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

m (Undo revision 66992 by Equationcrunchor (talk))
(Adding solution 2)
 
(10 intermediate revisions by 6 users not shown)
Line 5: Line 5:
 
<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 (Clean Counting)==
+
== Video Solution ==
 +
https://youtu.be/5UojVH4Cqqs?t=702
  
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>.
+
~ pi_is_3.14
  
--happiface
+
==Solution 1==
  
==Solution 2 (Casework)==
+
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>6</math> possible values for the number on each dice, there are <math>6^3=216</math> 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,  <cmath>(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)</cmath>
+
==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>.
  
There are <math>\frac{3!}{2}=3</math> ways to order the first, sixth, and ninth results, and there are <math>3!=6</math> ways to order the other results.
+
~MathFun1000
  
Therefore, there are a total of <math>3\times3+6\times6=45</math> ways to roll the dice such that 2 of the dice sum to the other die, so our answer is <cmath>\frac{45}{216}=\boxed{\textbf{(D)} \ \frac{5}{24}}</cmath>
+
==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>.
  
(Solution by bestwillcui1)
+
-aopspandy
  
 
==See Also==
 
==See Also==
Line 27: Line 28:
 
{{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

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)

$(1, 2, 3); (1, 3, 4); (1, 4, 5); (1, 5, 6); (2, 3, 5); (2, 4, 6)$ have $6$ ways to rearrange them for a total of $36$ ways. $(1, 1, 2); (2, 2, 4); (3, 3, 6)$ have $3$ ways to rearrange them for a total of $9$ ways. Adding them up, we get $45$ ways. We have to divide this values by $6^3$ because there are 3 dice. $\dfrac{45}{216}=\boxed{\dfrac{5}{24}}$.

~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 $6^3$ for 3 dice. $\dfrac{45}{216}$ is $\boxed{\textbf{(D)} \: \dfrac{5}{24}}$.

-aopspandy

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS