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?

$\dfrac{49}{512}$

$\dfrac{7}{64}$

$\dfrac{121}{1024}$

$\dfrac{81}{512}$

$\dfrac{9}{32}$

Answer: A

Concepts:basic probability symmetry casework

Difficulty rating: 1930

Small Hint:

Treat the four corners, the four edges, and the center as three independent groups

Big Hint:

A square ends black unless it is white and the square rotated into its position was also white; the center must start black

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.

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 $\frac{7}{16}.$ 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 Problem 15

Answer: A

Difficulty rating: 1930

Small Hint:

Big Hint:

Solution:

Problem 15 in Other Years