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?

$\dfrac16$

$\dfrac{13}{72}$

$\dfrac7{36}$

$\dfrac5{24}$

$\dfrac29$

Answer: D

Concepts:dice (probability)casework

Difficulty rating: 1540

Solution:

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.

2014 AMC 10A Problem 17

Answer: D

Difficulty rating: 1540

Solution:

Problem 17 in Other Years