Two squares are chosen at random on a chessboard (see figure). The probability that they have a side in common is :

Solution

Maths sir · 4 steps

Count the pairs of side-sharing squares row by row and column by column, then divide by all pairs.

1. Total outcomes. Choosing 2 squares out of 64, order does not matter: C(64, 2) = (64 × 63)/2 = 2016.
2. Horizontal neighbours. In one row of 8 squares there are 7 side-by-side pairs. With 8 rows, that gives 8 × 7 = 56 pairs.
3. Vertical neighbours. The same count works for columns: 8 columns with 7 pairs each, so another 56. In total 56 + 56 = 112 pairs share a side.

   P = 112/2016 = 1/18

4. So. The probability is 1/18, option (b).

The idea to keep. Keep the numerator and denominator counted the same way. If the total is unordered pairs C(n, 2), count each favourable pair only once too.

Answer from the NTA final answer key. Solution by PrepWiser.