2014 AMC 10A Problems/Problem 17

Revision as of 18:45, 7 February 2014 by Minimario (talk | contribs) (Solution)

Problem

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?

$\textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29$

Solution

Since there are $6$ possible values for the number on each dice, there are $6^3=216$ total possible rolls.

Note that the possible results of the 3 dice (without respect to order) are $(1, 1, 2), (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 2, 4), (2, 3, 5), (2, 4, 6) (3, 3, 6)$.

There are $3$ ways to order the first, sixth, and ninth results, while there are 6 ways to order the other results; therefore, there are a total of 45 ways to roll the dice s.t. 2 of the dice sum to the other, so our answer is $\frac{45}{216}=\boxed{\textbf{(D) \frac{5}{24}}}$ (Error compiling LaTeX. ! Missing $ inserted.)

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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