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

Answer: (C)

We have
$p=$ probability that a bulb is defective $=\frac{10}{100}=\frac{1}{10}$
$\therefore q=1-p=\frac{9}{10}$
Let $X$ be the number of defective bulbs in a sample of 5 bulbs.
$\therefore$ Required probability
$=P(x=0)+P(x=1)$
$={ }^{5} C_{0}\left(\frac{1}{10}\right)^{0}\left(\frac{9}{10}\right)^{5}+{ }^{5} C_{1}\left(\frac{1}{10}\right)^{1}\left(\frac{9}{10}\right)^{4}$
$=\left(\frac{9}{10}\right)^{5}+5 \times \frac{1}{10}\left(\frac{9}{10}\right)^{4}$
$=\left(\frac{9}{10}\right)^{5}+\frac{1}{2}\left(\frac{9}{10}\right)^{4}$