Question

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

Answer 1

Let $X$: be the number of defective eggs.

Picking eggs with replacement is a Bernoulli
Trial.
So, $X$ has binomial distribution
$P(X=x){=}^n{C}_x{q}^{n-x}{p}^x,x=0,1,2,\dots ,n$

Here,
$n=$ No. of eggs picked=$10$
$p=$ Probability of getting defective egg $=10\%$

$\Rightarrow p=\frac{10}{100}=\frac{1}{10}$

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

Putting the value of $n,p\&q,$ we get

$P(X=x){=}^{10}{C}_x{\left(\frac{1}{10}\right)}^x{\left(\frac{9}{10}\right)}^{10-x}$ ...$\left(1\right)$

Probability that at least one egg is defective

$=1-P\left(\text{getting}0\text{defective eggs}\right)$

$=1-P(X=0)$
Putting the value of $x=0\text{in}\left(1\right)$

$=1{-}^{10}{C}_0{\left(\frac{1}{10}\right)}^0{\left(\frac{9}{10}\right)}^{10-0}$

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

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

Therefore, probability that there is at least
one defective egg$=1-{\left(\frac{9}{10}\right)}^{10}$