Solution:

First, we compute the probability that the line through two random points in a triangle $ABC$ passes through segments $\overline{AB}$ and $\overline{AC}$. We can take an affine transform of the two random points and the triangle such that $ABC$ becomes equilateral. Since the distribution of the two points is still uniform and independent, the probability of the line intersecting any two given sides is $\frac{1}{3}$ by symmetry.

Next, we compute the probability that the line through two random points in a rectangle $ABCD$ passes through opposite edges $\overline{AB}$ and $\overline{CD}$.

If the line passes through $\overline{AB}$ and $\overline{BC}$, the points must both lie in triangle $ABC$, whose area is half that of $ABCD$. Given this, the probability the line passes through those two sides is $\frac{1}{3}$, as computed before. Thus the probability the line passes through $\overline{AB}$ and $\overline{BC}$ is $\left(\frac{1}{2}\right)^2 \cdot \frac{1}{3} = \frac{1}{12}$. The same goes for the other pairs of adjacent edges. By symmetry, the line is equally likely to pass through either pair of opposite edges, each with probability $\frac{1}{2}\left(1 - 4 \cdot \frac{1}{12}\right) = \frac{1}{3}$.

We now return to the original problem. If the line passes through a pair of opposite edges, then both points must be in the rectangle formed by these edges, which has area $\frac{2}{3}$ that of the hexagon. Given this, the probability the line passes through those two edges is $\frac{1}{3}$ as computed before. Thus, the probability that the line passes through the given pair of opposite edges is $\left(\frac{2}{3}\right)^2 \cdot \frac{1}{3} = \frac{4}{27}$. Hence, the probability the line passes through any of the three pairs of opposite edges is $3 \cdot \frac{4}{27} = \left[\frac{4}{9}\right]$.