Suppose that 90\% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

.: Let $p$ be the. probability of success that people is right handed

$\Rightarrow p = \cfrac{{90}}{{100}} = \cfrac{9}{{10}}$

and $q = 1 - p = 1 - \cfrac{9}{{10}} = \cfrac{1}{{10}}$

$\therefore$ $X$ has a binomial distribution with $n = 10,p = \cfrac{9}{{10}},q = \cfrac{1}{{10}}$

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

RequiredProbability $= P$ ( most 6 of 10 people are right handed)

$= P(X \le 6) = 1 - P(7 \le X \le 10)$

$= 1 - \sum\limits_{r = 7}^{10} {^{10}{C_r}} {\left( {\cfrac{9}{{10}}} \right)^r}{\left( {\cfrac{1}{{10}}} \right)^{10 - r}} = 1 - \sum\limits_{r = 7}^{10} {^{10}{C_r}} {(0.9)^r}{(0.1)^{10 - r}}$