Question

Charles 有两个六面骰子。其中一个是公平骰子，另一个有偏骰子掷出六点的概率为 $\frac{2}{3}$ ，其余五个面各自出现的概率为 $\frac{1}{15}$ 。Charles 随机选择这两个骰子中的一个并掷三次。已知前两次都掷出六点，第三次也掷出六点的概率为 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质正整数。求 $p + q$ 。

Charles has two six-sided dice. One of the dice is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3},$ and each of the other five sides has probability $\frac{1}{15}.$ Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

答案

答案：167

知识点：条件概率贝叶斯定理骰子（概率）

难度评级：2390

小提示：

所求概率等于“三次都是六点”的概率除以“前两次都是六点”的概率；这两个概率都要对两个等可能骰子取平均

The answer is the probability of three sixes divided by the probability of two sixes, each averaged over the two equally likely dice

大提示：

连续两次六点会让有偏骰子的可能性大大增加：先按每个骰子产生这两次六点的概率加权，再预测第三次

Two sixes in a row make the biased die much more likely: weight each die by its chance of producing the observed two sixes before predicting the third roll

解答：

所求条件概率为 $\begin{aligned} &\frac{\Pr(\text{三次都是六})}{\Pr(\text{前两次是六})} \\ &= \frac{\frac{1}{2}\left(\frac{2}{3}\right)^3 + \frac{1}{2}\left(\frac{1}{6}\right)^3} {\frac{1}{2}\left(\frac{2}{3}\right)^2 + \frac{1}{2}\left(\frac{1}{6}\right)^2} \end{aligned}\text{，}$ 因为每个骰子被选中的概率都是 $\frac{1}{2}$ ，公平骰子掷出六点的概率为 $\frac{1}{6}$ 。

分子为 $\frac{1}{2}\left(\frac{8}{27} + \frac{1}{216}\right) = \frac{65}{432}$ ，分母为 $\frac{1}{2}\left(\frac{4}{9} + \frac{1}{36}\right) = \frac{17}{72}$ ，所以概率为 $\frac{65}{432} \cdot \frac{72}{17} = \frac{65}{102}$ 。

因为 $65 = 5 \cdot 13$ ，而 $102 = 2 \cdot 3 \cdot 17$ ，二者没有公因数，所以 $p + q = 65 + 102 = 167$ 。

The desired conditional probability is $\begin{aligned} &\frac{\Pr(\text{three sixes})}{\Pr(\text{first two are sixes})} \\ &= \frac{\frac{1}{2}\left(\frac{2}{3}\right)^3 + \frac{1}{2}\left(\frac{1}{6}\right)^3} {\frac{1}{2}\left(\frac{2}{3}\right)^2 + \frac{1}{2}\left(\frac{1}{6}\right)^2}, \end{aligned}$ since each die is chosen with probability $\frac{1}{2}$ and the fair die shows a six with probability $\frac{1}{6}.$

The numerator is $\frac{1}{2}\left(\frac{8}{27} + \frac{1}{216}\right) = \frac{65}{432}$ and the denominator is $\frac{1}{2}\left(\frac{4}{9} + \frac{1}{36}\right) = \frac{17}{72},$ so the probability is $\frac{65}{432} \cdot \frac{72}{17} = \frac{65}{102}.$

Since $65 = 5 \cdot 13$ and $102 = 2 \cdot 3 \cdot 17$ share no factor, $p + q = 65 + 102 = 167.$

2014 AIME II 第 6 题

答案：167

难度评级：2390

小提示：

大提示：

解答：

其他年份的第 6 题