Probability of number of people in car park at any given time A building has 22 car spaces, each having a car parked within each spot in the morning. Each car is retrieved by its respective owner at some point (random time) between 7am and 9am (120minutes). Each owner takes 5 minutes in the car park to pack the car, get in, and drive out.
At any random point in time, what is the probability that there are:
a) 0 people present in the car park
b) 1 person present in the car park
c) 2 people present in the car park
So far I'm looking at this as  binomial distribution, but I'm not sure I'm taking the right approach, or if there is a better way. With the binomial: n = 22 people, k = number of people present (ie 0, 1, 2), and $p = \frac{5}{120} = 0.041667$. So I'd get $a = 0.392, b = 0.375$, and $c = 0.171$...but just have a feeling I'm overlooking something major.
EDIT - Tried to clarify a little bit of the wording. The reasoning being that the question is trying to establish what the probability would be that, at some random time of query between 7am and 9am, of there being 0, 1 or 2 people present in the car park. The reason it deals with people and the 5 minute info, is this presents the opportunity for overlap, that multiple people may be retrieving their car at similar times. The 2-hour information defines the bounds from when the there are 22 cars to when there are zero.
 A: Assume that the drivers arrive at times uniformly distributed over $[0,120-5]$ minutes. At time $t\in [5,120]$ (no driver can leave before $t=5$) the probability that a given driver has left is $p(t)=\frac{t-5}{115}$. 
Now the number who have left before $t>5$ is $n(t) \sim B(p(t),22)$, and the number remaining is $r(t)=22-n(t)$ 
A: Let $p$ denote the probability that any given person is present in the car park.

The probability that $0$ out of $22$ people are present in the car park is:
$$\binom{22}{0}\cdot(p^{0})\cdot(1-p)^{22-0}$$
The probability that $1$ out of $22$ people are present in the car park is:
$$\binom{22}{1}\cdot(p^{1})\cdot(1-p)^{22-1}$$
The probability that $2$ out of $22$ people are present in the car park is:
$$\binom{22}{2}\cdot(p^{2})\cdot(1-p)^{22-2}$$
A: S ince doors are opened  between 07.00 and 09.00 I'm assuming the arrival time $T$ of a particular owner is uniformly distributed in $[0,115]$. The probability $p_t$ that this owner is present at time $t\in[0,120]$ is then given by
$$p_t=\left\{\eqalign{ {\displaystyle{t\over115}}\qquad&(0\leq t\leq 5) \cr  
{\displaystyle{1\over23}}\qquad&(5\leq t\leq 115) \cr {\displaystyle{120-t\over115}}\quad&(115\leq t\leq120)\cr}\right.$$
The number $N_t$ of owners present at any given time $t$ is  binomially distributed, i.e. given by
$$P[N_t=k]={22\choose k}p_t^k\>(1-p_t)^{22-k}\ .$$
