AMC 10 · 2012 · #9
Grade 7 probabilityA pair of six-sided dice are labeled so that one die has only even numbers (two each of 2, 4, and 6), and the other die has only odd numbers (two of each 1, 3, and 5). The pair of dice is rolled. What is the probability that the sum of the numbers on the tops of the two dice is 7?
Pick an answer.
AMC 10 2012 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 special six-sided dice are rolled. One die shows only even numbers, with two faces each of 2, 4, and 6. The other die shows only odd numbers, with two faces each of 1, 3, and 5. Find the probability that the two top numbers add up to 7.
Givens: The even die has 6 faces: 2, 2, 4, 4, 6, 6; The odd die has 6 faces: 1, 1, 3, 3, 5, 5; The two dice are rolled together and we look at the sum of the top faces; Answer choices: (A) 1/6, (B) 1/5, (C) 1/4, (D) 1/3, (E) 1/2
Unknowns: The probability that the sum of the two top numbers equals 7
Understand
Restated: Two special six-sided dice are rolled. One die shows only even numbers, with two faces each of 2, 4, and 6. The other die shows only odd numbers, with two faces each of 1, 3, and 5. Find the probability that the two top numbers add up to 7.
Givens: The even die has 6 faces: 2, 2, 4, 4, 6, 6; The odd die has 6 faces: 1, 1, 3, 3, 5, 5; The two dice are rolled together and we look at the sum of the top faces; Answer choices: (A) 1/6, (B) 1/5, (C) 1/4, (D) 1/3, (E) 1/2
Plan
Primary tool: #2 Make a Systematic List
Secondary: #7 Identify Subproblems, #16 Change Focus / Count the Complement
Each die has 6 physical faces, so the two rolls give $6\times 6 = 36$ equally likely face-pairings, and the probability is (number of pairings that sum to 7)$/36$. Tool #2 (Make a Systematic List) counts those pairings cleanly. To keep the list short, Tool #7 (Identify Subproblems) first finds which value-pairs even+odd can sum to 7 — only $2+5$, $4+3$, $6+1$ — and then counts each value-pair using the doubled faces. Tool #16 (Change Focus) gives a fast independent check: fix the even die first and ask how many of the odd die's faces finish the sum.
Execute — Answer: D
7.SP.C.7 Step 1 Count all equally likely outcomes
- Each die has 6 faces, and even though some numbers are printed twice, the faces themselves are all equally likely to land on top.
- Rolling two dice independently gives $6 \times 6 = 36$ equally likely face-pairings.
- Because every pairing has the same chance, the probability of summing to 7 is just the number of pairings that sum to 7 divided by 36.
- The rest of the work is counting those winners.
💡 When all faces are equally likely, probability is a counting job: good pairings over all 36 pairings.
7.SP.C.8 Step 2 Find value-pairs that make 7
- The first die only shows even values 2, 4, 6 and the second only odd values 1, 3, 5.
- Match each even value with the odd value it needs to reach 7: $2$ needs $5$, $4$ needs $3$, and $6$ needs $1$.
- Every one of these partners is actually on the odd die, so there are exactly three value-pairs that sum to 7: $(2,5)$, $(4,3)$, and $(6,1)$.
💡 Each even number has exactly one odd partner that completes 7, so the winning pairs are easy to list.
7.SP.C.8 Step 3 Count faces for each value-pair
- Now use the doubled faces.
- The value $2$ sits on two faces of the even die and $5$ sits on two faces of the odd die, so the pair $(2,5)$ happens in $2 \times 2 = 4$ face-pairings.
- The same is true for $(4,3)$ and for $(6,1)$: each gives $4$ pairings.
- Adding the three value-pairs, there are $4 + 4 + 4 = 12$ face-pairings that sum to 7.
💡 Two matching faces on each die multiply, so every winning value-pair counts four times.
7.NS.A.3 Step 4 Form the probability
- Divide the 12 winning pairings by the 36 total pairings: $\tfrac{12}{36}$.
- Both numbers divide by 12, so this reduces to $\tfrac{1}{3}$.
- That matches choice (D), so the answer is (D).
💡 Reduce the count of winners over the count of all outcomes to a simple fraction.
7.SP.C.7 Each die has 6 faces, and even though some numbers are printed twice, the faces 7.SP.C.8 The first die only shows even values 2, 4, 6 and the second only odd values 1, 3 7.SP.C.8 Now use the doubled faces. The value $2$ sits on two faces of the even die and $ 7.NS.A.3 Divide the 12 winning pairings by the 36 total pairings: $\tfrac{12}{36}$. Both Review
Reasonableness: The value $\tfrac{1}{3}\approx 0.33$ is sensible: a sum of 7 is one of the more likely totals, so a probability near a third, well above the smallest choice $\tfrac{1}{6}$ and below $\tfrac{1}{2}$, fits. A structural check confirms it: for every one of the 6 faces of the even die, exactly one odd value (hence two odd faces) completes 7, so the winners are a clean $\tfrac{2}{6}=\tfrac{1}{3}$ of the odd die's faces. Choices like $\tfrac{1}{6}$ or $\tfrac{1}{4}$ would come from forgetting that each value appears on two faces.
Alternative: Fix the even die first (Tool #16, Change Focus). Whatever it shows — 2, 4, or 6 — there is exactly one odd number that makes 7, and that number sits on 2 of the odd die's 6 faces. So the odd die completes the sum with probability $\tfrac{2}{6}=\tfrac{1}{3}$, no matter what the even die did, giving $\tfrac{1}{3}$ directly without counting all 36 outcomes.
CCSS standards used (min grade 7)
7.SP.C.7Develop probability models and use them to find probabilities of events (Treating the 36 face-pairings as equally likely so the probability is the count of sum-7 pairings over 36.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Listing the value-pairs that sum to 7 and counting the doubled faces to reach 12 winning pairings.)7.NS.A.3Solve real-world problems involving the four operations with rational numbers (Reducing 12/36 to the final probability 1/3.)
⭐ Count the 36 equally likely face-pairings, find the 12 that add to 7, and divide to get 1/3.
⭐ Count the 36 equally likely face-pairings, find the 12 that add to 7, and divide to get 1/3.
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