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.

Answer 1

Let p denote the probability of drawing a defective pen. Then,
$$\begin{gathered} p=\frac{2}{20}=\frac{1}{10} \\ \Rightarrow q=1-p=1-\frac{1}{10}=\frac{9}{10} \end{gathered}$$
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)
$$\begin{gathered} ={}^{5}C_0{\left(\frac{1}{10}\right)}^0{\left(\frac{9}{10}\right)}^5+{}^{5}C_1{\left(\frac{1}{10}\right)}^1{\left(\frac{9}{10}\right)}^4+{}^{5}C_2{\left(\frac{1}{10}\right)}^2{\left(\frac{9}{10}\right)}^3 \\ ={\left(\frac{9}{10}\right)}^3\left(\frac{81}{100}+5\times \frac{9}{100}+\frac{10}{100}\right) \\ =\frac{729}{1000}\times \frac{136}{100} \\ =0.99144 \end{gathered}$$