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.

• 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(X \geq 1)=1-P(X=0) $
$ =1-{ }^{10} C_0\left(\frac{9}{10}\right)^{10}=1-\frac{9^{10}}{10^{10}}$