Question

A man and a woman agree to meet at a cafe about noon. If the man arrives at a time uniformly distributed between 11:30 and 12:10 and if the woman independently arrives at a time uniformly distributed between 11:55 and 12:35, what is the probability that the first to arrive waits no longer than 15 minutes?

Answer 1

Lets break the man's arrival time interval down into parts -

From 11:40 to 11:50,

Probability = 10/35,

on average there is a five minute interval for the woman to arrive less than 10 minutes later (0 minutes for 11:40 to 10 minutes for 11:50, average 5 minutes).

So it's 5/60 that one person (the man) waits no more than 10 minutes.

From 11:50 to 12:00,

Probability = 10/35,

on average there is a fifteen minute interval for the woman to arrive within 10 minutes of the man, 15/60 probability.

From 12:00 to 12:15,

Probability = 15/35,

There is always a 20 minute interval for the woman, 20/60 probability.

So,

Final Probability = (10/35)*(5/60) + (10/35)*(15/60) + (15/35)*(20/60)

= 500/2100

= 5/21.