Question

Jackie 和 Phil 有两枚公平硬币，还有一枚掷出正面的概率为 $\frac{4}{7}$ 的硬币。Jackie 抛这三枚硬币，然后 Phil 也抛这三枚硬币。设 $\frac{m}{n}$ 为 Jackie 和 Phil 得到相同正面数的概率，其中 $m$ 和 $n$ 是互质的正整数。求 $m + n$ 。

Jackie and Phil have two fair coins and a third coin that comes up heads with probability $\frac{4}{7}.$ Jackie flips the three coins, and then Phil flips the three coins. Let $\frac{m}{n}$ be the probability that Jackie gets the same number of heads as Phil, where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：515

知识点：基本概率独立事件分类讨论

难度评级：2340

小提示：

计算一个人抛出 $0$ 、 $1$ 、 $2$ 、 $3$ 个正面的概率；这四个概率的分母都是 $28$

Compute the probability that one player flips $0,$ $1,$ $2,$ $3$ heads; all four have denominator $28$

大提示：

两人的抛掷相互独立且分布相同，所以正面数相同的概率是这四个概率的平方和

The two players are independent and identical, so the probability they match is the sum of the squares of those four probabilities

解答：

设 $p(h)$ 为一个人抛出 $h$ 个正面的概率。按两枚公平硬币和一枚非均匀硬币分类， $\begin{aligned} p(0) &= \tfrac{1}{4} \cdot \tfrac{3}{7} = \tfrac{3}{28}, \\ p(1) &= \tfrac{2}{4} \cdot \tfrac{3}{7} + \tfrac{1}{4} \cdot \tfrac{4}{7} = \tfrac{10}{28}, \\ p(2) &= \tfrac{1}{4} \cdot \tfrac{3}{7} + \tfrac{2}{4} \cdot \tfrac{4}{7} = \tfrac{11}{28}, \\ p(3) &= \tfrac{1}{4} \cdot \tfrac{4}{7} = \tfrac{4}{28} \end{aligned}\text{。}$

Jackie 和 Phil 的抛掷独立且分布相同，所以正面数相同的概率为 $\small \begin{aligned} \sum_h p(h)^2 &= \frac{3^2 + 10^2 + 11^2 + 4^2}{28^2} \\ &= \frac{246}{784} = \frac{123}{392} \end{aligned}\text{。}$ 因此 $m + n = 123 + 392 = 515$ 。

Let $p(h)$ be the probability that one player flips $h$ heads. Splitting according to the two fair coins and the biased coin, $\begin{aligned} p(0) &= \tfrac{1}{4} \cdot \tfrac{3}{7} = \tfrac{3}{28}, \\ p(1) &= \tfrac{2}{4} \cdot \tfrac{3}{7} + \tfrac{1}{4} \cdot \tfrac{4}{7} = \tfrac{10}{28}, \\ p(2) &= \tfrac{1}{4} \cdot \tfrac{3}{7} + \tfrac{2}{4} \cdot \tfrac{4}{7} = \tfrac{11}{28}, \\ p(3) &= \tfrac{1}{4} \cdot \tfrac{4}{7} = \tfrac{4}{28}. \end{aligned}$

Jackie’s and Phil’s flips are independent with the same distribution, so the probability that their head counts agree is $\small \begin{aligned} \sum_h p(h)^2 &= \frac{3^2 + 10^2 + 11^2 + 4^2}{28^2} \\ &= \frac{246}{784} = \frac{123}{392}. \end{aligned}$ Thus $m + n = 123 + 392 = 515.$

2010 AIME I 第 4 题

答案：515

难度评级：2340

小提示：

大提示：

解答：

其他年份的第 4 题