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Question

A bag contains 5 white, 7 red and 4 black balls. If four balls are drawn one by one with replacement, what is the probability that none is white ?

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Solution

Let A, B, C and D denote the events of not getting a white ball in the first, second, third and fourth draw respectively. Since the balls are drawn with replacement. Therefore, A, B, C and D are independent events such that
P(A)=P(B)=P(C)=P(D)
There are 16 balls out of which 11 are not white. Therefore,
P(A)=1116

Now, required probability = P(ABCD)
=P(A)P(B)P(C)P(D)=(1116)4

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