1 vote

**Question**

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 ?

URL : https://www.assignmentexpert.com/homework-answers/Math-Answer-40654.pdf

0

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.

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

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

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).

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 (30^{2} -1/2*2*(20^{2})) / 30^{2} =5/9