Question

Three squares of normal chess board are chosen. The probability of getting two squares of one colour and the other of different colour is:
(a) $ \dfrac{16}{21} $ .
(b) $ \dfrac{8}{21} $
(c) $ \dfrac{8}{64\times 63\times 62} $
(d) $ \dfrac{7}{21} $

Answer 1

Hint:
We start solving the problem by recalling the fact that a normal chessboard has 64 squares with 32 squares of one color (black) and 32 squares of another color (white). We then recall the fact that the number of ways of choosing ‘r’ items out of ‘n’ $ \left( n\ge r \right) $ items is $ {}^{n}{{C}_{r}} $ ways. We then use this fact in the formula of probability $ \dfrac{\text{no}\text{. of ways to choose 2 black square and 1 white square + no}\text{. of ways to choose 1 black square and 2 white square}}{\text{no}\text{. of ways to choose 3 squares from the 64 squares on the chess board}} $ to get the required answer.
Complete step by step answer:
According to the problem, we are given that three squares of a normal chessboard are chosen. We have to find the probability of getting two squares of one color and the other of a different color.
We know that a normal chessboard has 64 squares with 32 squares of one color (black) and 32 squares of another colour (white).
So, we need to find the probability that two of the three chosen squares are black and the other is white or two of the three chosen squares are white and the other is black.
We know that the number of ways of choosing ‘r’ items out of ‘n’ $ \left( n\ge r \right) $ items is $ {}^{n}{{C}_{r}} $ ways.
So, the required probability is defined as $ \dfrac{\text{no}\text{. of ways to choose 2 black square and 1 white square + no}\text{. of ways to choose 1 black square and 2 white square}}{\text{no}\text{. of ways to choose 3 squares from the 64 squares on the chess board}} $.
  $ \Rightarrow probability=\dfrac{\left( {}^{32}{{C}_{2}}\times {}^{32}{{C}_{1}} \right)+\left( {}^{32}{{C}_{1}}\times {}^{32}{{C}_{2}} \right)}{{}^{64}{{C}_{3}}} $ .
We know that $ {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} $ and $ n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1 $ .
  $ \Rightarrow probability=\dfrac{\left( \left( \dfrac{32!}{2!30!} \right)\times \left( \dfrac{32!}{1!31!} \right) \right)+\left( \left( \dfrac{32!}{1!31!} \right)\times \left( \dfrac{32!}{2!30!} \right) \right)}{\left( \dfrac{64!}{3!61!} \right)} $ .
  $ \Rightarrow probability=\dfrac{\left( \left( \dfrac{32\times 31}{2\times 1} \right)\times \left( \dfrac{32}{1} \right) \right)+\left( \left( \dfrac{32}{1} \right)\times \left( \dfrac{32\times 31}{2\times 1} \right) \right)}{\left( \dfrac{64\times 63\times 62}{3\times 2\times 1} \right)} $ .
  $ \Rightarrow probability=\dfrac{\left( \left( 16\times 31 \right)\times \left( 32 \right) \right)+\left( \left( 32 \right)\times \left( 16\times 31 \right) \right)}{\left( 32\times 21\times 62 \right)} $ .
  $ \Rightarrow probability=\dfrac{\left( 16\times 31\times 32 \right)\times 2}{\left( 32\times 21\times 62 \right)} $ .
  $ \Rightarrow probability=\dfrac{16}{21} $ .
So, we have found the required probability as $ \dfrac{16}{21} $ .
∴ The correct option for the given problem is (a).

Note:
 Whenever we get this type of problem, we first find the favorable cases that satisfy the given condition. We should not make calculation mistakes while solving this problem. We should not forget to take the case of choosing two whites and one black as this is the most common mistake done by students. Similarly, we can expect problems to choose the three squares of the same color from all the squares of a normal chessboard.