2012 AMC 12A Problem 11

Question

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

\dfrac12,

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?

\dfrac{5}{72}

\dfrac{5}{36}

\dfrac{1}{6}

\dfrac{1}{3}

1

First find each player’s per-round win probability

Since

P(\text{Mel}) + P(\text{Chelsea}) = \dfrac12

and

P(\text{Mel}) = 2P(\text{Chelsea}),

multiply by the number

\dfrac{6!}{3!\,2!\,1!}

of arrangements

Since Alex wins with probability

\tfrac12,

the others share the remaining

\tfrac12.

With Mel twice as likely as Chelsea,

P(\text{Mel}) = \tfrac13

and

P(\text{Chelsea}) = \tfrac16.

The number of orderings of the wins

AAAMMC

is

\dfrac{6!}{3!\,2!\,1!} = 60.

The probability is

\begin{aligned} 60 \cdot \left(\tfrac12\right)^3 \left(\tfrac13\right)^2 \left(\tfrac16\right) &= \frac{60}{432} \\ &= \frac{5}{36}. \end{aligned}

Thus, the correct answer is B.

Problem 11 in other years