Question

Zou 和 Chou 正在练习 $100$ 米短跑，彼此比赛 $6$ 场。Zou 赢了第一场；之后，如果某人赢了上一场，那么他赢下一场的概率是 $\frac{2}{3}$ ，但如果他输了上一场，那么他赢下一场的概率只有 $\frac{1}{3}$ 。Zou 恰好赢 $5$ 场（共 $6$ 场）的概率为 $\frac{m}{n}$ ，其中 $m$ 与 $n$ 是互质正整数。求 $m + n$ 。

Zou and Chou are practicing their $100$ -meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac{2}{3}$ if they won the previous race but only $\frac{1}{3}$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：97

知识点：条件概率分类讨论

难度评级：2050

小提示：

Zou 必须在第 $2$ 到第 $6$ 场中恰好输一场。把输在第 $6$ 场的情况与更早输的情况分开处理

Zou must lose exactly one of races $2$ through $6.$ Treat a loss in race $6$ separately from an earlier loss.

大提示：

若较早输一场，会产生两次概率为 $\frac{1}{3}$ 的切换（进入输局和回到赢局）；若最后一场输，只产生一次切换

An earlier loss creates two probability- $\frac{1}{3}$ switches (into and out of the loss); a final-race loss creates just one

解答：

Zou 赢了第 $1$ 场，所以恰好赢 $5$ 场（共 $6$ 场），等价于他在第 $2$ 到第 $6$ 场中恰好输一场。第一场之后，每一场以概率 $\frac{2}{3}$ 重复上一场结果，以概率 $\frac{1}{3}$ 改变上一场结果。

如果输的是第 $6$ 场，那么五次转移是四次重复后接一次切换： $\left(\frac{2}{3}\right)^4 \cdot \frac{1}{3} = \frac{16}{243}$ 。如果输的是第 $i$ 场，其中 $2 \le i \le 5$ ，则有一次切换进入输局、一次切换回到赢局，另有三次重复： $\left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^2 = \frac{8}{243}$ 。这样的 $4$ 个位置共贡献 $\frac{32}{243}$ 。

总概率为 $\frac{16}{243} + \frac{32}{243} = \frac{48}{243} = \frac{16}{81}\text{，}$ 所以 $m + n = 16 + 81 = 97$ 。

Zou wins race $1,$ so winning exactly $5$ of the $6$ races means he loses exactly one of races $2$ through $6.$ Each race after the first repeats the previous outcome with probability $\frac{2}{3}$ and switches with probability $\frac{1}{3}.$

If the loss is race $6,$ the five transitions are four repeats followed by one switch: $\left(\frac{2}{3}\right)^4 \cdot \frac{1}{3} = \frac{16}{243}.$ If the loss is race $i$ for some $2 \le i \le 5,$ there is a switch into the loss and a switch back to winning, plus three repeats: $\left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^2 = \frac{8}{243}$ for each of the $4$ positions, contributing $\frac{32}{243}.$

The total is $\frac{16}{243} + \frac{32}{243} = \frac{48}{243} = \frac{16}{81},$ so $m + n = 16 + 81 = 97.$

2021 AIME I 第 1 题

答案：97

难度评级：2050

小提示：

大提示：

解答：

其他年份的第 1 题