jmc

We note that the points for which $x+y<3$ are those that lie below the line $x+y=3$, or $y=-x+3$. As the diagram below illustrates, these are all the points in the square except those in the triangle with vertices (2,1), (2,2), and (1,2).



Since this is a right triangle whose sides both of length 1, its area is $\frac{1}{2}\cdot 1^2=1/2$. Since the square in question has side length 2, its area is $2^2=4$, so the shaded region has area $4-1/2=7/2$. Our probability is therefore $\frac{7/2}{4}=\boxed{\frac{7}{8}}$.