Question

Two squares are chosen at random on a chessboard and let p denote the probability that they have exactly one side in common. Find 36p

• A. $2$
• B. $4$
• C. $5$
• D. $1$

Answer 1

The correct option is A $2$

In total, there are 64 squares on a chess board.

If we differentiate them with respect to the number of squares they are adjacent to, there are 3 types of squares:

Squares adjacent to 4 other squares

Out of the 64 squares, all the squares expect for the corner ones and edge ones are adjacent to 4 other squares.

There are a total of 36 of these squares.

P(selecting such a square) $=\frac{36}{64}$

Then, we need the probability of selecting a square that is adjacent to this square. Let's call this event A

$P\left(A\right)=463$

Squares adjacent to 3 other squares

The squares that lie on the edges of the chess board (neglecting the corner ones) are adjacent to 3 other squares.

There are 24 such squares.

P(selecting such a square) = $\frac{24}{64}$

Similarly, we need the probability of selecting a square that is adjacent to this square. Let this be event B

P(B)$=\frac{3}{63}$

Squares adjacent to 2 other squares

The corner squares are the ones.

There are 4 such squares.

P(selecting such a square) $=\frac{4}{64}$

Here also, let the event of selecting an adjacent square be C,

$P\left(C\right)=\frac{2}{63}$

Now, solving all the three cases and adding them will give us the answer.

Answer: P(selecting a square adjacent to 4 squares)*P(A) + P(selecting a square adjacent to 3 squares)*P(B) + P(selecting a square adjacent to 2 squares)*P(C)=($\frac{36}{64}$)$\ast$($\frac{4}{63}$)$+\left(\frac{24}{64}\right)\ast \left(\frac{3}{63}\right)+\left(\frac{4}{64}\right)\ast \left(\frac{2}{63}\right)=$$\frac{1}{18}$

therefore $36p=\frac{1}{18}\ast 36=2$