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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, $p=\frac{2}{20}=\frac{1}{10}\phantom{\rule{0ex}{0ex}}⇒q=1-p=1-\frac{1}{10}=\frac{9}{10}$ 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) $={}^{5}C_{0}{\left(\frac{1}{10}\right)}^{0}{\left(\frac{9}{10}\right)}^{5}+{}^{5}C_{1}{\left(\frac{1}{10}\right)}^{1}{\left(\frac{9}{10}\right)}^{4}+{}^{5}C_{2}{\left(\frac{1}{10}\right)}^{2}{\left(\frac{9}{10}\right)}^{3}\phantom{\rule{0ex}{0ex}}={\left(\frac{9}{10}\right)}^{3}\left(\frac{81}{100}+5×\frac{9}{100}+\frac{10}{100}\right)\phantom{\rule{0ex}{0ex}}=\frac{729}{1000}×\frac{136}{100}\phantom{\rule{0ex}{0ex}}=0.99144$

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