There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will not include more than 1 defective item?

$\frac{29}{20}\left(\frac{19}{20}\right)^9$

The correct answer is Option (2) → $\frac{29}{20}\left(\frac{19}{20}\right)^9$

Let E be the event of 'item is defective', then

$p = P(E) = 5\% =\frac{5}{100}=\frac{1}{20}$, so $q = 1-\frac{1}{20}=\frac{19}{20}$.

As a sample of 10 items is taken, so there are 10 trials i.e. $n = 10$.

Thus, we have a binomial distribution with $p=\frac{1}{20},q=\frac{19}{20}$ and $n=10$.

Required probability = P(not more than 1) = $P(X ≤ 1) = P(0) + P(1)$

$= {^{10}C}_0 q^{10} + {^{10}C}_1 p q^9 = (q + 10p) q^9$

$=\left(\frac{19}{20}+10.\frac{1}{20}\right)\left(\frac{19}{20}\right)^9=\frac{29}{20}\left(\frac{19}{20}\right)^9$