 6, 6$): two odd faces ($5, 5$) and four even faces. So it is odd with probability $\frac{1}{3}$ and even with probability $\frac{2}{3}$.
3. Case 1: first odd, second even. Probability $\frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9}$.
4. Case 2: first even, second odd. Probability $\frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}$.
5. These two cases cannot both happen, so add them: $\frac{4}{9} + \frac{1}{9} = \frac{5}{9}$.

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

For "is the sum odd or even?" questions, replace each number by just "odd" or "even." That turns each die into a simple two-outcome coin, but a lopsided one, so count the faces carefully. Unusual dice like these are made on purpose so that the guess $\frac{1}{2}$ is wrong.

Find the chance the sum is even, then subtract from $1$. Even happens when both dice are odd ($\frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9}$) or both are even ($\frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}$). That is $\frac{4}{9}$, so an odd sum has probability $1 - \frac{4}{9} = \frac{5}{9}$.

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