Question

一个 $3 \times 3$ 正方形被分成 $9$ 个单位正方形。每个单位正方形被涂成白色或黑色，两种颜色等可能且相互独立随机选择。然后将整个正方形绕中心顺时针旋转 $90^\circ$ ，并且凡是旋转后处在原来黑色正方形位置上的白色正方形，都被涂成黑色。所有其他正方形的颜色保持不变。此时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 that 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：基本概率对称性分类讨论

难度评级：1930

解答：

旋转时，四个角格构成一个循环，四个边格构成另一个循环，而中心格保持不动。这三组相互独立。

一个位置最终仍是白色，当且仅当它和旋转到该位置的方格原来都是白色。因此四个角最终全黑，恰好等价于它们的循环颜色串中没有两个相邻的白格。允许的颜色串包括全黑的一个、恰有一个白格的 $4$ 个，以及两个白格相对的 $2$ 个，共有 $7$ 种，而总数为 $2^4=16$ 。所以四个角全黑的概率是 $7/16.$ 四个边格同理。

中心格最终为黑色，只有它一开始就是黑色，概率为 $\dfrac12.$ 相乘可得整个方格全黑的概率为 $\frac12 \cdot \left(\frac{7}{16}\right)^2 = \frac{49}{512}.$

因此，正确答案是 A。

The four corners form one cycle under the rotation, the four edge squares form another, and the center is fixed. These three groups are independent.

A position remains white exactly when both it and the square rotated into it were originally white. Thus the corners end black exactly when their cyclic string has no adjacent pair of whites. The allowed strings are the all-black string, the $4$ strings with one white, and the $2$ strings with two opposite whites: $7$ of the $2^4=16$ possibilities. Hence the corner probability is $7/16.$ The same argument applies to the four edge squares.

The center is black at the end only if it started black, with probability $\dfrac12.$ Multiplying, the whole grid is black with probability $\frac12 \cdot \left(\frac{7}{16}\right)^2 = \frac{49}{512}.$

Thus, the correct answer is A.

2012 AMC 12A 第 15 题

答案：A

难度评级：1930

解答：

其他年份的第 15 题