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 the grid is now entirely black?

$\dfrac{49}{512}$

$\dfrac{7}{64}$

$\dfrac{121}{1024}$

$\dfrac{81}{512}$

$\dfrac{9}{32}$

Solution:

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 Problem 20

Solution:

Problem 20 in Other Years