Question

If two squares are chosen at random on a chess board, The probability that they have exactly one corner in common is equal to $\frac{k}{144}$.Find the value of $k$

Answer 1

Total number of ways $=64\times 63$
Now if first square is in one of the four corner, the second square can be chosen in just one way $=\left(4\right)\left(1\right)=4$
If the first square is one of the $24$ non-corner square along the side of the chess board, the second square can be chosen in two ways $=\left(24\right)\left(2\right)=48$
Now if the first square is any of the $36$ remaining square, the second square can be chosen in four ways $=\left(36\right)\left(4\right)=144$
Therefore favorable ways $=4+48+144=196$
Therefore required probability $=\frac{196}{64\times 63}=\frac{7}{144}$