Question

Ten eggs are drawn successively with replacement from a lot containing $10\%$ defective eggs. Then, the probability that there is atleast one defective egg is

• A. $1-\frac{7^{10}}{10^{10}}$
• B. $1+\frac{7^{10}}{10^{10}}$
• C. $1+\frac{9^{10}}{10^{10}}$
• D. $1-\frac{9^{10}}{10^{10}}$

Answer 1

Answer: (D)

Let $X$ denote the number of defective 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 $n=10$ and
$p=\frac{10}{100}=\frac{1}{10}$
Therefore, $q=1-p=\frac{9}{10}$
Now, $P$(atleast one defective egg)
$=P\left(X\ge 1\right)=1-P\left(X=0\right)$
$=1-{}^{10}{C}_0{\left(\frac{9}{10}\right)}^{10}=1-\frac{9^{10}}{10^{10}}$