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Question
A box has 20 pens of which 2 are defective. Calculate the probability that out of 5 pens drawn one by one with replacement, at most 2 are defective.
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Solution
Let p denote the probability of drawing a defective pen. Then,
\[p = \frac{2}{20} = \frac{1}{10}\]
\[ \Rightarrow q = 1 - p = 1 - \frac{1}{10} = \frac{9}{10}\]
Let X denote the number of defective pens drawn. Then, X is a binomial variate with parameter n = 5 and \[p = \frac{1}{10}\] .
Now, P(X = r) = Probability of drawing r defective pens = \[^{5}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^{5 - r} , r = 0, 1, 2, 3, 4, 5\]
∴ Probability of drawing at most 2 defective pens
= P(X ≤ 2)
= P(X = 0) + P(X = 1) + P(X = 2)
\[= ^{5}{}{C}_0 \left( \frac{1}{10} \right)^0 \left( \frac{9}{10} \right)^5 + ^{5}{C}_1 \left( \frac{1}{10} \right)^1 \left( \frac{9}{10} \right)^4 + ^{5}{C}_2 \left( \frac{1}{10} \right)^2 \left( \frac{9}{10} \right)^3 \]
\[ = \left( \frac{9}{10} \right)^3 \left( \frac{81}{100} + 5 \times \frac{9}{100} + \frac{10}{100} \right)\]
\[ = \frac{729}{1000} \times \frac{136}{100}\]
\[ = 0 . 99144\]
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