Let $x$ be a real number selected uniformly at random between 100 and 200. If $\lfloor \sqrt{x} \rfloor = 12$, the probability that $\lfloor \sqrt{100x} \rfloor = 120$ can be expressed in the form $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m$.
I got the answer $\frac{3}{25}$, but it was incorrect. What should I do?
Since $x$ is a real number, we have:
$$\lfloor \sqrt x \rfloor = 12 \iff x \in [12^2, 13^2) = [144, 169)$$
$$\lfloor \sqrt{100x} \rfloor = 120 \iff x \in [12^2, 12.1^2) = [144, 146.41)$$
The probability that $\lfloor \sqrt{100x} \rfloor = 120$ thus equals:
$$\frac{m}{n} = \frac{146.41 - 144}{169 - 144} = \frac{2.41}{25} = \frac{241}{2500}$$
We thus find $m = 241.$