Just one 0

There are:

a) 9 numbers between 1 and 99 with exactly one 0 digit (as only the second digit can be 0 and we can choose the first digit 9 ways, we get 10, 20, ... , 80, 90)

b) 162 numbers between 100 and 999 with exactly one 0 digit (as we can choose the first digit 9 ways, the place of the 0 digit 2 ways (either the second or the third digit can be 0) and the other non-zero digit 9 ways (from 1 to 9), 9 × 2 × 9 = 162)

c) 243 numbers between 1000 and 1999 with exactly one 0 digit (the reasoning is similar to b), but here we can choose the first digit 1 way (1), the place of the 0 digit 3 ways (either 2nd, 3rd or 4th) and the other two non-zero digit 9 ways (from 1 to 9) each, 3 × 9 × 9 = 243)

d) 6 numbers between 2000 and 2016 with exactly one 0 digit (from 2011 to 2016)

e) 1 one digit number: the 0 .

This means, that the total number of integers between 0 and 2016 with exactly one 0 digit is:

9 + 162 + 243 + 6 + 1 = 421

Since the number of integers between 0 and 2016 is 2017, the probability of getting one, which has exactly one 0 digit, when choosing randomly is:

$p = \frac {421}{2017}$

Because 421 and 2017 are coprimes, therefore a = 421, b = 2017 and

$a + b = 421 + 2017 = \boxed {2438}$