The city of Ithaca, New York, allows for two-hour parking in all downtown spaces. Methodical parking officials patrol the downtown area, passing the same point every two hours. When an official encounters a car, he marks it with chalk. If the car is still there two hours later, a ticket is written. Suppose that you park your car for a random amount of time that is uniformly distributed on (0, 4) hours. What is the probability you will get a ticket? (from Rick Durett)
When I park my car somewhere, it takes $2h-t_1$, until the official marks my car. ($t_1$ is the time, that has passed, since the official was the last time at this place.)
I suppose, that $t_1$ is uniformly distributed on the interval [0,2]. You get a ticket, if your parking time $t_2$ is longer than $4h-t_1$, so I have to compute the probability
$\mathbb{P}(t_2 \geq 4h-t_1)$.
This should be
$\mathbb{P}(t_1+t_2 \geq 4h) = 1-F(4h)$.
F is 0 if $x \notin [0,6]$, $x/8$, if $x \in [0,2]$, 1/4 if $x \in [2,4]$ and $6/8-x/8$, if $x \in [4,6]$. So I get
$\mathbb{P}(t_1+t_2 \geq 4h) = 1-1/4 = 3/4$ and this probability seems too much for me, so something must be missing.
Where is my mistake?
Thank you! 🙂
Best Answer
Like you said, lets have $t$ be the time since the officer's last visit, so $t\sim \mathcal{U}(0,2)$ Your parking time is $h\sim \mathcal{U}(0,4)$. Let $T:= \{4-t<h\}$ be the event that you get a ticket, as you correctly pointed out.Often the easiest way to tackle these problems is via conditioning.
$P(T|t=\tau)=P(h>4-\tau)=\frac{4-4+\tau}{4}=\frac{\tau}{4}$
This implies that $P(T)=\int\limits_{0}^{2}f_t(z)P(T|t=z)dz=\int\limits_{0}^{2}\left(\frac{1}{2}\right)\frac{z}{4}dz=\int\limits_{0}^{2}\frac{z}{8}dz=\frac{z^2}{16}|_{z=2} = \frac{1}{4}$
I think where you went awry is with calculating the pdf of the sum of two independent, dissimilar uniform distributions. Their joint pdf if simply $f_{h,t}=\frac{1}{8}$ over the domain $[0,4]\times[0,2]$, hence your probability is the area above the line $h+t=4$ on this rectangle times $\frac{1}{8}$, which is just the corner of the rectangle above the line connecting $(2,2)$ with $(4,0)$ which is just half the area of the rectangle formed by $[0,2]\times [2,4]$ which comes out to $2$, which when multiplied by $\frac{1}{8}$ you get the correct value of $\frac{1}{4}$