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
(A) $\frac{1}{5}$
(B) $\frac{1}{10}$
(C) $\frac{1}{9}$
(D) $\frac{19}{90}$
(E) $\frac{19}{89}$
From 10 to 99 inclusive, there is a total of 90 integers. (Note that $90=99-10+1$.)

If an integer in this range includes the digit 6, this digit is either the ones (units) digit or the tens digit.

The integers in this range with a ones (units) digit of 6 are 16, 26, 36, 46, 56, 66, 76, 86, 96 .

The integers in this range with a tens digit of 6 are $60,61,62,63,64,65,66,67,68,69$.

In total, there are 18 such integers. (Notice that 66 is in both lists and $9+10-1=18$.)

Therefore, the probability that a randomly chosen integer from 10 to 99 inclusive includes the digit 6 is $\frac{18}{90}=\frac{1}{5}$.

ANSWER: (A)