Question

许多州的标准车牌格式是三个字母后接三个数字。假设每一种三字母三数字的排列等可能出现，这种车牌至少含有一个回文（三个字母的排列或三个数字的排列从左到右读和从右到左读相同）的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：59

知识点：基本概率容斥原理回文数

难度评级：1890

解答：

三个字母的排列是回文，当且仅当第三个字母与第一个字母相同，所以字母部分为回文的概率是 $\frac{1}{26}$ 。同理，数字部分为回文的概率是 $\frac{1}{10}$ ，且这两个事件相互独立。

由容斥，至少有一个回文的概率为因此 $m + n = 7 + 52 = 59$ 。 $\begin{aligned} &\frac{1}{26} + \frac{1}{10} - \frac{1}{26} \cdot \frac{1}{10} \\ &= \frac{10 + 26 - 1}{260} \\ &= \frac{35}{260} \\ &= \frac{7}{52}. \end{aligned}$

A three-letter arrangement is a palindrome exactly when the third letter matches the first, so the probability of a letter palindrome is $\frac{1}{26}.$ Similarly, the probability of a digit palindrome is $\frac{1}{10},$ and the two events are independent.

By inclusion-exclusion, the probability of at least one palindrome is $\begin{aligned} &\frac{1}{26} + \frac{1}{10} - \frac{1}{26} \cdot \frac{1}{10} \\ &= \frac{10 + 26 - 1}{260} \\ &= \frac{35}{260} \\ &= \frac{7}{52}. \end{aligned}$ Thus $m + n = 7 + 52 = 59.$

2002 AIME I 第 1 题

答案：59

难度评级：1890

解答：

其他年份的第 1 题