How many times do two or more odd digits appear in a number when counting from $0$ to $999$?

Question

I'm looking for a short and quick way of solving this question. I could probably count the odd digits within smaller ranges (i.e. $0-9, 10-19, 20-29...$) but I was wondering if there was an easier (and faster) way of doing this question. So far I've done the smaller ranges I previously mentioned and from $0-99$, I've got $25$ odd numbers with two odd digits but when it comes to $100-199$ and so on I'm not sure how to go about doing that.

Answer 1

Either $n$ or $999-n$ has two or more odd digits

Answer 2

Since 0 is even, we can consider the three digit numbers from 000 to 999. For each digit P(Odd)=P(Even)=0.5. There are 8 possible combinations, each with equal probability: OOO, OOE, OEO, EOO, EEO, EOE, OEE, EEE. Since 4 of these have two or more odd digits then P(2 or more odd digits)=0.5, hence the answer is 500