0
$\begingroup$

I know that geometric probability works well when there are 2 or 3 variables involved. However, I am not sure how to use this method when there are more than 3 variables. For example:

*Five friends decide go meet at a restaurant at a random time between 7:00 p.m. and 8:00 a.m. Each person will only wait 20 minutes for all the others to arrive, and the meeting will be cancelled if not all of the participants show up. What is the probability that the meeting will take place? *

I have searched Google for a few minutes, and I haven't really found a similar problem.

How would one approach such a problem?

$\endgroup$
  • $\begingroup$ Assume that one of the friends show up at 7 PM. Do you consider the meeting cancelled if one of the remaining 4 don't appear until 7:20 PM? $\endgroup$ – zoli Oct 16 '15 at 20:40
  • $\begingroup$ Yes, that is what I mean. $\endgroup$ – Jed Oct 16 '15 at 20:42
  • $\begingroup$ What happens if $2$ friends show up at $7$ PM, $1$ at $7.19$, one at $7.38$ and the last friend at $7.50$, is then the meeting cancelled? $\endgroup$ – Hetebrij Oct 16 '15 at 20:55
0
$\begingroup$

In this Wikipedia article you can read about the joint distribution of the $i^{th}$ and $j^{th}$ member of an order statistic. The simplest case is when the random variables are all uniformly distributed over the interval $[0,1]$.

In our case there are $5$ independent random variables all uniformly distributed over the interval $[0,13]$, the time interval between $7$ PM and $8$ AM. Let's divide the arrival times (counted from $7$ PM) by $13$. This way we'll get $5$ independent random variables uniformly distributed over $[0,1]$. If we denote the elements of the order statistic of the arrival times by $U_1,U_2,...,U_5$ then we are interested in the joint distribution of $U_1 $ and $U_5$ the arrival times of the first friend and that of the last one.

The joint pdf of $U_1, U_5$ is (for $0\le u\le v\le 1$)

$$f_{U_1,U_5}(u,v)=\frac{5!}{3!}(v-u)^3=20(v-u)^3.$$

The question is the probability that $U_5-U_1 \le \frac1{39}$ ($\frac1{39}$ corresponds to $20$ minutes after shrinking the interval of $[0,13]$ to $[0,1]$. $13$ hours is $780$ minutes. So $20$ minutes corresponds to $\frac{20}{780}=\frac1{39}.$

$$P\left(U_5-U_1\le \frac1{39}\right)=\iint_A20(v-u)^3 \ dudv$$

where

$$A=\{(u,v): u\le v, v-u\le \frac1{39}\}$$ as shown below

enter image description here

So our probability is

$$20\int_0^{1-\frac1{39}}\int_u^{u+\frac1{39}} (v-u)^3\ dv\ du+20\int_{1-\frac1{39}}^1 \int_u^1 (v-u)^3\ dv\ du=\cdots$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.