A student is being late to catch the morning train in 3/10 of trials. If he runs of on time he will always catch the train. If he runs of late, he gets to the station with 5 min delay after regular train departure and so he catches the train if the train is at least 5 min delayed.
The train is at least 5 min delayed with probability 2/5. The student is at school on time in case that the train gets to his school with delay at most 30 min.
Probability that the delay of the train in the city he lived in is less than 30 minutes provided student at his station got on train on time (with delay less than 5 min) is 99/100.
Probability that the delay of the train in that city is less than 30 minutes provided student got on train with delay more than 5 min is 97/100.
What is the probability, that student is at school on time on a day when he runs of late?
I have read this problem for many time but i still can't figure out how to solve it, there are many infomation and it make me confuse, i don't know where to start and which infomation that i should be focus on.
Please show steps in how you solve it.
Maybe the problem author wanted to put a complex idea in it, but I see only the following simple scenario. This unlucky day was happened. The student is late. So there is no use to think what could happen if he was in time. It already gone. He have to think about future. At the first, he have to catch the train. “He catches the train if the train is at least 5 min delayed”. “The train is at least 5 min delayed with probability 2/5”. So probability of the success at the first step is $2/5$. The second step is that the delay of the train is less than 30 minutes. Since student got on train with delay more than 5 min, probability of the success at the second step is $97/100$. So the total success probability is $(2/5)(97/100)=97/250$.