Question

Two squares are chosen at random on a chessboard. What is the probability that they have a side in common?

• A. $\frac{1}{18}$
• B. $\frac{1}{17}$
• C. $\frac{1}{16}$
• D. $\frac{1}{15}$

Answer 1

Answer: (A)

$\frac{1}{18}$

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.

$P\left(selectingsuchasquare\right)=\frac{36}{64}$

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

$P\left(A\right)=\frac{4}{63}$

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.

$P\left(selectingsuchasquare\right)=\frac{24}{64}$

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

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

Squares adjacent to $2$ other squares:

The corner squares are the ones.

There are $4$ such squares.

$P\left(selectingsuchasquare\right)=\frac{4}{64}$

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.

$P\left(selectingasquareadjacentto4squares\right)\times P\left(A\right)+P\left(selectingasquareadjacentto3squares\right)\times P\left(B\right)+P\left(selectingasquareadjacentto2squares\right)\times P\left(C\right)=\frac{36}{64}\times \frac{4}{63}+\frac{24}{64}\times \frac{3}{63}+\frac{4}{64}\times \frac{2}{63}=\frac{1}{18}$

Thus the probability that they have a side in common is $\frac{1}{18}$.