imc

To be a three digit number in base-10: $10^2\le n\le 10^3-1$ $100\le n\le 999$ Thus there are $900$ three-digit numbers in base-10 To be a three-digit number in base-9: $9^2\le n\le 9^3-1$ $81\le n\le 728$ To be a three-digit number in base-11: $11^2\le n\le 11^3-1$ $121\le n\le 1330$ So, $121\le n\le 728$ Thus, there are $608$ base-10 three-digit numbers that are three digit numbers in base-9 and base-11. Therefore the desired probability is $\frac{608}{900}\approx 0.7\Rightarrow \boxed{0.7}$.