Question

一个 $3 \times 3$ 网格中的每个小方格随机填入右下方所示 $4$ 种阴影和非阴影瓷砖之一。

这个铺法在某个较小的 $2 \times 2$ 网格中包含一个大的阴影菱形的概率是多少？下面是这样一个铺法示例。

Each square in a $3 \times 3$ grid is randomly filled with one of the $4$ shaded-and-unshaded tiles shown below on the right.

What is the probability that the tiling will contain a large shaded diamond in one of the smaller $2 \times 2$ grids? Below is an example of such a tiling.

$\dfrac{1}{1024}$

$\dfrac{1}{256}$

$\dfrac{1}{64}$

$\dfrac{1}{16}$

$\dfrac{1}{4}$

答案：C

知识点：基本概率乘法原理

难度评级：1840

小提示：

先选择哪个 $2\times2$ 网格包含大菱形

Choose which $2\times2$ grid contains the large diamond

大提示：

一旦选定 $2\times2$ 网格，其中四块瓷砖方向就被确定

Once a $2\times2$ grid is chosen, its four tile orientations are forced

视频讲解：

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文字解答：

总铺法数为 $4^9$ 。可形成大阴影菱形的 $2\times2$ 小网格有 $4$ 个。

选定其中一个 $2\times2$ 小网格后，它内部四块瓷砖的方向就被确定了，其余 $5$ 个方格可以任意填。所以每选定一个这样的 $2\times2$ 小网格，都对应 $4^5$ 种铺法。

两个不同的 $2\times2$ 小网格不能同时形成大阴影菱形，因为重叠方格会要求不同方向。因此有利铺法数为 $4\cdot4^5=4^6$ 。

所求概率为 $\dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \dfrac{1}{64}\text{。}$

所以正确答案是 C。

There are $4^9$ possible tilings. There are $4$ possible $2\times2$ grids where a large shaded diamond could appear.

After one of these $2\times2$ grids is chosen, the four tile orientations inside it are forced, and the other $5$ squares can be filled in any way. This gives $4^5$ tilings for each chosen $2\times2$ grid.

Two different $2\times2$ grids cannot both contain a large shaded diamond, because their overlapping squares would require incompatible tile orientations. Therefore the number of favorable tilings is $4\cdot4^5=4^6.$

The desired probability is then $\dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \dfrac{1}{64}.$

Thus, C is the correct answer.

2023 AMC 8 第 23 题