Question

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 ?

Answer 1

A person can be either right-handed or left-handed.
It is given that 90% of the people are right-handed.
$\therefore p=P(right-handed)=\frac{9}{10}$
$q=P(left-handed)=1-\frac{9}{10}=\frac{1}{10}$
Using binomial distribution, the probability that more than 6 people are right-handed is given by,
${\sum }_{r=7}^{10}{}^{10}{C}_r{p}^r{q}^{n-r}={\sum }_{r=7}^{10}{}^{10}{C}_r{\left(\frac{9}{10}\right)}^r{\left(\frac{1}{10}\right)}^{10-r}$
Therefore, the probability that at most 6 people are right-handed
$=1-P(morethan6areright-handed)$
$=1-{\sum }_{r=7}^{10}{}^{10}{C}_r{\left(0.9\right)}^r{\left(0.1\right)}^{10-r}$