On a chess board if two squares are chosen at random, what is the probability that they haven't a side in common?

Explanation

Let S be the sample space and E the event that two square selected at random have a side in common.

Along the rows we have 7 pairs of squares which have a side in common,
viz, `A_1A_2, A_2A_3, A_2A_3, A_3A_4, ..........A_7A_8, A_8A_9.`
All these pairs can also be described as `A_iA_(i+1), A_(i+1) A_(i+2),i=1, 2, ....7.`
As there are 8 rows, no. of these squares having a side in common `=7xx8=56`
By the same argument, along the columns, number of squares having a side in common = 56.
`therefore` Total no. of squares `=2xx56=112`
The no. of ways of choosing two squares `=""^(64)C_2`
Probability `P=(n(E))/(n(S))=112/(""^64C_2)=(112xx2)/(64xx63)=7/126=1/18`
Required Probability=P'= `1-P= 1-1/18=17/18`