jmc

We first find the areas of the triangles under $l$ and $m$ and in the first quadrant. From $l$'s equation, we find that at $x=6$, $y=0$. From $m$, we find that at $x=\frac{3}{2}$, we find that $y=0$.

We then find the areas using the formula for a triangle: $\frac{1}{2}bh$. The area under $l$ is $\frac{1}{2}(6\times 6)=18$. The area under $m$ is $\frac{1}{2}\left(\frac{3}{2}\times 6\right)=\frac{9}{2}$. Thus, the probability that the point selected will fall between $l$ and $m$ has probability $\frac{18-\frac{9}{2}}{18}=\frac{27/2}{18}=\frac{3}{4}=0.75$. Thus, the probability is $\boxed{0.75}$.