jmc

We let the two numbers be $x$ and $y$. The set of all possible pairs $(x,y)$ satisfy the inequalities $0<x<2$ and $0<y<2$; we can graph these points as a square on the $x-y$ plane with vertices $(0,0),(2,0),(2,2)$ and $(0,2)$. This square has area $4.$



We want to find the area of the set of points which also satisfies $x^2+y^2\le 4$. The set of ordered pairs that satisfy this inequality is a circle with radius $2$ centered at the origin.

The overlap between the circle and the square is the quarter-circle in the first quadrant with radius $2.$ This region has area $\frac{1}{4}(2{)}^2\pi =\pi$.

Thus, the desired probability is $\boxed{\frac{\pi }{4}}$.