Question

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?

$\dfrac{49}{512}$

$\dfrac{7}{64}$

$\dfrac{121}{1024}$

$\dfrac{81}{512}$

$\dfrac{9}{32}$

Solution:

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.

For the four corners, checking the $2^4 = 16$ colorings shows that $7$ of them end all black, so the corners are all black with probability $\dfrac{7}{16}.$ The same holds for the four edges.

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 Problem 15

Solution:

Problem 15 in Other Years