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Question

A bag contains 7 red, 5 white and 8 black balls. If four balls are drawn one by one with replacement, what is the probability that
(i) none is white?
(ii) all are white?
(iii) any two are white?

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Solution

Let X be the number of white balls drawn when 4 balls are drawn with replacement.
X follows binomial distribution with n = 4.

p = Probability for a white ball = No of white ballsTotal no. of balls = 520=14and q = 1-p = 34P(X=r) = Cr414r344-r(i) Prob that none is white = P(X=0) =C04140344-0= 81256ii Prob that all are white = P(X=4) = C44144344-4= 1256iii Prob that any two are white = P(X=2) = C24142344-2= 54256=27128

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