AMC12 2012 A

AMC12 2012 A · Q11

AMC12 2012 A · Q11. It mainly tests Combinations, Probability (basic).

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?

Alex、Mel 和 Chelsea 玩一个有 $6$ 轮的游戏。每轮有一个获胜者，且各轮结果相互独立。每轮 Alex 获胜的概率为 $\frac{1}{2}$，并且 Mel 获胜的可能性是 Chelsea 的两倍。求 Alex 赢 $3$ 轮、Mel 赢 $2$ 轮、Chelsea 赢 $1$ 轮的概率。

(A) $\frac{5}{72}$ $\frac{5}{72}$

(B) $\frac{5}{36}$ $\frac{5}{36}$

(D) $\frac{1}{3}$ $\frac{1}{3}$

(E) $\frac{1}{2}$ $\frac{1}{2}$

Answer

Correct choice: (B)

正确答案：（B）

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

设 Mel 获胜的概率为 $m$，Chelsea 获胜的概率为 $c$，则 $m=2c$ 且 $m+c=\frac12$。由此得 $m=\frac13$，$c=\frac16$。若按顺序 Alex 赢三轮、Mel 赢两轮、Chelsea 赢一轮，则概率为 $\frac{1}{2^3\cdot3^2\cdot6}=\frac{1}{432}$。再乘以排列数（他们获胜顺序的种数）$\frac{6!}{3!2!1!}=60$。 \[\frac{1}{432}\cdot60=\frac{60}{432}=\boxed{\textbf{(B)}\ \frac{5}{36}}\]

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