Question

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

Answer 1

Hint: You can observe in a chessboard that there are types of squares when we are looking for common sides. The first type is with 4 adjacent squares, the second is with 3 and the third is with 4 adjacent squares. Find the probability that two squares have a common side each type and add.

Complete step-by-step answer:
We know from definition of probability that if there is $n\left( A \right)$ number of ways of event $A$ occurring and $n\left( S \right)$ is the size of the sample space then the probability of the event $A$ occurring is $\dfrac{n\left( A \right)}{n\left( S \right)}$.\[\]

We know there are 64 squares in a chess board alternatively in black and white. We have to select a square at random. We observe that there are three types of square with respect to adjacency in the chess board.\[\]
Type-I: \[\]
The first type of squares is squares who have 4 other squares adjacent to them. We see them they are not the edge of the chessboard. The total numbers of such square is 36. So probability of selecting first type of square is $\dfrac{36}{64}$.\[\]
After selection of one first type square the number squares left are 63. So the probability of selecting an adjacent square of for the selected first type is $\dfrac{4}{63}$.\[\]
So the probability of first type square being adjacent to another square is ${{P}_{1}}=\dfrac{36}{64}\times \dfrac{4}{63}$\[\]
Type-II: \[\]
The second type of squares is squares who have 3 other squares adjacent to them. We see them they are only on the edge of the chessboard but not in the corner. The total numbers of such square is 24. So probability of selecting second type of square is $\dfrac{24}{64}$.\[\]
We see that after selection of one second type square the number squares left are 63. So the probability of selecting an adjacent square for the selected second type is $\dfrac{3}{63}$.\[\]
So the probability of second type square being adjacent to another square is ${{P}_{2}}=\dfrac{24}{64}\times \dfrac{3}{63}$\[\]
Type-III
The third type of squares is squares who have 2 other squares adjacent to them. We see them they are only at the corner of the chess board. The total numbers of such squares is 4. So probability of selecting third type of square is $\dfrac{4}{64}$.\[\]
We see that after selection of one third type of square the number squares left are 63. So the probability of selecting an adjacent square for the selected third type is $\dfrac{2}{63}$.\[\]
So the probability of second type square being adjacent to another square is ${{P}_{2}}=\dfrac{4}{64}\times \dfrac{2}{63}$\[\]
So the sum total probability that two squares have a side common is \[ {{P}_{1}}+{{P}_{2}}+{{P}_{3}}=\dfrac{36}{64}\times \dfrac{4}{63}+\dfrac{24}{64}\times \dfrac{3}{63}+\dfrac{4}{64}\times \dfrac{2}{63}=\dfrac{1}{18} \]
 The required probability is $\dfrac{1}{18}$. \[\]

Note: The key to solving this question is distinguishing the different types of squares. It is also to be noted that the selection of three different types of square is mutually exclusive and that is the reason we could add the probabilities.