Question

Three squares of a chess board are chosen at random, the probability that two are of one color and one of another is

• A. $\frac{16}{21}$
• B. $\frac{8}{21}$
• C. $\frac{32}{12}$
• D. None of these

Answer 1

The correct option is A

$\frac{16}{21}$

Explanation for the correct answer:

There are $64$ squares on a chess board. $32$of which are white and $32$are black.

The number of ways of selecting three squares on a chess board are ${}^{64}{C}_3$

Now there are two ways of selecting $2$ squares of same color and $1$ square of the other square.

We can choose $2$ white squares and $1$ black square or $2$ black squares and $1$ white square.

This can be done in ${}^{32}{C}_2\times {}^{32}{C}_1+{}^{32}{C}_2\times {}^{32}{C}_1$ ways.

Hence, the $\mathbf{Required}\mathbf{}\mathbf{probability}\mathbf{=}\frac{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{favorable}\mathbf{}\mathbf{outcomes}}{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{total}\mathbf{}\mathbf{outcomes}}$

$\Rightarrow$ $\mathrm{Required}\mathrm{probability}=\frac{{}^{32}{C}_2\times {}^{32}{C}_1+{}^{32}{C}_2\times {}^{32}{C}_1}{{}^{64}{C}_3}$

$\Rightarrow$ $\mathrm{Required}\mathrm{probability}=\frac{2\times \left({\displaystyle \frac{32!}{2!30!}}\times {\displaystyle \frac{32!}{1!31!}}\right)}{{\displaystyle \frac{64!}{3!61!}}}$

$\Rightarrow$ $\mathrm{Required}\mathrm{probability}=\frac{\left({\displaystyle 32\times 31}\times {\displaystyle 32}\right)\times 6}{{\displaystyle 64\times 63\times 62}}$

$\Rightarrow$ $\mathrm{Required}\mathrm{probability}=\frac{{\displaystyle 16}}{{\displaystyle 21}}$

Hence, the probability of choosing $2$ squares of same color and $1$ square of the other square from a chessboard is $\frac{16}{21}$.

Hence, option A is the correct answer.