Question:
Given that $a$ and $b$ are real numbers such that $-3\leq a\leq1$ and $-2\leq b\leq 4$, and values for $a$ and $b$ are chosen at random, what is the probability that the product $a\cdot b$ is positive?  Express your answer as a common fraction.

Answer:
The probability that both $a$ and $b$ are positive is $\left(\frac{1}{4}\right)\left(\frac{2}{3}\right)=\frac{1}{6}$. The probability that $a$ and $b$ are both negative is $\left(\frac{3}{4}\right)\left(\frac{1}{3}\right)=\frac{1}{4}$.  Since $ab$ is positive if and only if one of these two events occur, the probability that $ab$ is positive is $\dfrac{1}{6}+\dfrac{1}{4}=\boxed{\dfrac{5}{12}}$.  Graphically, we can picture the possible outcomes for $(a,b)$ as a rectangle in a Cartesian plane.  The shaded rectangles are the regions in which $ab>0$. [asy]
size(5cm);
import graph;
defaultpen(linewidth(0.7)+fontsize(10));
dotfactor=5;
pair A=(-3,-2), B=(1,-2), C=(1,4), D=(-3,4);
fill((0,0)--(1,0)--(1,4)--(0,4)--cycle,gray);
fill((0,0)--(-3,0)--(-3,-2)--(0,-2)--cycle,gray);
draw(A--B--C--D--cycle,dashed);

draw((-5,0)--(2,0),Arrows(4));
draw((0,-4)--(0,6),Arrows(4));

int i;
for(i=-4;i<=1;++i)

{

draw((i,-0.3)--(i,0.3));

}
for(i=-3;i<=5;++i)

{

draw((-0.3,i)--(0.3,i));

}

label("$a$",(2.5,0));
label("$b$",(0,6.5));[/asy]