There are many ways to solve this problems including the ones using graphs or integral calculus. Although I've always got good grades for it, I hate advanced mathematics. At least I do not like it when I am at leisure, trying to solve really nice puzzles while stretching on a couch! So, the solution that I would offer would not include any calculus or graphs.
The answer to this puzzle is 7/8
The solution that I used is as follows. Let us find the probability of the man missing the train. Let us divide the total time interval in equal intervals of 5 minutes each:
09:55 to 10:00 T1
10:00 to 10:05 T2
10:05 to 10:10 T2
The man misses the train if he arrives within T2, and the train too leaves within T2, and the train leaves before the man arrives.
The probability of the man arriving within T2 is 1/2. Similarly, the probability of the train leaving within T2 is 1/2.
Now, we need the probability of the train leaving before the man arrives. Now, since the train's departure and the man's arrival are two independent events (with uniformly distributed probabilities), by symmetry it is equally probable that one event occurs before or after the other. Hence the probability of the train leaving before the man arrives (within T2) is 1/2.
So, the probability of the man missing the train is (1/2)*(1/2)*(1/2) = 1/8
The probability that he catches the train, hence, is 7/8.
