Question

A box has $100$ pens of which $10$ are defective. The probability that out of a sample of $5$ pens drawn one by one with replacement and atmost one is defective is

• A. $\frac{9}{10}$
• B. $\frac{1}{2}{\left(\frac{9}{10}\right)}^4$
• C. ${\left(\frac{9}{10}\right)}^5+\frac{1}{2}{\left(\frac{9}{10}\right)}^4$
• D. $\frac{1}{2}{\left(\frac{9}{10}\right)}^5$

Answer 1

The correct option is C ${\left(\frac{9}{10}\right)}^5+\frac{1}{2}{\left(\frac{9}{10}\right)}^4$

Out of total $100$ pens, $10$ are defective. Therfore, the probability that atmost $1$ pen is defective out of a sample of $5$ pens is sum of probabilities of no defective pen and $1$ defective pen.
$={\left(\frac{9}{10}\right)}^5+{\left(\frac{9}{10}\right)}^4\left(\frac{1}{10}\right){}^5{C}_1$
$={\left(\frac{9}{10}\right)}^5+{\left(\frac{9}{10}\right)}^4\left(\frac{1}{10}\right)\left(5\right)$
$={\left(\frac{9}{10}\right)}^5+\frac{1}{2}{\left(\frac{9}{10}\right)}^4$