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 1 | + | ==Solution 1== |
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>. | 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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==Solution 2 (Casework)== | ==Solution 2 (Casework)== | ||
Revision as of 13:37, 26 January 2015
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
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 
.
Solution 2 (Casework)
Since there are 
possible values for the number on each dice, there are 
total possible rolls.
The possible results of the 3 dice such that the sum of the values of two of the die is equal to the value of the third die are, without considering the order of the die, ![\[(1, 1, 2), (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 2, 4), (2, 3, 5), (2, 4, 6), (3, 3, 6)\]](http://latex.artofproblemsolving.com/f/3/c/f3cb5b1f6c5b3588c5082dae79d359bbd0a6862e.png)
There are 
ways to order the first, sixth, and ninth results, and there are 
ways to order the other results.
Therefore, there are a total of 
ways to roll the dice such that 2 of the dice sum to the other die, so our answer is ![\[\frac{45}{216}=\boxed{\textbf{(D)} \ \frac{5}{24}}\]](http://latex.artofproblemsolving.com/c/7/7/c771d6d8622c084a83d0137e49856819297384e4.png)
(Solution by bestwillcui1)
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. 
