# Probability that friends meet.

Two friends decide to meet between 1:00 PM and 2:00 PM on a given day. There is a condition that whoever arrives first will not wait for the other for more than $$15$$ minutes. What is the probability that they meet ?

How to solve this problem using Random Variables mathematically ?

• you need to figure out some boundary conditions - if A arrives at 1, then B has to arrive between 1 and 1:15, but if A arrives as 1:15 then B can arrive from 1:00 to 1:30 - the next boundary condition is at 1:45. You have to come up with a probability function, then integrate it
– Cato
Jan 26, 2017 at 17:49
• May 7, 2017 at 18:55
• related question solved using random variables math.stackexchange.com/questions/3449536/…
– ATK
Nov 25, 2019 at 9:41
• As a side note, it is quite easy to solve with geometric probability.
– mpnm
May 31, 2020 at 0:15
• Why isn't the answer accepted? Oct 10, 2020 at 11:24

One man must arrive before the other. The probability that man 1 arrives during the first $\frac34$ hour is $\frac34$. He'll then wait $\frac14$ hour.
The probability that he arrives during the last $\frac14$ hour is $\frac14$, and then (on average he'll wait) $\frac18$ hour.
So altogether the man 1 will wait $\frac34 × \frac14 + \frac14 × \frac18 = \frac7{32}$.
So the probability that man 2 arrives while the man 1 is waiting is $\frac7{32}$. Similarly if man 2 arrives before man 1. SO altogether, the probability of them meeting is $\frac7{32} + \frac7{32} = \frac7{16}$