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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

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Solution

(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)
= C13×C14C212+C13×C15C212+C14×C15C212
= 3×466+3×566+4×566

= 1266+1566+2056=4766

C18×C13×C19C3

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