Bunuel
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 the grid is now entirely black?
(A) 49/512
(B) 7/64
(C) 121/1024
(D) 81/512
(E) 9/32
Breaking Down the Info:First of all, there are two configurations for each square. Then the total amount of cases would be \(2^9 = 512\). This allows us to choose A or D as an educated guess.
The big square is rotated 90 degrees clockwise, the NW square can "spread" to the NE square, and the N square can spread to the E square for example.
To make the entire grid black, we first need the middle grid to be black. We will need at least 4 black squares on the outer side. We can have N, S, and NW, SE as an example and after rotating the grid will be entirely black.
We can separate into the NSEW side squares and the corner squares, we can note that we only need an opposite pair of black squares for each group in order to fill its group with black squares after rotating.
Take the NSEW group, we can have N+S, E+W, or more squares painted in black to fill this group in black. We can use \(2^4 = 16\) to find there are 16 ways to fill in the squares in this group, and remove 1 blank setup, remove 4 single black square cases, and finally remove 4 adjacent 2-black-square setups. Out of the 16 cases, we will accept 7 of them.
Similarly for the corners, there are 7 cases we accept. Finally, we can multiply 7 and 7 since the corners and sides are two independent groups.
The final answer would be 49/512.
Answer: A