A point $P$ is randomly selected from the rectangular region with vertices $(0,0), (2,0)$, $(2,1),(0,1)$. What is the probability that $P$ is closer to the origin than it is to the point $(3,1)$?
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 $\boxed{\frac{3}{4}}$.

[asy]
draw((-1,0)--(4,0),Arrow);
draw((0,-1)--(0,3),Arrow);
for (int i=0; i<4; ++i) {
draw((i,-0.2)--(i,0.2));
}
for (int i=0; i<3; ++i) {
draw((-0.2,i)--(0.2,i));
}
label("$x$",(3.7,0),S);
label("$y$",(0,2.7),W);
label("1",(1,-0.2),S);
label("2",(2,-0.2),S);
label("3",(3,-0.2),S);
label("1",(-0.2,1),W);
label("2",(-0.2,2),W);
draw((0,0)--(3,1),linewidth(0.7));
draw((1,2)--(2,-1),linewidth(0.7));
dot((1.5,0.5));
dot((3,1));
draw((1,0)--(1,1.3),dashed);
draw((1.5,0.5)--(1.7,1.5));
label("($\frac{3}{2}$,$\frac{1}{2}$)",(1.7,1.5),N);
draw((0,1)--(2,1)--(2,0),linewidth(0.7));
label("$(3,1)$",(3,1),N);
[/asy]