2008 AMC 10B Problem 20

P16-20~4 minPrint

Basic Probability Count the complement and subtract from the total Use an invariant, parity, or coloring argument

The faces of a cubical die are marked with the numbers $1$ , $2$ , $2$ , $3$ , $3$ , and $4$ . The faces of another die are marked with the numbers $1$ , $3$ , $4$ , $5$ , $6$ , and $8$ . Both dice are thrown. What is the probability that the sum of the top two numbers will be $5$ , $7$ , or $9$ ?

Show solution without answering

Key insight

Each die has six faces, so there are 36 equally likely outcomes. Count the face pairs for sums 5, 7 and 9, counting a repeated number once for each face.

Solution

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

Why this works

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.

Alternative approach

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.3)
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques