jmc

There are nine numbers with one digit: $1,2,...,9$, so the probability of selecting a one digit number is $\frac{9}{12}=\frac{3}{4}$. There are three numbers with two digits: $10,11,12$, so the probability of selecting a two digit number is $\frac{3}{12}=\frac{1}{4}$. The expected number of digits is therefore $E=\frac{3}{4}\cdot 1+\frac{1}{4}\cdot 2=\boxed{1.25}$.