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+1 vote


A subway train in a certain line runs after every half hour, between every midnight and 6 in the morning.
What is the probability that a man entering the station at random will have to wait at
least 20 minutes?

I'm stuck here...

It can be solved using uniform distribution method . But

1) I'm bit confused about the limits ?

  • X denote the RV waiting time of man for the next train given that man arrives at random time
  • So is probability has to be computed for P(X>20

2) What happens if man arrives of a day time after 6 in the morning ?

  • in this case what is the upper limit of the integral ?


in Probability by Boss (21.5k points) | 900 views
Applying simple logic and assuming resolution of time to be in minutes, shouldn't the answer be 0.3?....because say between 12:00 - 12:30 AM, there are 9 time instants which result in at least 20 minutes of waiting time...this results in total of 18 * 6 such time instants between 12:00 AM to 6:00 AM. Therefore, probability of waiting time exceeding 20 minutes = (18 * 6)/(6 * 60) = 3 / 10 = 0.3.

3 Answers

+1 vote
It is a Uniform distribution on [0,30]

a=0, b= 30, f(x) = 1/(b-a) = 1/30

here, if a person comes between 0 to 10, then he has to wait for at least 20 minutes.
so, P(X<=10) = $\int_{0}^{10}$(1/30)dx = 1/30(10-0) = 1/3
by (73 points)
edited by
0 votes

(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).

by Active (1.8k points)
0 votes

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 (302 -1/2*2*(202)) / 302 =5/9

by Veteran (118k points)

how u got (302 -1/2*2*(202)) / 302 ???

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