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

.: Let $p$ be the probability of defective item $= \cfrac{5}{{100}} = \cfrac{1}{{20}}$ .
$\therefore$ $q = 1 - \cfrac{1}{{20}} = \cfrac{{19}}{{20}}$

${\rm{X}}$ has a binomial distribution with $n = 10,p = \cfrac{1}{{20}},q = \cfrac{{19}}{{20}}$

$\therefore \quad P(X = r){ = ^n}{C_r}{q^{n - r}}{(p)^r}$

$\therefore P$ (not more than one defective item) $= P(X = 0) + P(X = 1)$

${ = ^{10}}{C_0}{q^{10}}{p^0}{ + ^{10}}{C_1}{q^9}{p^1} = (1){\left( {\cfrac{{19}}{{20}}} \right)^{10}}(1) + (10){\left( {\cfrac{{19}}{{20}}} \right)^9}\left( {\cfrac{1}{{20}}} \right)$

$= {\left( {\cfrac{{19}}{{20}}} \right)^9}\left[ {\cfrac{{19}}{{20}} + \cfrac{{10}}{{20}}} \right] = {\left( {\cfrac{{19}}{{20}}} \right)^9} \times \cfrac{{29}}{{20}}$