2023 AMC 8 第 23 题

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.