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> | ||
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==Solution 1== | ==Solution 1== | ||
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~JH. L | ~JH. L | ||
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| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/5UojVH4Cqqs?t=702 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
Latest revision as of 18:34, 11 September 2024
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?
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 (Bashy)
have 
ways to rearrange them for a total of 
ways. 
have 
ways to rearrange them for a total of 
ways. Adding them up, we get 
ways. We have to divide this values by 
because there are 3 dice. 
.
~MathFun1000
Solution 3 (Summary of Solution 2)
Start by listing all possible sums. Then find the ways to arrange them and sum them up and divide by for 3 dice. 
is 
.
-aopspandy
Solution 4 (enumeration and casework)
There are ways to pick the dice which is the sum. From here proceed with "casework":
- If the sum is 1, there are
ways because the minimum sum is 
. - If the sum is 2, there is
way (namely 
). - If the sum is 3, there are ways (namely
and 
). - If the sum is 4, there are
ways (namely 
, 
, and 
).
ways because the minimum sum is 
.
way (namely 
).
and 
).
, 
, and 
).We start noticing that for a sum of 
there are 
cases, but let's keep going to verify.
- If the sum is 5, there are
ways (namely 
, 
, 
, and 
). - If the sum is 6, there are
ways (namely 
, 
, 
, 
, and 
).
ways (namely 
, 
, 
, and 
).
, 
, 
, 
, and 
).We have exhausted all cases because there is no way a dice can roll a 
. Summing, there are 15 ways here.
To summarize: there are ways to pick the dice that is the sum, and 
ways to achieve the sum, for a total of configurations. There are 
total configurations, and the probability is 
.
~JH. L
Video Solution by OmegaLearn
https://youtu.be/5UojVH4Cqqs?t=702
~ pi_is_3.14
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. 
