Competition · AMC preparation · step 4 of 4
AMC 8 · 2019 · #18
Grade 7 probabilitycountingPick an answer.
AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Sort the faces by parity
Sort the labels into even {2, 8} and odd {1, 3, 5, 7}: each die has 2 even faces and 4 odd faces.
Telling whether a whole number is even or odd is a Grade 2 skill — just look at the ones digit.
2.OA.C.3Make A Systematic ListRecall the parity rule
Parity rule: a sum is even only when both are even or both are odd — split into those two sub-cases.
The even-plus-odd parity rules are still Grade 2 even-and-odd reasoning, just applied to a sum.
The two top faces add to an even number exactly when both dice show the same parity — both even or both odd.
▸ Why?
An even total shows up only in the matched-parity cases: even-plus-even and odd-plus-odd both land on an even number, while even-plus-odd lands on an odd number.
▸ Why?
Each even face already splits into two equal whole groups, so joining two even faces gives a bigger pile that still splits into two equal whole groups with nothing left over — the sum is even.
▸ Why?
Each odd face splits into two equal groups with one lonely unit left over, and the two lonely units pair up, so the combined pile again splits into two equal whole groups with nothing left over — the sum is even.
▸ Why?
The even face splits into two equal groups with nothing left over while the odd face keeps one lonely unit, so the combined pile still has that single unit that cannot split into two equal groups — the sum is odd.
Count all outcomes
Each die's 6 faces pair with the other die's 6, so the sample space has 6 × 6 = 36 equally likely outcomes.
Listing all pairs from two dice as an organized 6 × 6 grid is the Grade 7 "compound events with organized lists" idea.
7.SP.C.8Make A Systematic ListCount the even-even rolls
Both even: 2 even faces on each die give 2 × 2 = 4 even-even pairs.
Multiplying 2 × 2 to count pairs is Grade 3 multiplication within 100.
3.OA.C.7Identify SubproblemsCount the odd-odd rolls
Both odd: 4 × 4 = 16 odd-odd pairs; adding the two cases gives 4 + 16 = 20 favorable outcomes.
Multiplying 4 × 4 and then adding the two sub-case counts stays within Grade 3 arithmetic.
3.OA.C.7Identify SubproblemsForm and simplify the probability
Divide favorable by total and simplify: = , which matches choice (C).
Writing the probability as (favorable outcomes) / (total outcomes) is the Grade 7 probability-model definition.
7.SP.C.7Make A Systematic ListThis AMC 8 problem only needs Grade 7 probability — favorable outcomes divided by total outcomes — you already know!
- Sort the faces by parity
- Recall the parity rule
- Count all outcomes
- Count the even-even rolls
- Count the odd-odd rolls
- Form and simplify the probability
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