Since $90\%$ of people are right handed, P (righthanded) $ = p = \large\frac{90}{100}=\frac{9}{10}$

$q=1-p=1-\large\frac{9}{10}=\frac{1}{10}$

Out of 10 selected persons, We want at most 6 right handed persons.

P (at most 6 right handed person) = $p(X\leq 6)$ =$1-P(X\geq 6)$ =$1-[p(X=7)+p(X=8)+p(X=10)]$

=$ 1 - \large\sum_{7}^{10} C^{10}_{r} $$\large\frac{9}{10}^{r} \frac{1}{10}^{10-r}$

=$1-[\large C^{10}_{7} p^{7}q^{3}+c^{10}_{8}p^{9}q^{1}+c^{10}_{10}p^{10}q_{10}]$

=$1-\large [C^{10}_{7}\left(\frac{9}{10}\right)^{7}\left(\frac{1}{10}\right)^{3}+C^{10}_{8}\left(\frac{9}{10}\right)^{8}\left(\frac{1}{10}\right)^{2}+C^{10}_{9}\left(\frac{9}{10}\right)^{9}\left(\frac{1}{10}\right)^{1}+C^{10}_{10}\left(\frac{9}{10}\right)^{10}\left(\frac{1}{10}\right)^{0}]$

=$1-\large[$$120 \large\frac{9^{7}}{10^{10}}$$+45\large \frac{9^{8}}{10^{10}}$$+10\large\frac{9^{9}}{10^{10}}$$+1\large\frac{9^{10}}{10^{10}}]$

=$1-\large\frac{9^{7}}{10^{10}}$$(120+45\times 9+10\times 9^{2}+9^{3})$

$=1-\large\frac{9^{7}}{10^{10}}$$[1578]$