\(\displaystyle P(x \leq 3 | x \geq 2)=\frac{0.01(3^2-2^2)}{0.01(10^2-2^2)}=\frac{5}{96} \\) renders correctly while

\(\displaystyle P(x \geq 5 | 3 \leq x \leq 8)=\frac{0.01(8^2-5^2)}{0.01(8^2-3^2)}=\frac{39}{55}\)

does not?

You may have to quote and preview to see what I mean.