AMC 10 · 2012 · #9

Grade 7 probability
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
probability-basicsystematic-enumeration systematic-enumeration ↑ Prerequisites: probability-basic
📏 Medium solution 💡 2 insights
Problem
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.

Pick an answer.

(A)
$\frac{1}{6}$
(B)
$\frac{1}{5}$
(C)
$\frac{1}{4}$
(D)
$\frac{1}{3}$
(E)
$\frac{1}{2}$

AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

Each die has 6 physical faces, so the two rolls give 6 × 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.

1STEP 1

Count all equally likely outcomes

Every face is equally likely even where a number repeats, so the two dice give 6 × 6 = 36 equally likely pairings.

total outcomes = 6 × 6 = 36
2STEP 2

Find value-pairs that make 7

Each even value has exactly one odd partner reaching 7 — 2+5, 4+3, 6+1 — and all are on the odd die, so 3 value-pairs work.

2+5=7, 4+3=7, 6+1=7
3STEP 3

Count faces for each value-pair

Each value sits on two faces, so every winning pair gives 2 × 2 = 4 face-pairings: 4 + 4 + 4 = 12 in all.

2 × 2 = 4 each; 4+4+4 = 12
4STEP 4

Form the probability

Divide winners by total: 12/36, which reduces to 1/3 — choice (D).

P(sum=7) = 12/36 = 1/3 → (D)
Answer
1/3
The value 1/3≈ 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 1/6 and below 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 2/6=1/3 of the odd die's faces. Choices like 1/6 or 1/4 would come from forgetting that each value appears on two faces.
💡Key takeaway

Count the 36 equally likely face-pairings, find the 12 that add to 7, and divide to get 1/3.

  • Count all equally likely outcomes
  • Find value-pairs that make 7
  • Count faces for each value-pair
  • Form the probability

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