An integer from 100 through 999, inclusive, is to be chosen at random. what is the probability that the number chosen will have 0 as at least 1 digit?

So, we know that for ever 100 digits, there are 10 digits that have a 0 in the one's place (100, 110, 120, 130, 140, 150, 160, 170, 180, and 190), and 9 different numbers with a 0 in the ten's place (101, 102, 103, 104, 105, 106, 107, 108, and 109). You do not count any number twice. That's 19/100. Now, that's an answer for ever 100, until you get to the 900's, since there's only 99 in that place.

We know there are 899 options, so that's our denominator. Then, 19+19+19+19+19+19+19+19+19=171.

The answer is 171/899, I believe.