4

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 ?

Spectre
  • 1,579
spaul
  • 356
  • 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 '17 at 17:49
  • 1
    related https://math.stackexchange.com/questions/103015/chance-of-meeting-in-a-bar – Henry May 07 '17 at 18:55
  • related question solved using random variables https://math.stackexchange.com/questions/3449536/what-is-the-probability-that-the-witches-will-meet-at-the-coffee-shop/3449556#3449556 – ATK Nov 25 '19 at 09:41
  • As a side note, it is quite easy to solve with geometric probability. – explicitEllipticGroupAction May 31 '20 at 00:15
  • Why isn't the answer accepted? – ViktorStein Oct 10 '20 at 11:24
  • What kind of distribution is assumed for their inter-arrival time? – vvg Oct 10 '20 at 11:31

1 Answers1

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

Kanwaljit Singh
  • 8,820