Numbers may begin with 0, so 00000000 is valid.
How many of these numbers contain at least two 2's?
What is the probability of a number containing at least two 2's?
I got as far as deducting the permutations that definitely don't meet the criteria. Which I think would be $9^{8}+9^{7}\cdot10$
There are $10^{8}$ numbers possible.
To count the numbers with at least $2$ $2$'s, we will count the number with less than $2$ and subtract from $10^{8}$ (this method is called complementary counting). There are $9^{8}$ numbers that do not contain a $2$, and $8\cdot 9^{7}$ numbers containing exactly $1$ $2$. (Note that we must pick which digit will be a $2$, and then the rest of the digits must not be $2$'s.) Thus, there are $\boxed{10^{8}-9^{8}-8\cdot 9^{7}}$ numbers containing at least $2$ $2$'s.
To get the probability, we must divide by the total number, for an answer of $\boxed{\frac{10^{8} - 9^{8}-8\cdot 9^{7}}{10^{8}}.}$