jmc

The area of the rectangular region is 2. Hence the probability that $P$ is closer to $(0,0)$ than it is to $(3,1)$ is half the area of the trapezoid bounded by the lines $y=1$, the $x$- and $y$-axes, and the perpendicular bisector of the segment joining $(0,0)$ and $(3,1)$. The perpendicular bisector goes through the point $(3/2,1/2)$, which is the center of the square whose vertices are $(1,0),(2,0),(2,1),\text{ and }(1,1)$. Hence, the line cuts the square into two quadrilaterals of equal area $1/2$. Thus the area of the trapezoid is $3/2$ and the probability is $\boxed{\frac{3}{4}}$.