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> | ||
| − | == Video Solution | + | == Video Solution == |
https://youtu.be/5UojVH4Cqqs?t=702 | https://youtu.be/5UojVH4Cqqs?t=702 | ||
Revision as of 21:32, 24 January 2021
Contents
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?
Video Solution
https://youtu.be/5UojVH4Cqqs?t=702
~ pi_is_3.14
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 
.
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 | ||
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