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

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) $\frac{9}{10}$

Solution:

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

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

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

Clearly, $\mathrm{X}$ has a binomial distribution with $n=5$ and $p=\frac{1}{10}$

$\therefore \mathrm{P}(\mathrm{X}=x)={ }^{n} \mathrm{C}_{x} q^{n-\mathrm{x}} p^{x}={ }^{5} \mathrm{C}_{x}\left(\frac{9}{10}\right)^{5-x}\left(\frac{1}{10}\right)^{x}$

P (none of the bulbs is defective) = P(X = 0)

$={ }^{5} \mathrm{C}_{0} \cdot\left(\frac{9}{10}\right)^{5}$

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

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