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

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

$D$ → defective blub

$\bar D$ → Bulb is not defective

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

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

For 5 blubs P(none is defective) = ${^5C}_5(P(\bar D))^5$

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