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Q.
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
Probability - Part 2
Solution:
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)
$= P\left(X \ge 1\right) = 1-P\left(X = 0\right)$
$= 1-\,{}^{10}C_{0}\left(\frac{9}{10}\right)^{10} = 1-\frac{9^{10}}{10^{10}}$