Difference between revisions of "2014 AMC 10A Problems/Problem 17"
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<math> \textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29 </math> | <math> \textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29 </math> | ||
| − | ==Solution== | + | ==Solution 1 (Clean Counting)== |
| + | |||
| + | First, we note that there are <math>1, 2, 3, 4,</math> and <math>5</math> ways to get sums of <math>2, 3, 4, 5, 6</math> respectively--this is not too hard to see. With any specific sum, there is exactly one way to attain it on the other die. This means that the probability that two specific dice have the same sum as the other is <cmath>\dfrac16 \left( \dfrac{1+2+3+4+5}{36}\right) = \dfrac{5}{72}.</cmath> Since there are <math>\dbinom31</math> ways to choose which die will be the one with the sum of the other two, our answer is <math>3 \cdot \dfrac{5}{72} = \boxed{\textbf{(D)} \: \dfrac{5}{24}}</math>. | ||
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| + | --happiface | ||
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| + | ==Solution 2 (Bashy Casework)== | ||
Since there are <math>6</math> possible values for the number on each dice, there are <math>6^3=216</math> total possible rolls. | Since there are <math>6</math> possible values for the number on each dice, there are <math>6^3=216</math> total possible rolls. | ||
Revision as of 23:11, 7 February 2014
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?
Solution 1 (Clean Counting)
First, we note that there are 
and 
ways to get sums of 
respectively--this is not too hard to see. With any specific sum, there is exactly one way to attain it on the other die. This means that the probability that two specific dice have the same sum as the other is ![\[\dfrac16 \left( \dfrac{1+2+3+4+5}{36}\right) = \dfrac{5}{72}.\]](http://latex.artofproblemsolving.com/0/2/3/0232e2cd81c4f2dc3f7c46c1875a3ee3578cbe7e.png)
Since there are 
ways to choose which die will be the one with the sum of the other two, our answer is 
.
--happiface
Solution 2 (Bashy Casework)
Since there are 
possible values for the number on each dice, there are 
total possible rolls.
Note that the possible results of the 3 dice (without respect to order) are 
.
There are 
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 
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
