AMC 8 · 2019 · #18

Grade 7 probabilitycounting

probability-basicparitycombinations-basic caseworksystematic-enumeration ↑ Prerequisites: probability-basicparity

📏 Medium solution 💡 2 insights

Problem

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?

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

Pick an answer.

(A)

$\frac{4}{9}$

(B)

$\frac{1}{2}$

(C)

$\frac{5}{9}$

(D)

$\frac{3}{5}$

(E)

$\frac{2}{3}$

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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

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 Make a Systematic List 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.

$$\text{even} = \{2, 8\}\;(2\text{ faces}),\quad \text{odd} = \{1, 3, 5, 7\}\;(4\text{ faces})$$

💡 Telling whether a whole number is even or odd is a Grade 2 skill — just look at the ones digit.

#7 Identify Subproblems 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.

$$\text{Even}+\text{Even}=\text{Even},\quad \text{Odd}+\text{Odd}=\text{Even},\quad \text{Even}+\text{Odd}=\text{Odd}$$

💡 The even-plus-odd parity rules are still Grade 2 even-and-odd reasoning, just applied to a sum.

#2 Make a Systematic List 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.

$$6 \times 6 = 36 \text{ total outcomes}$$

💡 Listing all pairs from two dice as an organized $6 \times 6$ grid is the Grade 7 "compound events with organized lists" idea.

#7 Identify Subproblems 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.

$$2 \times 2 = 4 \text{ even-even pairs}$$

💡 Multiplying $2 \times 2$ to count pairs is Grade 3 multiplication within $100$.

#7 Identify Subproblems 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: $4 + 16 = 20$.

$$4 \times 4 = 16, \quad 4 + 16 = 20 \text{ favorable outcomes}$$

💡 Multiplying $4 \times 4$ and then adding the two sub-case counts stays within Grade 3 arithmetic.

#2 Make a Systematic List 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).

$$P(\text{even sum}) = \dfrac{20}{36} = \dfrac{5 \times 4}{9 \times 4} = \dfrac{5}{9} \;\Rightarrow\; \textbf{(C)}$$

💡 Writing the probability as (favorable outcomes) / (total outcomes) is the Grade 7 probability-model definition.

[1] #2 2.OA.C.3 Sort the six face labels $\{1, 2, 3, 5, 7, 8\}$ into the "even" and "odd" bucket

[2] #7 2.OA.C.3 Recall the parity rule for addition: Even+Even and Odd+Odd both give an even sum

[3] #2 7.SP.C.8 Count the total number of equally likely outcomes when two dice are rolled. Each

[4] #7 3.OA.C.7 Count the "both even" sub-case. Die 1 has $2$ even faces and die 2 has $2$ even

[5] #7 3.OA.C.7 Count the "both odd" sub-case the same way. Die 1 has $4$ odd faces and die 2 ha

[6] #2 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.3 Determine 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.7 Fluently 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.7 Develop 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.8 Find 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 7)

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