A bag contains 3 red, 4 white and 5 blue balls. All balls are different. Two balls are drawn at random. The probability that they are of different colour is

The correct option is A

$\frac{47}{66}$

Out of 12 balls, two balls can be drawn in ${}^{12}{C}_2$ ways.

$\therefore$ Total number of elementary events, n(S) = ${}^{12}{C}_2=66$

We observe that at least one ball of each colour can be drawn in one of the following mutually exclusive ways :

(i) 1 red and 1 white

(ii) 1 red and 1 blue

(iii) 1 white and 1 blue

Thus, if we define three events, A. B and C as follows :

A = drawing 1 red and 1 white

B = drawing 1 red and 1 blue

C = drawing 1 white and 1 blue

Then, A, B and C are mutually exclusive events.

$\therefore$ Required probability $=P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)$

$=\frac{{}^3{C}_1{\times }^4{C}_1}{{}^{12}{C}_2}+\frac{{}^3{C}_1{\times }^5{C}_1}{{}^{12}{C}_2}+\frac{{}^4{C}_1{\times }^5{C}_1}{{}^{12}{C}_2}$

$=\frac{3\times 4}{66}+\frac{3\times 5}{66}+\frac{4\times 5}{66}=\frac{47}{66}$