Trains headed for destination A leave the train station at 15-minute intervals s

Question

Trains headed for destination A leave the train station at 15-minute intervals starting at 7 A. M. (i.e., trains leave at 7:00, 7:15, 7:30,7:45, 8:00, etc), whereas trains headed for destination B leave at 15-minute intervals starting at 7:05 A. M. If a certain passenger arrives at the station at a time uniformly distributed between 7 and 8 A. M. and then gets on the first train that leaves, what is the probability that he goes to destination A? What if the passenger arrives at a time uniformly distributed between 7:10 and 8:10 A. M.? Will this probability change?

Explanation / Answer

a)

For this random person to randomly wander onto the train going to destination A, he/she must arrive right after the train to B leaves, but before the next train to A leaves. This means the person must arrive some time between 7:05-7:15, or between 7:20-7:30 or between 7:35-7:45 or between 7:50-8:00. The person must arrive during one of these 10 minute windows, and there are 4 of these non-overlapping windows. Thus,
the probability that this random person will arrive at location A is,

P (A) = 40/60 = 2/3

b) All of the 10 minute windows in the previous part still apply, except for the rst. To randomly jump onto a train headed to location A, the person must arrive between 7:10-7:15, 7:20-7:30, 7:35-7:45, 7:50-8:00 or between 8:10-8:15. So we have a total of 3 windows of 10 minutes each, and 2 windows of 5 minutes each. There are a total of 40 minutes (still) during which a person could arrive and end up at location A. Therefore,

P (A) = 40/60 = 2/3

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