2012 AMC 12A Problems/Problem 11

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 $\frac{6!}{3!2!1!}=60.$

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

See Also

2012 AMC 12A (ProblemsAnswer KeyResources)
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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