Calculus and Probability?

Let $AF = x$ and $AE = y$ be our two independent, random variables, where each variable has the domain $[0,1]$ .

The sample space can be represented by another unit square with lower left corner $(0,0)$ and upper right corner $(1,1)$ . The area of $\Delta AEF$ will exceed $\dfrac{1}{3}$ when

$(\dfrac{1}{2})xy \ge \dfrac{1}{3} \Longrightarrow xy \ge \dfrac{2}{3} \Longrightarrow y \ge \dfrac{2}{3x}.$

The region $R$ of intersection of the sample space and the region $y \ge \dfrac{2}{3x}$ is bounded above by the line $y = 1$ going from $x = \dfrac{2}{3}$ to $x = 1$ , below by the hyperbola $y = \dfrac{2}{3x}$ and to the right by the line $x = 1$ . The area of region $R$ is then

$\displaystyle\int_{\frac{2}{3}}^{1} (1 - \frac{2}{3x}) dx = x - \frac{2}{3}\ln(x)$

evaluated from $x = \dfrac{2}{3}$ to $x = 1$ , which comes out to

$(1 - 0) - (\dfrac{2}{3} - \dfrac{2}{3}\ln(\dfrac{2}{3})) = \dfrac{1}{3} + \dfrac{2}{3}\ln(\dfrac{2}{3}) = \boxed{0.063}$

to 3 decimal places. Note that since the sample space has an area of $1$ the desired probability is just the area of region $R$ .