2016 Pascal Problem 16

An integer from 10 to 99 inclusive is randomly chosen so that each such integer is equally likely to be chosen. The probability that at least one digit of the chosen integer is a 6 is

\textbf{(A)}\ \frac{1}{5}\quad \textbf{(B)}\ \frac{1}{10}\quad \textbf{(C)}\ \frac{1}{9}\quad \textbf{(D)}\ \frac{19}{90}\quad \textbf{(E)}\ \frac{19}{89}

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Pascal contests here.