Question
I'm stuck here...
It can be solved using uniform distribution method . But
1) I'm bit confused about the limits ?
2) What happens if man arrives of a day time after 6 in the morning ?
URL : https://www.assignmentexpert.com/homework-answers/Math-Answer-40654.pdf
(1). I think it should be $P(30 > X \geq 20)$, because questions asks for atleast 20 minutes and train arrives every 30 minutes, so $X < 30$.
(2). I suspect this possibility is out of scope of this problem because train arrives only between 12:00 AM to 6:00 AM, but maybe I am missing your point here.
As described in the referred solution, this problem is reduced to the problem in 30 minutes(instead of its original duration between 12:00 - 6:00) because of its periodic nature(train arriving every 30 minutes).
As here it is mentioned between midnight and 6 am in the morning, we only consider that period
Now man entering the station have to wait minimum 20 mins.
So, the man have to enter maximum between 0 to 10 mins. after leaving the first train
Probability of it have to wait atleast 20 mins (30^{2} -1/2*2*(20^{2})) / 30^{2} =5/9
how u got (30^{2} -1/2*2*(20^{2})) / 30^{2 ???}