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