Two squares are chosen from the squares of a ordinary chess board. It is given that the selected squares do not belong to the same row or column. The probability that the chose squares are of same colour, is equal to

$\frac{25}{49}$

Let the first chosen square is white. In this case total number of squares that don’t belong the row or column associated with the selected square is 49. Out of these 25 are white and 24 are black.

Thus, corresponding probability

$=\frac{1}{2} \cdot \frac{25}{49}$

Similarly, if the first square is black, then the corresponding probability

$=\frac{1}{2} \cdot \frac{25}{49}$

Thus required probability

$=\frac{1}{2}\left(\frac{25}{49}+\frac{25}{49}\right)=\frac{25}{49}$