A Challenging Probability Question

Because we are working with numbers in the triple digits, our numbers with at least one $0$ will have that $0$ in either the units digit or the tens digit. The first $10$ numbers of $200, 300, 400, 500, 600, 700, 800,$ and $900$ will be counted, which gives us $90$ possibilities so far. Now, find the rest that has a zero in $100$ , not including itself, as we have already counted it: $110, 120, 130, 140, 150, 160, 170, 180, 190.$ If we do this to every number, we get $81$ more possibilities ( $9$ x $9$ ). Add the two sets together ( $90$ and $81$ ) to give $171$ out of the $900$ possibilities total (there are 900 numbers between $100$ and $899$ ), and this gives $171/900$