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

$\begin{array}{1 1}(A)\;\bigg(\frac{9}{10}\bigg)^5\quad(B)\;\frac{1}{2}\bigg(\frac{9}{10}\bigg)^4\quad (C)\;\frac{1}{2}\bigg(\frac{9}{10}\bigg)^5\quad(D)\;\bigg(\frac{9}{10}\bigg)^5+\frac{1}{2}\bigg(\frac{9}{10}\bigg)^4\end{array}$

## 1 Answer Comment
A)
Toolbox:
• A random variable $X$ following Bianomial distribution with parameters $n$ and $p$ its probability distribution is given by probability of $r$ success
• $p(X=r)=c_{r}^{n} p^{r}q^{n-r}$
• Where $p$ is probability of success
• $q=1-p$ and $r=0,1,2,\dots,n$
In a sample of $100$ pens $10$ are defective.
$p$(defective pen)$=p=\large\frac{10}{100}=\frac{1}{10}$
$q=1-p=1-\large\frac{1}{10}=\frac{9}{10}$
There is sample of $5$ pens so $n=5$
$p$(Atmost one defective)$=p(X\leq 1)$
$p(X\leq 1)=p(X=0)+p(X=1)$
=$\large c^{5}_{0}\left(\frac{9}{10}\right)^{0} \left(\frac{9}{10}^{5}\right)+ c^{5}_{1}\left(\frac{1}{10}\right)^{1} \left(\frac{9}{10}^{4}\right)$
=$\large\left(\frac{9}{10}\right)^{5}+\frac{1}{2}\left(\frac{9}{10}\right)^{4}$

$D$ option is correct