smc

Since $\lfloor \sqrt{x}\rfloor =12$, $12\le \sqrt{x}<13$ and thus $144\le x<169$. The successful region is when $120\le 10\sqrt{x}<121$ in which case $12\le \sqrt{x}<12.1$ Thus, the successful region is when $$144\le x<146.41$$ The successful region consists of a 2.41 long segment, while the total possibilities region is 25 wide. Thus, the probability is $$\frac{2.41}{25}=\boxed{\frac{241}{2500}};\boxed{\frac{241}{2500}}.$$