2008 AMC 10B Problem 20

There are $6 \cdot 6 = 36$ equally likely (face, face) outcomes; repeated numbers are separate faces.

All three target sums are odd, so start with parity. The first die has three odd faces ($1, 3, 3$) and three even faces ($2, 2, 4$); the second has three odd ($1, 3, 5$) and three even ($4, 6, 8$). The sum is odd when the parities differ: $3 \cdot 3 + 3 \cdot 3 = 18$ outcomes.

The odd sums range from $1 + 1 = 2$ up to $4 + 8 = 12$, so the only odd sums other than $5, 7, 9$ are $3$ and $11$. Sum $3$ occurs only as $2 + 1$, and there are two faces marked $2$: $2$ outcomes. Sum $11$ occurs only as $3 + 8$, with two faces marked $3$: $2$ outcomes.

Favorable outcomes: $18 - 2 - 2 = 14$, so the probability is $\tfrac{14}{36} = \tfrac{7}{18}$.

The answer is $\boxed{\textbf{(B)}\ 7/18}$.

When the target set is "all odd sums except a few," parity gives the bulk count instantly and the exceptions are found at the extremes. Repeated faces are the real trap: probability must be computed over faces, not over distinct numbers, so a $2$ on the first die is twice as likely as a $1$.

Direct count: sum $5$ from $(1,4), (2,3)\times 2, (4,1)$ gives $4$; sum $7$ from $(1,6), (2,5)\times2, (3,4)\times2, (4,3)$ gives $6$; sum $9$ from $(1,8), (3,6)\times2, (4,5)$ gives $4$. Total $14$ of $36$.

• Treating the repeated faces (two 2s, two 3s) as single outcomes, or forgetting to exclude the odd sums 3 and 11.
• Stopping at "odd sum has probability 1/2" without noticing $\tfrac12$ is not even offered as a choice.