jmc

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 $\frac{1}{6}+\frac{1}{4}=\boxed{\frac{5}{12}}$. 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 $ab>0$.