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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