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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
Options
47/66
10/33
1/3
1
Solution
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)
=\[\frac{^{3}{}{C}_1 \times ^{4}{}{C}_1}{^{12}{}{C}_2} + \frac{^{3}{}{C}_1 \times ^{5}{}{C}_1}{^{12}{}{C}_2} + \frac{^{4}{}{C}_1 \times ^{5}{}{C}_1}{^{12}{}{C}_2}\]
= \[\frac{3 \times 4}{66} + \frac{3 \times 5}{66} + \frac{4 \times 5}{66}\]
= \[\frac{12}{66} + \frac{15}{66} + \frac{20}{56} = \frac{47}{66}\]
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