Probability

This is a solution to a problem posed in Probability by Salim Dewani (H#3 02-06), mainly because blogspot comments don't allow insertion of images.

[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 :) ]