Competition problem — Combinatorics

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Combinatorics Difficulty 2.6 Junior Find the answer

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?

Pick one

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 $\frac{6!}{3!2!1!}=60.$
$\frac{1}{432}\cdot60=\frac{60}{432}=\boxed{\textbf{(B)}\ \frac{5}{36}}$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.