Suppose that 90% of people are right-handed.

A person can be either right-handed or left-handed.

It is given that 90% of the people are right-handed.

$\therefore p=\mathrm{P}($ right-handed $)=\frac{9}{10}$

$q=\mathrm{P}($ 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,

$\sum_{r=7}^{10}{ }^{10} C_{r} p^{\prime} q^{n-r}=\sum_{r=7}^{10}{ }^{10} C_{r}\left(\frac{9}{10}\right)^{r}\left(\frac{1}{10}\right)^{10-r}$

Therefore, the probability that at most 6 people are right-handed

= 1 − P (more than 6 are right-handed)

$=1-\sum_{r=7}^{10}{ }^{10} C_{r}(0.9)^{r}(0.1)^{10-r}$