AMC 8 · 2019 · #18

Grade 7 probabilitycounting
Archetype Overlap / Complement Counting └ Sub-archetype Parity Partition: Complement via Odd-Even Split
probability-basicparitycombinations-basic caseworksystematic-enumeration ↑ Prerequisites: probability-basicparity
📏 Medium solution 💡 2 insights
Problem
Two fair dice each have the same six face labels — 1, 2, 3, 5, 7, 8 — instead of the usual 1 through 6. Both dice are rolled and the two numbers on top are added. What is the probability that this sum is an even number?

Pick an answer.

(A)
$\frac{4}{9}$
(B)
$\frac{1}{2}$
(C)
$\frac{5}{9}$
(D)
$\frac{3}{5}$
(E)
$\frac{2}{3}$

AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

An even sum happens only in two clean cases: both dice show even numbers, or both dice show odd numbers (Even+Odd is always odd). Tool #2 (Systematic List) is used to organize the 6 × 6 = 36 outcomes by parity — sorting each face into the "even" or "odd" bucket — instead of listing all 36 pairs by hand. Tool #7 (Identify Subproblems) breaks the count into two independent sub-counts (even-even pairs and odd-odd pairs) that we add at the end.

1STEP 1

Sort the labels into even {2, 8} and odd {1, 3, 5, 7}: each die has 2 even faces and 4 odd faces.

even = {2, 8} (2 faces), odd = {1, 3, 5, 7} (4 faces)
2STEP 2

Parity rule: a sum is even only when both are even or both are odd — split into those two sub-cases.

Even+Even=Even, Odd+Odd=Even, Even+Odd=Odd
3STEP 3

Each die's 6 faces pair with the other die's 6, so the sample space has 6 × 6 = 36 equally likely outcomes.

6 × 6 = 36 total outcomes
4STEP 4

Both even: 2 even faces on each die give 2 × 2 = 4 even-even pairs.

2 × 2 = 4 even-even pairs
5STEP 5

Both odd: 4 × 4 = 16 odd-odd pairs; adding the two cases gives 4 + 16 = 20 favorable outcomes.

4 × 4 = 16, 4 + 16 = 20 favorable outcomes
6STEP 6

Divide favorable by total and simplify: 2036\frac{20}{36} = 59\frac{5}{9}, which matches choice (C).

P(even sum) = 2036\frac{20}{36} = (5×4)(9×4)\frac{(5 × 4)}{(9 × 4)} = 59\frac{5}{9} → (C)
Answer
59\frac{5}{9}
On a standard die you have 3 even and 3 odd faces, so the chance of an even sum is exactly 12\frac{1}{2}. Here each die has more odd faces (4) than even (2), so odd-odd pairs are more common than they would be on a standard die — that should push the probability of an even sum above 12\frac{1}{2}. Our answer 59\frac{5}{9} ≈ 0.556 is indeed just above 12\frac{1}{2}, which makes sense. It also matches choice (C) exactly.
💡Key takeaway

This AMC 8 problem only needs Grade 7 probability — favorable outcomes divided by total outcomes — you already know!

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