Question

Azar、Carl、Jon 和 Sergey 是单打网球锦标赛剩下的四名选手。他们被随机分配半决赛对手，半决赛的胜者再进行决赛以决出冠军。当 Azar 对 Carl 时，Azar 以 $\frac{2}{3}$ 的概率获胜。当 Azar 或 Carl 对 Jon 或 Sergey 中的任意一人时，Azar 或 Carl 以 $\frac{3}{4}$ 的概率获胜。假设不同比赛的结果相互独立。Carl 赢得锦标赛的概率为 $\frac{p}{q}$ ，其中 $p$ 和 $q$ 是互质的正整数。求 $p + q$ 。

Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability $\frac{2}{3}.$ When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability $\frac{3}{4}.$ Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

答案

答案：125

知识点：条件概率独立事件分类讨论

难度评级：2180

解答：

四名选手的三种配对方式等可能，所以 Carl 在半决赛对 Azar 的概率为 $\frac{1}{3}$ 。在这种情况下， Carl 以 $\frac{1}{3}$ 的概率战胜 Azar，再以 $\frac{3}{4}$ 的概率战胜 Jon 和 Sergey 中的胜者，所以 Carl 赢得锦标赛的概率为 $\frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}$ 。

否则（概率为 $\frac{2}{3}$ ），Carl 对 Jon 或 Sergey，并以 $\frac{3}{4}$ 的概率获胜。他在决赛中的对手以 $\frac{3}{4}$ 的概率是 Azar（此时 Carl 以 $\frac{1}{3}$ 的概率获胜），以 $\frac{1}{4}$ 的概率是 Jon 或 Sergey（此时 Carl 以 $\frac{3}{4}$ 的概率获胜）。所以在这种情况下， Carl 赢得锦标赛的概率为 $\begin{aligned} &\frac{3}{4}\left(\frac{3}{4} \cdot \frac{1}{3} + \frac{1}{4} \cdot \frac{3}{4}\right) \\ &= \frac{3}{4} \cdot \frac{7}{16} \\ &= \frac{21}{64}. \end{aligned}$

总概率为 $\frac{1}{3} \cdot \frac{1}{4} + \frac{2}{3} \cdot \frac{21}{64} = \frac{1}{12} + \frac{7}{32} = \frac{29}{96}$ ，所以 $p + q = 29 + 96 = 125$ 。

The three ways to pair the four players are equally likely, so Carl plays Azar in the semifinal with probability $\frac{1}{3}.$ In that case Carl beats Azar with probability $\frac{1}{3}$ and then beats the Jon–Sergey winner with probability $\frac{3}{4},$ so Carl wins the tournament with probability $\frac{1}{3} \cdot \frac{3}{4} = \frac{1}{4}.$

Otherwise (probability $\frac{2}{3}$ ) Carl plays Jon or Sergey and wins with probability $\frac{3}{4}.$ His opponent in the final is Azar with probability $\frac{3}{4}$ (Carl then wins with probability $\frac{1}{3}$ ) and is Jon or Sergey with probability $\frac{1}{4}$ (Carl then wins with probability $\frac{3}{4}$ ). So in this case Carl wins the tournament with probability $\begin{aligned} &\frac{3}{4}\left(\frac{3}{4} \cdot \frac{1}{3} + \frac{1}{4} \cdot \frac{3}{4}\right) \\ &= \frac{3}{4} \cdot \frac{7}{16} \\ &= \frac{21}{64}. \end{aligned}$

The total probability is $\frac{1}{3} \cdot \frac{1}{4} + \frac{2}{3} \cdot \frac{21}{64} = \frac{1}{12} + \frac{7}{32} = \frac{29}{96},$ so $p + q = 29 + 96 = 125.$

2022 AIME II 第 2 题

答案：125

难度评级：2180

解答：

其他年份的第 2 题