Question

一个 $3$ 乘 $3$ 的单位正方形网格中，每个单位正方形要被涂成蓝色或红色。对每个小方格，两种颜色被使用的概率相等。得到一个不含 $2$ 乘 $2$ 红色正方形的网格的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数。求 $m + n$ 。

Each unit square of a $3$ -by- $3$ unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a $2$ -by- $2$ red square is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：929

知识点：容斥原理对立事件概率

难度评级：2710

解答：

对四个可能位置使用容斥法，计算网格中含有一个全红 $2$ 乘 $2$ 方块的概率。一个方块强制 $4$ 个格子为红色；两个方块若共边，则强制 $6$ 个格子为红色，共有 $4$ 对；另外两对对角位置各强制 $7$ 个格子为红色。任意三个方块强制 $8$ 个格子，四个方块强制全部 $9$ 个格子。

每种强制红色格子的配置概率为 $\left(\frac{1}{2}\right)^{\text{cells}}$ ，所以至少有一个全红方块的概率是 $\begin{aligned} &4 \cdot \frac{1}{16} - \left(4 \cdot \frac{1}{64} + 2 \cdot \frac{1}{128}\right) \\ &\quad {}+ 4 \cdot \frac{1}{256} - \frac{1}{512} \\ &= \frac{128 - 40 + 8 - 1}{512} = \frac{95}{512}. \end{aligned}$

所求概率为 $1 - \frac{95}{512} = \frac{417}{512}$ ，且 $417 = 3 \cdot 139$ 与 $512$ ，互质，因此 $m + n = 417 + 512 = 929$ 。

Compute the probability that the grid does contain an all-red $2$ -by- $2$ block by inclusion-exclusion over the four possible positions. One block forces $4$ cells; two blocks sharing an edge force $6$ cells ( $4$ such pairs), while the two diagonal pairs force $7;$ any three blocks force $8$ cells, and all four force all $9.$

Each configuration of forced red cells has probability $\left(\frac{1}{2}\right)^{\text{cells}},$ so the probability of at least one red block is $\begin{aligned} &4 \cdot \frac{1}{16} - \left(4 \cdot \frac{1}{64} + 2 \cdot \frac{1}{128}\right) \\ &\quad {}+ 4 \cdot \frac{1}{256} - \frac{1}{512} \\ &= \frac{128 - 40 + 8 - 1}{512} = \frac{95}{512}. \end{aligned}$

The desired probability is $1 - \frac{95}{512} = \frac{417}{512},$ and $417 = 3 \cdot 139$ is coprime to $512,$ so $m + n = 417 + 512 = 929.$

2001 AIME II 第 9 题

答案：929

难度评级：2710

解答：

其他年份的第 9 题