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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 any two are white ?
Solution
Let X be the number of white balls drawn when 4 balls are drawn with replacement.
X follows binomial distribution with n = 4.
\[p = \text { Probability for a white ball } = \frac{\text{ No of white balls }} {\text{ Total no . of balls}} \]
\[ = \frac{5}{20}\]
\[ = \frac{1}{4}\]
\[\text{ and } q = 1 - p = \frac{3}{4}\]
\[P(X = r) =^{4}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{4 - r} \]
\[\text{ Prob that any two are white} = P(X = 2) = ^{4}{}{C}_2 \left( \frac{1}{4} \right)^2 \left( \frac{3}{4} \right)^{4 - 2} = \frac{54}{256} = \frac{27}{128}\]
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