If two squares are chosen at random on chess board , find the probabilty that they have a side in common

• 1/9
• 4/7
• 3/7
• 1/18

PLZ explain with the reaon

There total 64 squares on a chess board, arranged in 8 × 8 form 
There are 4 corner squares where only 2 other squares share a side, so total number of ways for such a square = (4 × 2) 
There are 24 squares touching the border of the chess board (excluding 4 corner squares) where 3 other squares share a side,

So total number of ways of selecting such type of square = 24 × 3 
There are remaining 36 squares where 4 other squares share a side, so total number of ways of selection = 36 × 4 
So, total number of ways in which a square can be selected with one side shared is = (4 × 2) + (24 × 3) + (36 × 4) = 224 
And total number of ways in which a square can be selected = 64 × 63 
Therefore, the probability of selecting one square such that one of it sides is in common = 224/(64 × 63) = 1/18

Hence option (d) is correct.