Question

Three squares of a chessboard being chosen at random, what is the chance that two are of one color and one of another?

Answer 1

We find sample size $n\left( S \right)$ by finding the number of ways we can get 3 squares out 64 if we randomly choose. We denote the event of getting two squares of one colour and one of another as $A$. We see that $A$ can occur in two ways either we get 2 black squares and 1 white square $n\left( 2B,1W \right)$ and or we get 2 white squares and 1 black square $n\left( 2W,1B \right)$. We add the number of ways for both the cases to get the number of ways $A$ can occur as $n\left( A \right)$. The required probability is $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$.$$$$

Complete step-by-step solution:
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 given by
$P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$
We are given the question of 3 squares of the chess board being chosen at random. We know that there are a total of 64 squares in the chess board out of which 32 are in black colour and 32 are in white colour. We can select 3 squares out of 64 squares in ${}^{64}{{C}_{3}}$ ways which is the size of sample space. So we have
$n\left( S \right)={}^{64}{{C}_{3}}$
We are asked in the question to find the chance or probability of two the selected squares are of one colour and one of another colour. Let us denote the event of getting two squares of one colour and one square of another colour as $A$. We see that $A$ can occur in 2 two ways either we get 2 squares of black colour and 1 square of white colour (2B, 1 W) or we get 2 squares of white colour and 1 square of black colour (2W, 1B). $$$$
Case-1: We can get 2 squares of black colour out 32 in ${}^{32}{{C}_{2}}$ ways and we can get the 1 square of white out of 32 in ${}^{32}{{C}_{1}}$ ways. So by the rule of product the numbers of ways we can get outcomes for case-1 is,
$n\left( 2B,1W \right)={}^{32}{{C}_{2}}\times {}^{32}{{C}_{1}}$
Case-2: We can get 2 squares of white colour out 32 in ${}^{32}{{C}_{2}}$ ways and we can get the 1 square of black out of 32 in ${}^{32}{{C}_{1}}$ ways. So by the rule of product the numbers of ways we can get outcomes for case-2 is,
$n\left( 2W,1B \right)={}^{32}{{C}_{2}}\times {}^{32}{{C}_{1}}$
We use rule of sum and get the number of ways the event $A$ can occur as
$n\left( A \right)=n\left( 2B,1W \right)+n\left( 2W,1B \right)={}^{32}{{C}_{2}}\times {}^{32}{{C}_{1}}+{}^{32}{{C}_{2}}\times {}^{32}{{C}_{1}}=2\times {}^{32}{{C}_{2}}\times {}^{32}{{C}_{1}}$
So the probability that the event $A$ can happen is
$P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{2\times {}^{32}{{C}_{2}}\times {}^{32}{{C}_{1}}}{{}^{64}{{C}_{3}}}=\dfrac{2\times \dfrac{32\times 31}{2\times 1}\times 32}{\dfrac{64\times 63\times 62}{3\times 2\times 1}}=\dfrac{16}{21}$
The above probability is the required probability.

Note: We note that we could use combination because the positions of all squares in the chess board are distinct not identical and we have used the formula for combination${}^{n}{{C}_{r}}=\dfrac{n\left( n-1 \right)...\left( n-r+1 \right)}{r\left( r-1 \right)...1}$. We can alternatively find the probabilities for case-1 $P\left( 2B,1W \right)$ and case-2 $P\left( 2W,1B \right)$ and then add them to get the required probability $P\left( A \right)$ as they are mutually exclusive events.