An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colors is

Solution :

Total number of cases = \(^9C_3\) = 84

Number of favourable cases = \(^3C_1\).\(^4C_1\).\(^2C_1\) = 24

\(\therefore\)  P = \(24\over 84\) = \(2\over 7\)

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