Question

Alex、Mel 和 Chelsea 玩一个有 $6$ 轮的游戏。每一轮只有一个获胜者，且各轮结果相互独立。每一轮 Alex 获胜的概率是 $\dfrac12$ ，Mel 获胜的可能性是 Chelsea 的两倍。Alex 赢三轮、Mel 赢二轮、Chelsea 赢一轮的概率是多少？

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?

$\dfrac{5}{72}$

$\dfrac{5}{36}$

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$1$

答案：B

知识点：基本概率独立事件多重集排列

难度评级：1540

解答：

Alex 获胜的概率为 $\tfrac12$ ，其余两人共享剩下的 $\tfrac12$ 。因为 Mel 获胜的可能性是 Chelsea 的两倍， $P(\text{Mel}) = \tfrac13$ ， $P(\text{Chelsea}) = \tfrac16$ 。

获胜顺序 $AAAMMC$ 的排列数为 $\dfrac{6!}{3!\,2!\,1!} = 60$ 。概率为 $\begin{aligned} 60 \cdot \left(\tfrac12\right)^3 \left(\tfrac13\right)^2 \left(\tfrac16\right) &= \frac{60}{432} \\ &= \frac{5}{36}. \end{aligned}$

因此，正确答案是 B。

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 第 11 题

答案：B

难度评级：1540

解答：

其他年份的第 11 题