From a group of 10 persons consisting of 5 lawyers, 3 doctors and 2 engineers, four persons are selected at random. The probability that selection contains one of each category is

Solution :

n(S) = \(^{10}C_4\) = 210

n(E)= \(^5C_2 \times ^3C_1 \times ^2C_1\) + \(^5C_1 \times ^3C_2 \times ^2C_1\) + \(^5C_1 \times ^3C_1 \times ^2C_2\) = 105

\(\therefore\) P(E) = \(105\over 210\) = \(1\over 2\)

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