2014 AMC 10A Problem 17

Question

Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?

\dfrac16

\dfrac{13}{72}

\dfrac7{36}

\dfrac5{24}

\dfrac29

List possible ordered dice pairs by their sum

Remember to count ordered triples of dice rolls

Note that if one die is the sum of the other two dice, then it is strictly greater than the other two dice.

There are

3

ways to choose which of the dice is the sum of the other two, which makes it the greatest.

This die cannot be

1,

since there is no way to sum two positive integers to get

1.

There is a

\dfrac{1}{6}

chance that this die is any of the other numbers.

There is

1

way to get a sum of

2,

2

ways for

3,

3

for

4,

4

for

5,

and

5

for

6.

We take these numbers of ways out of a total of

6^2 = 36

possibilities. The desired probability is then

3 \cdot \dfrac{1}{6} \cdot \dfrac{1 + 2 + 3 + 4 + 5}{36} = \dfrac{5}{24}.

Thus, D is the correct answer.

Problem 17 in other years