A bag contains 4 red and 4 blue balls. Four balls are drawn one by one from the bag, then find the probability that the drawn balls are in alternate color.

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 \(\times\) \(4\over 8\) \(\times\) \(4\over 7\) \(\times\) \(3\over 6\) \(\times\) \(3\over 5\) = \(6\over 35\)


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