Question

一个 $3 \times 3$ 网格的 $9$ 个小方格独立随机染成黑色或白色。

将整个网格顺时针旋转 $90^{\circ}$ 后，把每个现在位于原来黑格位置上的白格染成黑色，其余方格不变。最终整个网格全为黑色的概率是多少？

A $3 \times 3$ square is partitioned into $9$ unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random.

The square is then rotated $90^{\circ}$ clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability the grid is now entirely black?

$\dfrac{49}{512}$

$\dfrac{7}{64}$

$\dfrac{121}{1024}$

$\dfrac{81}{512}$

$\dfrac{9}{32}$

答案：A

知识点：基本概率独立事件对称性分类讨论

难度评级：1980

解答：

中心格必须一开始就是黑色，贡献概率 $\dfrac12$ 。四个角在旋转下形成一个循环，四个边中点也形成另一个相同的循环。

对一个 $4$ 循环，除非有一个白格被旋转到原本也是白格的位置，否则最后四个位置都会变黑。成功的初始颜色序列为 $BBBB$ 、 $BBBW$ 的四种旋转，以及 $BWBW$ 的两种旋转，共 $7$ 种，占 $16$ 种。

边中点循环有相同的独立计数。因此概率为 $\dfrac12\left(\dfrac7{16}\right)^2=\dfrac{49}{512}$ 。

所以正确答案是 A。

The center square must initially be black, contributing probability $\dfrac12$ . The four corner squares form one cycle under the rotation, and the four edge-middle squares form another identical cycle.

For one $4$ -cycle, the final four positions are all black unless a white square is rotated into a position that was also white. The successful initial colorings are $BBBB$ , the four rotations of $BBBW$ , and the two rotations of $BWBW$ , for $7$ colorings out of $16$ .

The same count applies to the edge-middle cycle, independently. Therefore the probability is $\dfrac12\left(\dfrac7{16}\right)^2=\dfrac{49}{512}$ .

Thus, A is the correct answer.

2012 AMC 10A 第 20 题

答案：A

难度评级：1980

解答：

其他年份的第 20 题