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

\(\frac{10^{10}-9^{10}}{10^{10}}\)

Probability of defective eggs =10%

$P=\frac{10}{100}=\frac{1}{10}$

Let the (Probability of good eggs) = q

$q=1-p=1-\frac{10}{100}=\frac{100-10}{100}=\frac{90}{100}=\frac{9}{10}$

(Probability of at least one egg defective out of 10) ⇒ p(1) + p(2) + p(3)+...

$=[p(0)+p(1)+p(2)+...+p(10)−p(0)$

$=1-P(0)=1-(\frac{9}{10})^{10}⇒\frac{10^{10}-9^{10}}{10^{10}}$.

So option 1 is correct.