imc

In order for the divisor chosen to be a multiple of $11$, the original number chosen must also be a multiple of $11$. Among the first $100$ positive integers, there are 9 multiples of 11; 11, 22, 33, 44, 55, 66, 77, 88, 99. We can now perform a little casework on the probability of choosing a divisor which is a multiple of 11 for each of these 9, and see that the probability is 1/2 for each. The probability of choosing these 9 multiples in the first place is $\frac{9}{100}$, so the final probability is $\frac{9}{100}\cdot \frac{1}{2}=\frac{9}{200}$, so the answer is $\boxed{\frac{9}{200}}.$ $$\begin{gathered} 11=1,11\Rightarrow \frac{1}{2} \\ 22=2\times 11:1,2,11,22\Rightarrow \frac{1}{2} \\ 33=3\times 11:1,3,11,33\Rightarrow \frac{1}{2} \\ 44=2^2\times 11:1,2,4,11,22,44\Rightarrow \frac{1}{2} \\ 55=5\times 11:1,5,11,55\Rightarrow \frac{1}{2} \\ 66=2\times 3\times 11:1,2,3,6,11,22,33,66\Rightarrow \frac{1}{2} \\ 77=7\times 11:1,7,11,77\Rightarrow \frac{1}{2} \\ 88=2^3\times 11:1,2,4,8,11,22,44,88\Rightarrow \frac{1}{2} \\ 99=3^2\times 11:1,3,9,11,33,99\Rightarrow \frac{1}{2} \end{gathered}$$