(a) 47/66
Out of 12 balls, two balls can be drawn in 12C2 ways.
∴ Total number of elementary events, n(S) = 12C2 = 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.
∴ Required probability = P(A ∪ B ∪ C)
= P(A) + P(B) + P(C)
=
=
=
C18×C13×C19C3