Question

In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is

(a) ${10}^{-1}$

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

Answer 1

The repeated selection of defective bulbs from a box are Bernoulli trails. Let X denote the number of defective bulbs out of a sample of 5 bulbs.

Now, probability of getting a defective bulb, p= $\frac{10}{100}=\frac{1}{10}$

and q=1-p=1- $\frac{1}{100}=\frac{9}{10}$

Clearly, X binomial distribution with n=5,

p= $\frac{1}{100}andq=\frac{9}{10}$

$\therefore P(X=r)={}^n{C}_r{p}^r{q}^{n-r}={}^5{C}_r.{\left(\frac{1}{10}\right)}^r{\left(\frac{9}{10}\right)}^{5-r}$

P (none of the bulbs is defective) = $P(X=0)={}^5{C}_0.{\left(\frac{1}{10}\right)}^0{\left(\frac{9}{10}\right)}^5$

=$1\times 1\times {\left(\frac{9}{10}\right)}^5={\left(\frac{9}{10}\right)}^5$

Here, (c) is the correct option.