Question

两个由三个字母组成的字符串 $aaa$ 和 $bbb$ 通过电子方式传输，每个字符串逐字母发送。由于设备故障，六个字母中的每一个都有 $\frac{1}{3}$ 的概率被错误接收：本应是 $b$ 时被接收为 $a$ ，或本应是 $a$ 时被接收为 $b$ 。不过，每个字母接收正确与否都独立于其他字母的接收情况。

设发送 $aaa$ 时接收到的三字母字符串为 $S_a$ ，发送 $bbb$ 时接收到的三字母字符串为 $S_b$ 。设 $S_a$ 按字母顺序排在 $S_b$ 之前的概率为 $p$ 。将 $p$ 写成最简分数后，其分子是多少？

Two three-letter strings, $aaa$ and $bbb,$ are transmitted electronically. Each string is sent letter by letter. Due to faulty equipment, each of the six letters has a $\frac{1}{3}$ chance of being received incorrectly, as an $a$ when it should have been a $b,$ or as a $b$ when it should be an $a.$ However, whether a given letter is received correctly or incorrectly is independent of the reception of any other letter.

Let $S_a$ be the three-letter string received when $aaa$ is transmitted and let $S_b$ be the three-letter string received when $bbb$ is transmitted. Let $p$ be the probability that $S_a$ comes before $S_b$ in alphabetical order. When $p$ is written as a fraction in lowest terms, what is its numerator?

答案

答案：532

知识点：独立事件等比数列基本概率

难度评级：2200

小提示：

两个接收字符串第一次出现不同字母的位置决定了它们的先后顺序。

The ordering is decided at the first position where the two received strings differ

大提示：

对一个位置，分别计算两个字母相同的概率，以及 $S_a$ 中收到 $a$ 且 $S_b$ 中收到 $b$ 的概率。

At one position, compute the probabilities of equality and of receiving $a$ in $S_a$ and $b$ in $S_b$

解答：

在任意一个位置，接收到的两个字母相同的概率为 $2\left(\frac23\right)\left(\frac13\right)=\frac49\text{。}$ 对排序有利的第一次不同，即 $S_a$ 收到 $a$ 且 $S_b$ 收到 $b$ ，其概率为 $(\frac{2}{3})^2=\frac{4}{9}$ 。它可以在零次、一次或两次相同之后，分别出现在第一个、第二个或第三个位置。因此 $\begin{aligned}p&=\frac49\left(1+\frac49+\left(\frac49\right)^2\right)\\&=\frac{532}{729}\end{aligned}\text{。}$ 这个分数已经最简，所以其分子为 $532$ 。

At any position, the received letters agree with probability $2\left(\frac23\right)\left(\frac13\right)=\frac49.$ The favorable first difference, $S_a$ receiving $a$ and $S_b$ receiving $b,$ has probability $(\frac{2}{3})^2=\frac{4}{9}.$ It can occur in the first, second, or third position after zero, one, or two agreements. Hence $\begin{aligned}p&=\frac49\left(1+\frac49+\left(\frac49\right)^2\right)\\&=\frac{532}{729}.\end{aligned}$ This fraction is in lowest terms, so its numerator is $532.$

1991 AIME 第 10 题

答案：532

难度评级：2200

小提示：

大提示：

解答：

其他年份的第 10 题