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Difference between revisions of "2012 AMC 12A Problems/Problem 11"

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<cmath>\frac{1}{432}\cdot60=\frac{60}{432}=\boxed{\textbf{(B)}\ \frac{5}{36}}</cmath>
 
<cmath>\frac{1}{432}\cdot60=\frac{60}{432}=\boxed{\textbf{(B)}\ \frac{5}{36}}</cmath>
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==Video Solution by TheBeautyofMath==
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I preview the concept that will be used in the first part of this video, and show why it's important "Don't Memorize, Understand". I adapt the solution method of an older problem to inform the solution of this one: https://youtu.be/PO3XZaSchJc
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~IceMatrix
  
 
== See Also ==
 
== See Also ==

Revision as of 19:40, 20 October 2021

Problem

Alex, Mel, and Chelsea play a game that has $6$ rounds. In each round there is a single winner, and the outcomes of the rounds are independent. For each round the probability that Alex wins is $\frac{1}{2}$, and Mel is twice as likely to win as Chelsea. What is the probability that Alex wins three rounds, Mel wins two rounds, and Chelsea wins one round?

$\textbf{(A)}\ \frac{5}{72}\qquad\textbf{(B)}\ \frac{5}{36}\qquad\textbf{(C)}\ \frac{1}{6}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ 1$

Solution

If $m$ is the probability Mel wins and $c$ is the probability Chelsea wins, $m=2c$ and $m+c=\frac12$. From this we get $m=\frac13$ and $c=\frac16$. For Alex to win three, Mel to win two, and Chelsea to win one, in that order, is $\frac{1}{2^3\cdot3^2\cdot6}=\frac{1}{432}$. Multiply this by the number of permutations (orders they can win) which is $\frac{6!}{3!2!1!}=60.$

\[\frac{1}{432}\cdot60=\frac{60}{432}=\boxed{\textbf{(B)}\ \frac{5}{36}}\]

Video Solution by TheBeautyofMath

I preview the concept that will be used in the first part of this video, and show why it's important "Don't Memorize, Understand". I adapt the solution method of an older problem to inform the solution of this one: https://youtu.be/PO3XZaSchJc

~IceMatrix

See Also

2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 12 Problems and Solutions

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