Question

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
(a) 47/66
(b) 10/33
(c) 1/3
(d) 1

Answer 1

(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)
= $\frac{{}^{3}C_1\times {}^{4}C_1}{{}^{12}C_2}+\frac{{}^{3}C_1\times {}^{5}C_1}{{}^{12}C_2}+\frac{{}^{4}C_1\times {}^{5}C_1}{{}^{12}C_2}$
= $\frac{3\times 4}{66}+\frac{3\times 5}{66}+\frac{4\times 5}{66}$

= $\frac{12}{66}+\frac{15}{66}+\frac{20}{56}=\frac{47}{66}$

C18×C13×C19C3