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

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

D → defective bulb

$\overline{D}$ → bulb is not defective 

P(D) = $\frac{10}{100} = \frac{1}{10}$

$P(\overline{D}) = 1 - P(D) = \frac{9}{10}$

for 5 bulbs P(none is defective) = ${}^5 C_5 (P(\overline{D}))^5$

$=\frac{9^5}{10^5}$