# 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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