Question

A commuter is accustomed to leaving his house between 7:30 and 8:00 AM, the drive to the station taking between 20 and 30 minutes. It is assumed that the departure time and length of trip are independent random variables, uniformly distributed over their respective intervals. There are two trains which he can take: the first leaves precisely at 8:05 AM and takes exactly 35 minutes for the trip, and the second leaves at 8:25 AM and takes 30 minutes. (a) Assuming that he makes one of these trains, what is his average arrival time at the destination? (b) What is the probability that he misses both trains?

Answer 1

(a)

P(makes 1st train) = P(leaves before 7:35) + P(leaves between 7:35 and 7:45 and makes 1st train)

P(leaves between 7:35 and 7:45 and makes 1st train) =

P(makes 1st train)= 1/6 + 1/6 = 1/3

{

P(leaves between 7:35 and 7:45 and doesn't make 1st train) = 1/3 * 1/2

}

P(makes 2nd train) = P(leaves between 7:35 and 7:45 and doesn't make 1st train) + P(leaves between 7:45 and 7:55) + P(leaves between 7:55 and 8:00 and makes 2nd train)

P(leaves between 7:55 and 8:00 and makes 2nd train)

P(makes 2nd train) = P(leaves between 7:35 and 7:45 and doesn't make 1st train) + P(leaves between 7:45 and 7:55) + P(leaves between 7:55 and 8:00 and makes 2nd train)

= 1/6 + 10/30 + 1/12

= 7/12

avg time at destination assuming he makes one of the train (x)= 8 : x

x = (40*P(1st train) + 55*P(2nd train) )/ (P(1st train) + P(2nd train))

x= (40*1/3 + 55*7/12) / (11/12) = 49.54 = 49 min 32.4 sec

avg time at destination assuming he makes one of the train = 8 : 49 min 32.4 sec

(b)P(misses both train) = P(leaves between 7:55 and 8:00 and doesn't make 2nd train) = 1/6 * 1/2 = 1/12

P(misses both train) = 1/12