[I shall hope to post something interesting soon. In the meantime, I wrote something.]
If you have had your try, I present a solution.
Consider a rectangle with the horizontal axis representing the time the dad leaves work and the vertical axis the time you leave home. Each point in the rectangle has an equi-probability of occurring. The points in the rightmost region represent starting times of the duo in which they can't meet Similarly in the left most triangle, the father has crossed D before the son, hence the path chosen after D doesn't matter.
Net probability = 0.5(Fraction of area covered by leftmost triangle - ADEFC route)
+ 0.5(Fraction of area covered by the leftmost triangle and the middle area - ADGFC route)
= 0.5(32/300) + 0.5(210/300)
= 342/600 = 0.57
[It feels so nice to be doing maths :) ]
