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 $\frac{1}{4}$
First we do it the way you proposed. For this problem, it is a little inefficient, but the idea will be needed at other times.
There are $\binom{40}{5}$ ways to choose $5$ tickets from $40$.
We find the number of $5$-ticket hands that win exactly $1$ prize. There are $\binom{3}{1}$ ways to choose the "good" ticket, and for each of these ways there are $\binom{37}{4}$ ways to choose the "bad" tickets, for a total of $\binom{3}{1}\binom{37}{4}$ one prize hands.
Thus the probability of winning exactly one prize is $\dfrac{\binom{3}{1}\binom{37}{4}}{\binom{40}{5}}$.
Similarly, the number of two prize hands is $\binom{3}{2}\binom{37}{3}$.
Thus the probability of winning exactly two prizes is $\dfrac{\binom{3}{2}\binom{37}{3}}{\binom{40}{5}}$.
Finally, the probability of winning exactly three prizes is $\dfrac{\binom{3}{3}\binom{37}{2}}{\binom{40}{5}}$.
Add up.
Another way: We find the probability $p$ of winning no prize. Then the probability of winning at least one is $1-p$.
There are $\binom{37}{5}$ hands in which all the tickets are bad. It follows that
$$p=\dfrac{\binom{37}{5}}{\binom{40}{5}}.$$
Still another way: Again, we first find the probability of no prize. Suppose she checks her tickets one by one to see whether she has won something.
The probability the first ticket is bad is $\frac{37}{40}$. Given that the first ticket was bad, the probability the second ticket is bad is $\frac{36}{39}$, for there are only $36$ bads left. So the probability the first two tickets are bad is $\frac{37}{40}\cdot \frac{36}{39}$. Given the first two were bad, the probability the third is bad is $\frac{35}{38}$. And so on. Thus
$$p=\frac{37}{40}\cdot \frac{36}{39}\cdot \frac{35}{38}\cdot \frac{34}{37}\cdot \frac{33}{36}.$$
To evaluate, it can be useful to do some cancellation.
Best Answer
Hint: Apply the reflection principle in a similar way to deal with the general case.
This problem is equivalent to starting at $(0, 2k)$ with steps of $(1,1)$ and $(1, -1)$. Without restriction, there are $2^{2n}$ possible paths.
How many ways are there which do not go below the line $y=0$? That's the number of ways that do not touch the line $y=-1$. Let's count the complement, namely the number of lines which touch $y=-1$.
For each such path, using the reflection principle, we reflect the path after the last point of contact with the line $y=-1$. This gives us a path from $(0, 2k)$ which ends with $y \leq -1$. Now show that this is a bijection.
The number of such paths is thus clearly $\sum_{i=0}^j { 2n \choose i}$ for a certain value of $j$ which you should determine.
Let's say we took $a$ up steps and $2n-a$ down steps. Since we started out at $(0,2k)$, we would end up at a y-coordinate of $2k + a - (2n-a) = 2k+2a - 2n$. We want this to be $\leq -1$, and hence $2a \leq 2n - 2k - 1 \Rightarrow a \leq n - k - 1$ (since they are all integers). This is our value of $j$.