Question

Three squares of normal chess board are chosen. The probability of getting two getting two squares of one colour and the other of different colour is .

• A. $\frac{16}{21}$
• B. $\frac{8}{21}$
• C. $\frac{8}{64\times 63\times 62}$
• D. $\frac{7}{21}$

Answer 1

The correct option is A $\frac{16}{21}$

In a chess board, there are 64 squares of which 32 are white and 32 are black.

Since 2 of one colour and 1 of other can be 2W, 1B, or 1W, 2B,

the number of ways is (${{32}_C}_1$ × ${{32}_C}_{12}$) × 2 and also, the number of ways of choosing any 3 boxes is ${{64}_C}_3$.

Hence, the required probability =$\frac{{{32}_C}_1\times {{32}_C}_2\times 2}{{{64}_C}_3}=\frac{16}{21}$