In a sample of $100$ pens $10$ are defective.
$p$(defective pen)$=p=\large\frac{10}{100}=\frac{1}{10}$
$q=1-p=1-\large\frac{1}{10}=\frac{9}{10}$
There is sample of $5$ pens so $n=5$
$p$(Atmost one defective)$=p(X\leq 1)$
$p(X\leq 1)=p(X=0)+p(X=1)$
=$\large c^{5}_{0}\left(\frac{9}{10}\right)^{0} \left(\frac{9}{10}^{5}\right)+ c^{5}_{1}\left(\frac{1}{10}\right)^{1} \left(\frac{9}{10}^{4}\right)$
=$\large\left(\frac{9}{10}\right)^{5}+\frac{1}{2}\left(\frac{9}{10}\right)^{4}$