Ten bulbs are drawn successively with replacement from a lot containing 10% defective bulbs. The probability that there is at least one defective bulb is :

$1-(9^{10}/10^{10})$

The correct answer is Option (3) → $1-(9^{10}/10^{10})$

P(atleast one defective bulb) = 1 - P(no defective bulb)

$=1-{^{10}C}_0×(100\%-10\%)^{10}$

$=1-(\frac{9}{10})^{10}$