Question

Mark the correct alternative in the following question:

A box contains 100 pens of which 10 are defective. What is the probability that out of a sample of 5 pens drawn one by one with replacement at most one is defective?

$\left(\mathrm{a}\right){\left(\frac{9}{10}\right)}^5\left(\mathrm{b}\right)\frac{1}{2}{\left(\frac{9}{10}\right)}^4\left(\mathrm{c}\right)\frac{1}{2}{\left(\frac{9}{10}\right)}^5\left(\mathrm{d}\right){\left(\frac{9}{10}\right)}^5+\frac{1}{2}{\left(\frac{9}{10}\right)}^4$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ p=\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{defective}\mathrm{pen}\mathrm{in}\mathrm{a}\mathrm{trial}=\frac{10}{100}=\frac{1}{10}\mathrm{and} \\ q=\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{good}\mathrm{pen}\mathrm{in}\mathrm{a}\mathrm{trial}=1-p=1-\frac{1}{10}=\frac{9}{10} \\ \mathrm{Let}X\mathrm{denote}\mathrm{a}\mathrm{success}\mathrm{of}\mathrm{getting}\mathrm{a}\mathrm{defective}\mathrm{pen}.\mathrm{Then}, \\ X\mathrm{follows}\mathrm{a}\mathrm{binomial}\mathrm{distribution}\mathrm{with}\mathrm{parameters}n=5\mathrm{and}p=\frac{1}{10} \\ \therefore \mathrm{P}\left(X=r\right)={}^5\mathrm{C}_r{p}^r{q}^{\left(5-r\right)}={}^5\mathrm{C}_r{\left(\frac{1}{10}\right)}^r{\left(\frac{9}{10}\right)}^{\left(5-r\right)}={}^5\mathrm{C}_r{\left(\frac{1}{10}\right)}^5{9}^{\left(5-r\right)}=\frac{{}^5\mathrm{C}_r{9}^{\left(5-r\right)}}{{10}^5},\mathrm{where}r=0,1,2,3,4,5 \\ \mathrm{Now}, \\ \mathrm{Required}\mathrm{probability}=\mathrm{P}\left(X\le 1\right) \\ =\mathrm{P}\left(X=0\right)+\mathrm{P}\left(X=1\right) \\ =\frac{{}^5\mathrm{C}_0{9}^5}{{10}^5}+\frac{{}^5\mathrm{C}_1{9}^4}{{10}^5} \\ =\frac{{}^5\mathrm{C}_0{9}^5}{{10}^5}+\frac{{}^5\mathrm{C}_1{9}^4}{{10}^5} \\ ={\left(\frac{9}{10}\right)}^5+\frac{5\times 9^4}{{10}^5} \\ ={\left(\frac{9}{10}\right)}^5+\frac{9^4}{2\times {10}^4} \\ ={\left(\frac{9}{10}\right)}^5+\frac{1}{2}{\left(\frac{9}{10}\right)}^4 \end{gathered}$$

Hence, the correct alternative is option (d).