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Difference between revisions of "2014 AMC 10A Problems/Problem 17"

(→‎Video Solution (Meta-Solving Technique))
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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 (Bashy)==
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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>.
  
 
==See Also==
 
==See Also==

Revision as of 18:45, 15 August 2021

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?

$\textbf{(A)}\ \dfrac16\qquad\textbf{(B)}\ \dfrac{13}{72}\qquad\textbf{(C)}\ \dfrac7{36}\qquad\textbf{(D)}\ \dfrac5{24}\qquad\textbf{(E)}\ \dfrac29$

Video Solution

https://youtu.be/5UojVH4Cqqs?t=702

~ pi_is_3.14

Solution 1

First, we note that there are $1, 2, 3, 4,$ and $5$ ways to get sums of $2, 3, 4, 5, 6$ 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 \[\dfrac16 \left( \dfrac{1+2+3+4+5}{36}\right) = \dfrac{5}{72}.\] Since there are $\dbinom31$ ways to choose which die will be the one with the sum of the other two, our answer is $3 \cdot \dfrac{5}{72} = \boxed{\textbf{(D)} \: \dfrac{5}{24}}$.

Solution 2 (Bashy)

Start by listing all possible sums. Then find the ways to arrange them and sum them up and divide by $6^3$ for 3 dice. $\dfrac{45}{216}$ is $\boxed{\textbf{(D)} \: \dfrac{5}{24}}$.

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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