Two friends decide to meet between 1PM and 2PM 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 '17 at 17:49
  • $\begingroup$ related math.stackexchange.com/questions/103015/… $\endgroup$ – Henry May 7 '17 at 18:55
  • $\begingroup$ related question solved using random variables math.stackexchange.com/questions/3449536/… $\endgroup$ – Lambda Nov 25 '19 at 9:41
  • $\begingroup$ As a side note, it is quite easy to solve with geometric probability. $\endgroup$ – David Dong May 31 at 0:15

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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