2021 AMC Fall 10A Problem 9

P1-10~2 minPrint

Basic Probability Set up the equation/formula and compute; no special trick needed

When a certain unfair die is rolled, an even number is $3$ times as likely to appear as an odd number. The die is rolled twice. What is the probability that the sum of the numbers rolled is even?

Show solution without answering

Key insight

Each roll is even with probability 3/4 and odd with probability 1/4. The sum is even when both rolls match, so the chance is 9/16 + 1/16 = 5/8.

Solution

Only whether each roll is even or odd matters, so think of each roll as just "even" or "odd".

1. Chances for one roll. Say the chance of odd is some amount $q$ . Even is $3$ times as likely, so its chance is $3q$ . These two cover everything, so $q + 3q = 1$ , which gives $q = \frac14$ .
- Chance of even: $\frac34$ .
- Chance of odd: $\frac14$ .
2. When is the sum even? Look at examples: $2 + 4 = 6$ (even + even is even), $3 + 5 = 8$ (odd + odd is even), $2 + 3 = 5$ (even + odd is odd). So the sum is even exactly when both rolls are even or both are odd.
3. Both even. The rolls do not affect each other, so multiply: $\frac34 \times \frac34 = \frac{9}{16}$ .
4. Both odd. $\frac14 \times \frac14 = \frac{1}{16}$ .
5. Add the two cases. $\frac{9}{16} + \frac{1}{16} = \frac{10}{16} = \frac58$ .

The answer is $\boxed{\textbf{(E)}\ \frac{5}{8}}$ .

Why this works

The actual numbers on the die do not matter, only even or odd. Once you shrink six faces down to two outcomes with chances $\frac34$ and $\frac14$ , the problem is two independent rolls: multiply for "and", add for separate cases.

Alternative approach

Find the chance the sum is odd, then subtract from $1$ . The sum is odd when one roll is even and the other odd. That can happen as even-then-odd, $\frac34 \times \frac14 = \frac{3}{16}$ , or odd-then-even, also $\frac{3}{16}$ . Together that is $\frac{6}{16} = \frac38$ . So the sum is even with chance $1 - \frac38 = \frac58$ . Be careful: $\frac38$ is choice (A), waiting for anyone who stops one step early.

Concise solution

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

Only parity matters. On one roll, let the probability of an odd number be $q$ ; then the probability of an even number is $3q$ , and $q + 3q = 1$ gives $q = \tfrac14$ . So each roll is even with probability $\tfrac34$ and odd with probability $\tfrac14$ .

The sum of two rolls is even exactly when the rolls have the same parity. The two rolls are independent, so
$P(\text{both even}) + P(\text{both odd}) = \left(\tfrac34\right)^2 + \left(\tfrac14\right)^2 = \tfrac{9}{16} + \tfrac{1}{16} = \tfrac{10}{16} = \tfrac{5}{8}.$

The answer is $\boxed{\textbf{(E)}\ \frac{5}{8}}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 7: Introduction to Probability
Introduction to Counting & Probability, Chapter 8: Basic Probability Techniques