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?

$\dfrac{5}{72}$

$\dfrac{5}{36}$

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$1$

Answer: B

Concepts:basic probability independent events multiset permutations

Difficulty rating: 1540

Solution:

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.

2012 AMC 12A Problem 11

Answer: B

Difficulty rating: 1540

Solution:

Problem 11 in Other Years