Easy word prob but still need hlep!

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

I dont remember how to do this the easy, right way
the answr is 171/900
please explain

There are 9*10=90 0's in the first (units) digit, and 9*10=90 0's in the in the second (tens) digit. There are 9 cases [100,200,...,900] of double 0's. By inclusion-exclusion principle, the probability of having 0 at least one digit is

90/900 + 90/900 - 9/900 = 171/900.

we have 9 times 19 "0" in the respecified range, so we have a total of 9 times 19 =171 numbers that
have a 0 at least in 1 digit and there are 900 total numbers so the probability is
171/900=19%