AMC 8 · 2019 · #18
Grade 7 probabilitycountingProblem
The faces of each of two fair dice are numbered 1, 2, 3, 5, 7, and 8. When the two dice are tossed, what is the probability that their sum will be an even number?
Pick an answer.
AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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?
Givens: Each die has faces labeled $\{1, 2, 3, 5, 7, 8\}$; Both dice are fair, so every face is equally likely; Both dice are rolled once and the two top faces are added; Answer choices: (A) $\tfrac{4}{9}$, (B) $\tfrac{1}{2}$, (C) $\tfrac{5}{9}$, (D) $\tfrac{3}{5}$, (E) $\tfrac{2}{3}$
Unknowns: The probability that the sum of the two top faces is even
Understand
Restated: 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?
Givens: Each die has faces labeled $\{1, 2, 3, 5, 7, 8\}$; Both dice are fair, so every face is equally likely; Both dice are rolled once and the two top faces are added; Answer choices: (A) $\tfrac{4}{9}$, (B) $\tfrac{1}{2}$, (C) $\tfrac{5}{9}$, (D) $\tfrac{3}{5}$, (E) $\tfrac{2}{3}$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems
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 $6 \times 6 = 36$ 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.
Execute — Answer: C
2.OA.C.3 Step 1 - Sort the six face labels $\{1, 2, 3, 5, 7, 8\}$ into the "even" and "odd" buckets.
- The even bucket holds $\{2, 8\}$ and the odd bucket holds $\{1, 3, 5, 7\}$, so 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.3 Step 2 - Recall the parity rule for addition: Even+Even and Odd+Odd both give an even sum, while Even+Odd gives an odd sum.
- So the favorable outcomes split into exactly two sub-cases — "both even" or "both odd" — which is the Tool #7 (Subproblems) move.
💡 The even-plus-odd parity rules are still Grade 2 even-and-odd reasoning, just applied to a sum.
7.SP.C.8 Step 3 - Count the total number of equally likely outcomes when two dice are rolled.
- Each of the $6$ faces on die 1 can pair with each of the $6$ faces on die 2, so there are $6 \times 6 = 36$ ordered outcomes — the full sample space.
💡 Listing all pairs from two dice as an organized $6 \times 6$ grid is the Grade 7 "compound events with organized lists" idea.
3.OA.C.7 Step 4 - Count the "both even" sub-case.
- Die 1 has $2$ even faces and die 2 has $2$ even faces, so by the multiplication principle there are $2 \times 2 = 4$ even-even pairs.
💡 Multiplying $2 \times 2$ to count pairs is Grade 3 multiplication within $100$.
3.OA.C.7 Step 5 - Count the "both odd" sub-case the same way.
- Die 1 has $4$ odd faces and die 2 has $4$ odd faces, giving $4 \times 4 = 16$ odd-odd pairs.
- Add the two sub-cases for the total favorable count: $4 + 16 = 20$.
💡 Multiplying $4 \times 4$ and then adding the two sub-case counts stays within Grade 3 arithmetic.
7.SP.C.7 Step 6 - Form the probability as favorable / total and simplify.
- $\tfrac{20}{36}$ shares a common factor of $4$ in both numerator and denominator, so $\tfrac{20}{36} = \tfrac{5}{9}$, which matches choice (C).
💡 Writing the probability as (favorable outcomes) / (total outcomes) is the Grade 7 probability-model definition.
2.OA.C.3 Sort the six face labels $\{1, 2, 3, 5, 7, 8\}$ into the "even" and "odd" bucket 2.OA.C.3 Recall the parity rule for addition: Even+Even and Odd+Odd both give an even sum 7.SP.C.8 Count the total number of equally likely outcomes when two dice are rolled. Each 3.OA.C.7 Count the "both even" sub-case. Die 1 has $2$ even faces and die 2 has $2$ even 3.OA.C.7 Count the "both odd" sub-case the same way. Die 1 has $4$ odd faces and die 2 ha 7.SP.C.7 Form the probability as favorable / total and simplify. $\tfrac{20}{36}$ shares Review
Reasonableness: On a standard die you have $3$ even and $3$ odd faces, so the chance of an even sum is exactly $\tfrac{1}{2}$. 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 $\tfrac{1}{2}$. Our answer $\tfrac{5}{9} \approx 0.556$ is indeed just above $\tfrac{1}{2}$, which makes sense. It also matches choice (C) exactly.
Alternative: Use Tool #16 (Change Focus / Complement). An even sum means "both same parity", so the complement is "mixed parity" (one even, one odd). Mixed-parity pairs $= 2 \times 4 + 4 \times 2 = 16$, so $P(\text{odd sum}) = \tfrac{16}{36} = \tfrac{4}{9}$, and therefore $P(\text{even sum}) = 1 - \tfrac{4}{9} = \tfrac{5}{9}$. Same answer (C), reached by counting the opposite event.
CCSS standards used (min grade 7)
2.OA.C.3Determine whether a group of objects has an odd or even number (Sorting the six face labels into even faces $\{2, 8\}$ and odd faces $\{1, 3, 5, 7\}$, and using the Even/Odd addition rules to see that an even sum needs matching parity.)3.OA.C.7Fluently multiply and divide within 100 (Computing the sub-case counts $2 \times 2 = 4$ (both even) and $4 \times 4 = 16$ (both odd), and adding them to $20$ favorable outcomes.)7.SP.C.7Develop probability models and use them to find probabilities of events (Defining the probability as (favorable outcomes) / (total outcomes) $= \tfrac{20}{36} = \tfrac{5}{9}$ for the equally likely sample space.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Treating the two dice as a compound event whose $6 \times 6 = 36$ outcomes form an organized grid that we partition by parity.)
⭐ This AMC 8 problem only needs Grade 7 probability — favorable outcomes divided by total outcomes — you already know!
⭐ This AMC 8 problem only needs Grade 7 probability — favorable outcomes divided by total outcomes — you already know!
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