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 ?

  • $\begingroup$ 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 $\endgroup$
    – Cato
    Jan 26, 2017 at 17:49
  • 1
    $\begingroup$ related math.stackexchange.com/questions/103015/… $\endgroup$
    – Henry
    May 7, 2017 at 18:55
  • $\begingroup$ related question solved using random variables math.stackexchange.com/questions/3449536/… $\endgroup$
    – ATK
    Nov 25, 2019 at 9:41
  • $\begingroup$ As a side note, it is quite easy to solve with geometric probability. $\endgroup$
    – mpnm
    May 31, 2020 at 0:15
  • $\begingroup$ Why isn't the answer accepted? $\endgroup$ Oct 10, 2020 at 11:24

1 Answer 1


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 $\frac7{32} + \frac7{32} = \frac7{16}$


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