Question

Two squares are chosen at random on a chess board. If p denotes the probability that they have exactly one side in common, find 144p

Answer 1

The first square can be chosen in 64 ways, after which the second can be chosen in 63 ways. Now, if the first square is in one of the four corners, the second square can be chosen in just one way. If the first square is one of the 24 non-corner squares along the sides of the chessboard, the second square can be chosen in two ways. Finally, if the first square is any of the 36 remaining squares, the second square can be chosen in four ways. Therefore the number of favourable choices is
$\left(4\right)\left(1\right)+\left(24\right)\left(2\right)+\left(36\right)\left(4\right)=196$,
So that the required probability is
$p=\frac{196}{64\times 63}=\frac{7}{144}\Rightarrow 144p=7$