jmc

A little casework seems like the simplest approach. First, if Paco spins a 1 or 2, it does not matter what Manu spins; the product is less than 30 regardless. If Paco spins a 3, the product will be 30 or greater only if Manu spins a 10, and both of these will be true with probability $\frac{1}{5}\cdot \frac{1}{10}=\frac{1}{50}$. If Paco spins a 4, Manu's spins of 8, 9 or 10 will tip us over the 30 barrier, and this with probability $\frac{1}{5}\cdot \frac{3}{10}=\frac{3}{50}$. If Paco spins a 5, Manu will break the 30 threshold with a 6, 7, 8, 9 or 10, probabilities being $\frac{1}{5}\cdot \frac{5}{10}=\frac{1}{10}$. The total probability for these three cases is $\frac{1+3+5}{50}=\frac{9}{50}$. But, we want the probability that the product is less than 30, so we subtract our fraction from 1 and get $\boxed{\frac{41}{50}}$.