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 | + | == Video Solution == |
+ | https://youtu.be/5UojVH4Cqqs?t=702 | ||
− | + | ~ pi_is_3.14 | |
− | + | ==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>. | |
− | |||
− | Since there are <math> | ||
− | + | ==Solution 2 (Bashy)== | |
+ | <math>(1, 2, 3); (1, 3, 4); (1, 4, 5); (1, 5, 6); (2, 3, 5); (2, 4, 6)</math> have <math>6</math> ways to rearrange them for a total of <math>36</math> ways. <math>(1, 1, 2); (2, 2, 4); (3, 3, 6)</math> have <math>3</math> ways to rearrange them for a total of <math>9</math> ways. Adding them up, we get <math>45</math> ways. We have to divide this values by <math>6^3</math> because there are 3 dice. <math>\dfrac{45}{216}=\boxed{\dfrac{5}{24}}</math>. | ||
− | + | ~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 <math>6^3</math> for 3 dice. <math>\dfrac{45}{216}</math> is <math>\boxed{\textbf{(D)} \: \dfrac{5}{24}}</math>. | ||
− | + | -aopspandy | |
==See Also== | ==See Also== | ||
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{{AMC10 box|year=2014|ab=A|num-b=16|num-a=18}} | {{AMC10 box|year=2014|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Introductory Combinatorics Problems]] |
Latest revision as of 11:14, 7 September 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 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
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.