Question

Two squares are chosen at random on a chessboard, the probability that they have a side in common is:

• A. $\frac{3}{32}$
• B. $\frac{1}{32}$
• C. $\frac{1}{18}$
• D. $noneofthese$

Answer 1

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

Two different squares can be chosen in 64 $\times$ 63 ways.
For each of the four corner squares, the favourable number of cases is 2.
For each of the 24 non-corner squares on all the four sides of the chessboard, the favourable number of cases is 3.
For each of the 36 remaining squares, the favourable number of cases is 4.
Thus, the total number of favourable cases
$=4\times 2+24\times 3+36\times 4=224$
Hence, the required probability = $\frac{224}{64\times 63}=\frac{1}{18}$