Probability

There are $19$ numbers containing at least one zero from $100-199$ inclusive. That means that we have the same number in each hundred from $x00-x99$ where $x$ ranges from $2-9$ .

Hence, the total number of integers from $100-999$ inclusive that have at least one zero is $9(19)=171$ .

The total number of integers in the set $100-999$ is $999-100+1=900$ .

Therefore, the probability is $\color{#D61F06}\boxed{\dfrac{171}{900}}.$

There are $900$ positive intgerer with three digits. So if there is $x$ number of three digit positive intgeres which not contain $0$ , then the answer is: $\dfrac{900-x}{900}$ .

Now we find the value of $x$ . It is clear that each digit of these numbers is from ${1, 2, 3, 4, 5, 6, 7, 8, 9}$ , so $x=9^3=729$ .

From that the probality is: $\dfrac{900-729}{900}=\dfrac{171}{900}$