Probability of events Here is a probability problem: you observe .5 cars on average passing in front of you every 5 minutes on a road. What is the probability of seeing at least 1 car in 10 minutes?
I'm trying to solve this in 2 ways. The first way is to say: P(no car in 5 minutes) = 1 - .5 = .5. P(no car in first 5 minutes and no car in second 5 minutes) = P(no car in first 5 minutes) * P(no car in second 5 minutes) by independence. Therefore P(at least 1 car in 10 minutes) = 1 - .5*.5 = .75.
However, if I try the same, with a Poisson distribution with rate lambda = .5 per unit of time, for 2 units of time, I get: P(at least 1 car in 2 units of time) = 1 - exp(-2*lambda) = .63.
Am I doing something wrong? If not, what explains the discrepancy?
 A: There is actually a connection between your two calculations. As the commenters pointed out, the key is in interpreting the statement of "0.5 cars on average every 5 minutes". It could mean that in every 5 minute interval there is either 0 or 1 car passing, with P(1 car)=0.5. In that case your first calculation is correct.  
But you could get that average in lots of other ways. For example, if every minute there is ether 0 or 1 car passing, with P(1 car)=0.1, that is there are still 0.5 cars on average every 5 minutes, then a very similar calculation would get you P(at least 1 car in 10 minutes)$=1-0.9^{10} = 0.651$. Note how much closer it is to the Poisson result.
As you divide your 5 minute interval into more and more pieces (say, $k\rightarrow\infty$), and assume that only 0 or 1 car can pass during those $5/k$ minutes with probability $0.5/k$, that is still at 0.5 cars on average during 5 minutes, the probability from the first calculation will be $1 - (1 - 0.5/k)^{2k} \rightarrow 1-e^{-1}=0.632 $, which is the result of the Poisson-based calculation.
In fact, this construction is essentially the definition of a Poisson process. The other properties, like exponential inter-arrival time, are just consequences.
