11. Randomly select a three-digit number $I$. Then the probability that the number $I$ contains the factor 5 is $\qquad$ .
11. $\frac{1}{5}$.

From the problem, we know that this is a classical probability model, because all the events in the experiment are three-digit numbers, totaling 999 $-99=900$ numbers. The events that meet the condition are numbers $I$ that contain the factor 5, i.e., numbers $I$ that are multiples of 5. Among these, the multiples of 5 are $\mathrm{C}_{9}^{1} \mathrm{C}_{10}^{1} \mathrm{C}_{2}^{1}=180$.
Thus, $p=\frac{180}{900}=\frac{1}{5}$.