AMC8 2023

AMC8 2023 · Q23

AMC8 2023 · Q23. It mainly tests Inclusion–exclusion (basic), Probability (basic).

Each square in a $3 \times 3$ grid is randomly filled with one of the $4$ gray and white tiles shown below on the right. What is the probability that the tiling will contain a large gray diamond in one of the smaller $2 \times 2$ grids? Below is an example of such tiling.

一个 $3 \times 3$ 网格中的每个方格随机填充右边所示的 $4$ 种灰白瓦片之一。\n\n求该铺砖包含至少一个较小的 $2 \times 2$ 网格中有一个大灰色菱形的概率？下面是一个这样的铺砖示例。

(A) \frac{1}{1024} \frac{1}{1024}

(B) \frac{1}{256} \frac{1}{256}

(D) \frac{1}{16} \frac{1}{16}

(E) \frac{1}{4} \frac{1}{4}

Answer

Correct choice: (C)

正确答案：（C）

Solution

There are $4$ cases that the tiling will contain a large gray diamond in one of the smaller $2 \times 2$ grids, as shown below: There are $4^5$ ways to decide the $5$ white squares for each case, and the cases do not have any overlap. So, the requested probability is \[\frac{4\cdot4^5}{4^9} = \frac{4^6}{4^9} = \frac{1}{4^3} = \boxed{\textbf{(C) } \frac{1}{64}}.\]

有 $4$ 种情况会使铺砖在某个小的 $2 \times 2$ 网格中包含大灰色菱形，如下所示：\n\n每种情况有 $4^5$ 种方法填充 $5$ 个白色方格，且这些情况没有重叠。\n\n因此所需概率为 \[\frac{4\cdot4^5}{4^9} = \frac{4^6}{4^9} = \frac{1}{4^3} = \boxed{\textbf{(C) } \frac{1}{64}}.\]

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