AMC10 2012 A

AMC10 2012 A · Q20

AMC10 2012 A · Q20. It mainly tests Probability (basic), Transformations.

A 3 × 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° 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?

一个3×3方格被分成9个单位方格。每个单位方格独立随机涂成白色或黑色，两种颜色等概率。方格绕中心顺时针旋转90°，每个原先被黑色方格占据位置的白色方格涂成黑色，其他方格颜色不变。求现在整个格子全黑的概率。

(A) \frac{49}{512} \frac{49}{512}

(B) \frac{7}{64} \frac{7}{64}

(D) \frac{81}{512} \frac{81}{512}

(E) \frac{9}{32} \frac{9}{32}

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): There are $2^4=16$ possible initial colorings for the four corner squares. If their initial coloring is $BBBB$, one of the four cyclic permutations of $BBBW$, or one of the two cyclic permutations of $BWBW$, then all four corner squares are black at the end. If the initial coloring is $WWWW$, one of the four cyclic permutations of $BWWW$, or one of the four cyclic permutations of $BBWW$, then at least one corner square is white at the end. Hence all four corner squares are black at the end with probability $\frac{7}{16}$. Similarly, all four edge squares are black at the end with probability $\frac{7}{16}$. The center square is black at the end if and only if it was initially black, so it is black at the end with probability $\frac{1}{2}$. The probability that all nine squares are black at the end is $\frac{1}{2}\cdot\left(\frac{7}{16}\right)^2=\frac{49}{512}$.

答案（A）：四个角格的初始染色共有 $2^4=16$ 种可能。若初始染色为 $BBBB$、$BBBW$ 的四种循环排列之一，或 $BWBW$ 的两种循环排列之一，则最终四个角格都为黑色。若初始染色为 $WWWW$、$BWWW$ 的四种循环排列之一，或 $BBWW$ 的四种循环排列之一，则最终至少有一个角格为白色。因此，最终四个角格都为黑色的概率是 $\frac{7}{16}$。同理，最终四个边格都为黑色的概率也是 $\frac{7}{16}$。中心格最终为黑当且仅当它初始为黑，所以中心格最终为黑的概率为 $\frac{1}{2}$。因此九个格子最终全为黑色的概率是 $\frac{1}{2}\cdot\left(\frac{7}{16}\right)^2=\frac{49}{512}$。

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