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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