Question

Four unit squares are chosen at random on a chessboard. What is the probability that three of them are of one colour and fourth is of opposite colour?

• A. $\frac{620}{1281}$
• B. $\frac{320}{1281}$
• C. $\frac{640}{1281}$
• D. None of the above

Answer 1

Answer: (C)

$\frac{640}{1281}$

The number of ways of choosing $4$ squares from $64$ is $64C_4$.

The possible split ups as far as colour is concerned and the number of ways of selecting the square are listed below:

White $=0$, Black $=4$, No. of Ways $=32C_0\times 32C_4$

White $=1$, Black $=3$, No. of Ways $=32C_1\times 32C_3$

White $=2$, Black $=2$, No. of Ways $=32C_2\times 32C_2$

White $=3$, Black $=1$, No. of Ways $=32C_3\times 32C_1$

White $=4$, Black $=0$, No. of Ways $=32C_4\times 32C_0$

The number of ways in which $3$ squares are of one colour and fourth is of opposite colour is:

$P\left(A\right)=2\times 32C_1\times 32C_3$

$=2\times 32\times \frac{32\times 31\times 30}{3\times 2\times 1}=317440$

The total number of ways of selecting $4$ squares is: $P\left(B\right)=64C_4=\frac{64\times 63\times 62\times 61}{4\times 3\times 2\times 1}=635376$

The required probability$=\frac{P\left(A\right)}{P\left(B\right)}$

$=\frac{317440}{635376}$

$=\frac{640}{1281}$

Therefore the probability that three of them are of one colour and fourth is of opposite colour is $\frac{640}{1281}$.