Question

Three squares of a chess board are selected at random. What is the probability of getting 2 squares of one colour and one of different colour ?

Answer 1

favourable ways = Two black and one white + two white and one black

Answer 2

Answer: The answer is $\dfrac{16}{21}.$

Step-by-step explanation: We know what is a chess board. It is a square-shaped board consisting of 64 squares, out of which 32 are black in colour and 32 are white in colour.

We are to select three cards, 2 of one colour and one of different colour. So, we can either choose 2 black and one white or 2 white and one black card.

Therefore, no. of ways in which we can do so is given by

$m=2\times \dfrac{32!}{2!(32-2)!}\times \dfrac{32!}{1!(32-1)!}\\\\\Rightarrow m=2\times \dfrac{32\times 31}{2\times 1}\times 32\\\\\Rightarrow m=32\times 31\times 32.$

Also, the total no. of ways is

$n=\dfrac{64!}{3!(64-3)!}\\\\\Rightarrow n=\dfrac{64\times 63\times 62}{3\times 2\times 1}=32\times 21\times 62.$

Therefore, required probability is

$p=\dfrac{m}{n}=\dfrac{32\times 31\times 32}{32\times 21\times 62}=\dfrac{16}{21}.$

Thus, the probability is $\dfrac{16}{21}.$