2008 AMC 10B Problem 16

P16-20~3 minPrint

Basic Probability Conditional Probability & States Split into exhaustive cases and handle each

Two fair coins are to be tossed once. For each head that results, one fair die is to be rolled. What is the probability that the sum of the die rolls is odd? (Note that if no die is rolled, the sum is $0$ .)

Show solution without answering

Key insight

Once at least one die is rolled, the sum is odd half the time. Only two tails is different: no dice, and 0 is even.

Solution

What happens depends on how many heads come up, so split into cases and add.

1. The coins give $0$ , $1$ or $2$ heads. Out of the four equally likely results HH, HT, TH, TT:
- $0$ heads (TT): probability $\tfrac14$ .
- $1$ head (HT or TH): probability $\tfrac24 = \tfrac12$ .
- $2$ heads (HH): probability $\tfrac14$ .
2. $0$ heads. No die is rolled and the sum is $0$ , which is even. The chance of an odd sum here is $0$ .
3. $1$ head. One die is rolled. Three of its six faces ( $1, 3, 5$ ) are odd, so the chance is $\tfrac12$ . This case adds $\tfrac12 \times \tfrac12 = \tfrac14$ .
4. $2$ heads. Two dice are rolled. The sum is odd when one die is odd and the other is even. Whatever the first die shows, the second die has a $\tfrac12$ chance of being the opposite type. So the chance is $\tfrac12$ , and this case adds $\tfrac14 \times \tfrac12 = \tfrac18$ .
5. Add the cases:
$0 + \frac14 + \frac18 = \frac38.$

The answer is $\boxed{\textbf{(A)}\ \frac{3}{8}}$ .

Why this works

When an experiment has two stages, split by what happens in the first stage, find the chance in each case, and weight it by how likely that case is. The handy fact here: as soon as at least one die is rolled, the last die rolled decides odd or even with a fifty-fifty chance. So the answer is just (chance of rolling at least one die) $\times \tfrac12 = \tfrac34 \times \tfrac12 = \tfrac38$ .

Concise solution

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

Split by the number of heads.

Zero heads (probability $\tfrac14$ ): no die is rolled, the sum is $0$ , which is even. Contributes $0$ .
One head (probability $\tfrac12$ ): one die, odd with probability $\tfrac12$ . Contributes $\tfrac12 \cdot \tfrac12 = \tfrac14$ .
Two heads (probability $\tfrac14$ ): two dice. The sum is odd exactly when the dice have different parities, which happens with probability $\tfrac12$ (whatever the first die shows, the second matches its parity half the time). Contributes $\tfrac14 \cdot \tfrac12 = \tfrac18$ .

Total:
$0 + \frac14 + \frac18 = \frac38.$

The answer is $\boxed{\textbf{(A)}\ \frac{3}{8}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques (§8.5)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Intermediate Counting & Probability, Chapter 11: Conditional Probability