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

The repeated selections of defective bulbs from a box are Bernoulli trials. Let X denotes the number of defective bulbs out of a sample of 5 bulbs.
Probability of getting a defective bulb,
$\begin{array}{l}\mathrm{p}=\frac{10}{100}\\ =\frac{1}{10}\\ \therefore \mathrm{q}=1-\mathrm{p}\\ =1-\frac{1}{10}\\ =\frac{9}{10}\end{array}$
Clearly, X has a binomial distribution with n=5 and p=$\frac{1}{10}$.
$\begin{array}{l}\therefore \mathrm{P}(\mathrm{X}=\mathrm{x}){=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{x}}{\mathrm{q}}^{\mathrm{n}-\mathrm{x}}{\mathrm{p}}^{\mathrm{x}}\\ {=}^5{\mathrm{C}}_{\mathrm{x}}{\left(\frac{9}{10}\right)}^{5-\mathrm{x}}{\left(\frac{1}{10}\right)}^{\mathrm{x}}\end{array}$
P (none of the bulbs is defective) = P(X=0)
$\begin{array}{l}{=}^5{\mathrm{C}}_0\cdot {\left(\frac{9}{10}\right)}^5\\ =1\cdot {\left(\frac{9}{10}\right)}^5\\ ={\left(\frac{9}{10}\right)}^5\end{array}$