Question

Two squares are chosen from the squares of an ordinary chess board. It is given that the selected squares do not belong to the same row or column. The probability that they have a side in common

• A. $\frac{25}{49}$
• B. $\frac{32}{49}$
• C. $\frac{1}{18}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{1}{18}$

In $64$ squares, there are:

(1)$4$ at-corner squares, each has ONLY 2 squares each having a side in common with...

(2) $6\times 4=24$ side squares, each has ONLY 3 squares such that each has a side in common with...

(3) $6\times 6=36$ inner squares, each has 4 squares such that each has a side in common with...

So we have the calculation:

$P=\frac{4}{64}\times \frac{2}{63}+\frac{24}{64}\times \frac{3}{63}+\frac{36}{64}\times \frac{4}{63}$

$P=\frac{1}{18}$