Question

Ten eggs are drawn succesively with replacement from a lot of $100$ eggs containing $10$ rotten eggs. Then the probability that there is at least one rotten egg is:

• A. $\frac{9^9}{{10}^{10}}$
• B. ${\left(\frac{9}{10}\right)}^{10}$
• C. $1-{\left(\frac{9}{10}\right)}^{10}$
• D. $1-{\left(\frac{1}{10}\right)}^{10}$

Answer 1

The correct option is C $1-{\left(\frac{9}{10}\right)}^{10}$

Let $X$ denotes the number of rotten eggs in the $10$ eggs drawn. Since the drawing is done with replacement, the trials are Bernoulli trials. Clearly, $X$ has the binomial distribution with $n=100\text{and}p=\frac{10}{100}=\frac{1}{10}$
Therefore $q=1-p=\frac{9}{10}$
Now $P$(at least one rotten egg) $=P(X\ge 1)=1-P(X=0)=1{-}^{10}{C}_0{\left(\frac{9}{10}\right)}^{10}=1-\frac{9^{10}}{{10}^{10}}$