2008 AMC 10B Problem 20

Each die has $6$ faces, so there are $6 \times 6 = 36$ equally likely ways the pair can land. A number printed on two faces counts twice, because it is twice as likely to come up. So count face pairs.

1. First die faces: $1, 2, 2, 3, 3, 4$. Second die faces: $1, 3, 4, 5, 6, 8$.
2. Sum $5$. Go through the first die and ask what the second die needs.
- $1$ needs $4$: yes. ($1$ way)
- $2$ needs $3$: yes, and there are two faces marked $2$. ($2$ ways)
- $3$ needs $2$: no.
- $4$ needs $1$: yes. ($1$ way)

Total: $4$.
3. Sum $7$.
- $1$ needs $6$: yes. ($1$)
- $2$ needs $5$: yes, two faces. ($2$)
- $3$ needs $4$: yes, two faces. ($2$)
- $4$ needs $3$: yes. ($1$)

Total: $6$.
4. Sum $9$.
- $1$ needs $8$: yes. ($1$)
- $2$ needs $7$: no.
- $3$ needs $6$: yes, two faces. ($2$)
- $4$ needs $5$: yes. ($1$)

Total: $4$.
5. Altogether $4 + 6 + 4 = 14$ of the $36$ outcomes work, so the probability is
$$\frac{14}{36} = \frac{7}{18}.$$

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

Probability counts equally likely outcomes, and for dice those are the faces, not the distinct numbers. A number printed twice is twice as likely. Walking through the first die's faces one by one and asking "what does the other die need?" is a tidy way to make sure nothing is missed.

Use odd and even. The targets $5, 7, 9$ are all odd. Each die has three odd faces and three even faces, and a sum is odd exactly when one die is odd and the other even: $3 \times 3 + 3 \times 3 = 18$ outcomes. The smallest possible sum is $2$ and the largest is $12$, so the only other odd sums are $3$ and $11$. Sum $3$ happens only as $2 + 1$ ($2$ ways, two faces marked $2$), and sum $11$ only as $3 + 8$ ($2$ ways). That leaves $18 - 2 - 2 = 14$ outcomes, the same as before.

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