What is the chance to get a parking ticket in half an hour if the chance to get a ticket is 80% in 1 hour? This sounds more like a brain teaser, but I had some kink to think it through :( Suppose you're parking at a non-parking zone, the probability to get a parking ticket is 80% in 1 hour, what is the probability to get a ticket in half an hour? Please show how you deduce the answer. Thanks!
 A: It really depends on what model is assumed. However, if the idea is that no matter how long you leave your car there, you have a $20$% chance of getting through any given hour unscathed, you can treat it as an exponential decay problem. Let $p(t)$ be the probability that you do not get a ticket in the first $t$ hours. Then $p(1)=0.2$, $p(2)=0.2^2$ (a $20$% chance of making it through the first hour times a $20$% chance of making it through the second), and in general $p(t)=0.2^t$. The probability of not getting a ticket in the first half hour is then $p(1/2)=0.2^{1/2}=\sqrt{0.2}\approx 0.4472$, and the probability that you do get a ticket in the first half hour is about $1-0.4472=0.5528$.
A: Not enough data to come up with an answer. You would have to know the pattern that the parking warden takes to get the correct answer (first half hour, second half hour), also this depends on which part of the time unit the parking warden would take up, for example 1pm-2pm, 1.05pm-2.05pm etc
80% can reasonable be assumed for an hour, but for half an hour less so.
I would therefore conclude that there is an 80% chance that you will get a ticket as there is no way to distinguish the half hour from the full hour because of lack of data for the movements of the parking warden.
A: If my probability of getting a parking ticket in a half hour is T, then
$$T+(1-T)T$$
are the odds I got one in 1 hour. The first term says that I got it in the 1st half hour (so it doesn't matter what happens after that), and the second term says that I got it in the 2nd half hour (so not in the first). Solving and taking the sensible solution:
$$2T-T^2=.8\implies T^2-2T+.8$$
$$T=1-\sqrt{.2}$$
Which are roughly $55.3\%$ odds in a half hour.
A: Another way of looking at the exponential model: If $P$ is the probability you get lucky and don't get ticketed in the first 30 minutes, then $P^2$ is the chance you luck out twice in a row and don't get ticketed in the first hour. You know $P^2 = 0.2$. So $P$ is the square root of that or $0.447$. So the chance you are unlucky in the first 30 minutes and get a ticket is $1 - 0.447 = 0.553$ or $55.3$ percent.
A: This sounds like a Poisson Process to me... so, assuming that:
Since we don't have additional data, I will asume that the expected probability of being spotted once per hour is $p=0.8$. This is a Bernoulli experiment (either you get spotted or not). I will model the probability of being spotted by a machine, person, or some other authority that can issue a ticket ;-)
If $\lambda$ is the expected times that you get spotted every hour, then $\theta=\frac{1}{\lambda}\ $ is the expected time between events. 
The number of events given in $t$ can be modeled by a random variable $N \sim Poisson(\lambda·t)$ where $t$ is the amount of time (in hours) elapsed, and the time between events can be modeled by a random variable $X\sim Exponential(\theta)$.
So, the probability of getting $n$ tickets in $t$ hours, with an expected probability of $\lambda$ that you get a ticket in one hour is: 
$$Pr\{N=n\ |\ \lambda, t\}=\frac{e^{-\lambda·t}·(\lambda·t)^{n}}{n!} $$
Since we need to calculate the probability of being spotted at least once, we can calculate $1-Pr\{N=0\}$. Then, for $n=0$, $\lambda=0.8$ and $t=0.5$, then:
$$1-Pr\{N=0\ |\ \lambda=0.8, t=0.5\}=1-\frac{e^{-0.4}·(0.4)^{0}}{0!}=1-e^{-0.4}\approx0.4918 $$
You should notice that, even if you "get spotted" more than once per day, this model still holds, since this process is memoryless.
Further reference: Ross, Sheldon, "Introduction to probability models", Ninth edition, Chapter  5.
(I really hope this is not your homework)
A: A simple but not unreasonable model is that the ticket person cycles through her/his round in roughly constant time.  Suppose that you park illegally for exactly an hour at a randomly chosen spot in the round, and at a start time uniformly distributed with respect to the cycle. Since your probability of getting a ticket in an hour is $80\%$, the cycle length is $1.25$ hours. Thus your probability of getting a ticket if you park illegally for half an hour is $40\%$.
A: There is another way to look at this problem.
Most of the answers of ~55.3% (including the currently accepted answer) assume the "meter reader" is checking cars COMPLETELY at random.  In this model, there is an extremely small but non-zero chance you could leave a car in the same spot for a month and not get a ticket (although it's up there in "lottery" winner and struck-by-lightning chances).  In fact, the chance of getting a ticket using this model is always less than 100% for a finite amount of time.  
However, meter readers typically work in a linear pattern starting at one point and then looking at ALL the cars in a line down one side of a block.  If we consider the realistic pattern that cars are checked in a linear fashion instead of cars being checked randomly, the answer becomes much simpler.  Therefore, when they completely cover an area (in a finite amount of time), your car has a 100% chance of getting a ticket.  If you are 100% likely to get a ticket within a finite amount of time, then the p(t)=1-0.2^t answer for getting a ticket is invalid.
So lets consider linear coverage.
If you have an 80% chance of getting a ticket in one hour, the meter readers in your city walk 4 out of 5 blocks per hour (in 60 minutes).  Assuming they follow a pattern which has complete coverage and fairly evenly sweeps the whole parking area, they could have 100% coverage of all parking spots in 75 minutes.  That means no matter where you park, within 75 minutes you will get a ticket.  Under the linear coverage model, p(t)=0.8t.  So in half an hour, you have a 40% chance of getting a ticket.
This is a case of not trying to get too mathematically clever because the simplest answer could actually be the correct one.
The real answer depends whether your meter people ticket cars randomly or sweep areas linearly with even coverage (and gets more complicated with real life where they sweep some areas much more often than others but still get 100% coverage over a finite amount of time).
A: I agree that it depends on the model. Otherwise some fair of independence is presume in suggesting either 'half hours are equi-likely'
Although if you specify that is its an arbitrary hr, and any arbitrary half hour segment (rather than the first half of every hr) that might be different. Otherwise in every hr one inspector might give a zero percent chance for each first half hour, and 80 percent for the second 'half hour'. 
But once its specified that
' to be arbitrary hr, or any arbitrary half hr segment'or the average or expected probability value".
Or that 
" that the model or probability, value does  not pertain a one shot case (ie one is not considering a singular case scenario)- where it only applies to a singular occasion for some specific hr ."
Then I presume that this, would not be  be a counter-example to the claim (the second half - hour, is also the first half  of the full next hour segment from t/2 to 1.5 t where t = 1 hrs), and that the conventional answer sounds reasonable.
