Question

爱丽丝、鲍勃和卡萝尔轮流反复掷一个骰子。爱丽丝先掷；爱丽丝之后总是轮到鲍勃；鲍勃之后总是轮到卡萝尔；卡萝尔之后总是轮到爱丽丝。求卡萝尔第一个掷出六点的概率。（每次掷出六点的概率为 $\frac16$ ，且与其他任何一次投掷的结果相互独立。）

Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob; and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is $\frac16,$ independent of the outcome of any other toss.)

$\frac13$

$\frac29$

$\frac5{18}$

$\frac{25}{91}$

$\frac{36}{91}$

答案：D

知识点：骰子（概率）几何分布独立事件

难度评级：1680

小提示：

要让卡萝尔在某一轮获胜，爱丽丝和鲍勃必须先失败，然后卡萝尔成功

For Carol to win in a given round, Alice and Bob must fail before Carol succeeds

大提示：

如果三人都失败，则过程以相同的概率重新开始

If all three fail, the process restarts with the same probabilities

解答：

卡萝尔在第一轮获胜的概率为 $(\frac{5}{6})^2(\frac{1}{6})=\frac{25}{216}$ 。完整一轮都没有出现六点的概率为 $(\frac{5}{6})^3=\frac{125}{216}$ 。因此所求的等比级数为 $\frac{\frac{25}{216}}{1-\frac{125}{216}} =\frac{25}{91}\text{。}$

所以正确答案是 D。

Carol wins in the first round with probability $(\frac{5}{6})^2(\frac{1}{6})=\frac{25}{216}.$ A complete round with no six has probability $(\frac{5}{6})^3=\frac{125}{216}.$ Therefore the desired geometric series is $\frac{\frac{25}{216}}{1-\frac{125}{216}} =\frac{25}{91}.$

Therefore, the correct answer is D.

1981 AMC 12 第 26 题

答案：D

难度评级：1680

小提示：

大提示：

解答：

其他年份的第 26 题