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2012 AMC 10A Problem 9

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2012 AMC 10A Problem 9AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 2 of 5 · about 2 min · position P1-10

A pair of six-sided dice are labeled so that one die has only even numbers (two each of 22 , 44 , and 66 ), and the other die has only odd numbers (two each of 11 , 33 , and 55 ). The pair of dice is rolled. What is the probability that the sum of the numbers on the tops of the two dice is 77 ?

  1. A)

    16\frac{1}{6}

  2. B)

    15\frac{1}{5}

  3. C)

    14\frac{1}{4}

  4. D)

    13\frac{1}{3}

  5. E)

    12\frac{1}{2}

Answer

D

Key insight

Each die shows one of three values with equal probability, so there are 9 equally likely value pairs, and exactly three of them (2+5, 4+3, 6+1) sum to 7.

Solution

Because each value appears on two of the six faces, the even die shows 22, 44 or 66 each with probability 13\frac13, and the odd die shows 11, 33 or 55 each with probability 13\frac13. So the 3×3=93 \times 3 = 9 value pairs are equally likely.

List the pairs whose sum is 77:
(2,5),(4,3),(6,1).(2, 5), \quad (4, 3), \quad (6, 1).
Every even value has exactly one odd partner, so there are 33 favorable pairs out of 99.

The probability is 39=13\frac{3}{9} = \frac{1}{3}.

The answer is (D) 13\boxed{\textbf{(D)}\ \frac{1}{3}}.

Why this works

Doubled labels do not change the probabilities of the values, so the dice behave like three-sided dice. Reducing the sample space to equally likely value pairs makes the count tiny. Equivalently: whatever the even die shows, exactly one of the three odd values completes a 77, so the probability is 13\frac13 regardless.

Alternative approach

Use all 3636 face pairs: each favorable value pair corresponds to 2⋅2=42 \cdot 2 = 4 face pairs, so 3⋅4=123 \cdot 4 = 12 favorable out of 3636, again 13\frac13.

The trap

Using 36 outcomes but counting only 3 favorable ones (each favorable value pair actually covers 4 of the 36 face pairs).

Common mistakes

  • Using 36 outcomes but counting only 3 favorable ones (each favorable value pair actually covers 4 of the 36 face pairs).
  • Forgetting the pair (6,1)(6, 1) or including impossible pairs like (1,6)(1, 6) from the wrong dice.

Techniques

Organized listing / direct enumeration