## Solution :

\(E_1\) : Event that first drawn ball is red, second is blue and so on.

\(E_2\) : Event that first drawn ball is blue, second is red and so on.

\(\therefore\) P(\(E_1\)) = \(4\over 8\) \(\times\) \(4\over 7\) \(\times\) \(3\over 6\) \(\times\) \(3\over 5\) and

\(\therefore\) P(\(E_2\)) = \(4\over 8\) \(\times\) \(4\over 7\) \(\times\) \(3\over 6\) \(\times\) \(3\over 5\)

P(E) = P(\(E_1\)) + P(\(E_2\)) = 2 \(\times\) \(4\over 8\) \(\times\) \(4\over 7\) \(\times\) \(3\over 6\) \(\times\) \(3\over 5\) = \(6\over 35\)