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?
Thanks!
Your first calculation is incorrect. An average .5 cars / 5 minutes does not imply P(no car in 5 minutes) = 0.5. Consider for instance a process where every five minute, you see either no car with probability 90%, or 5 cars with probability 10%. On average you will see 0.5 cars every five minute, but the probability you see 0 cars in the next 5 minutes is clearly not 50%.
I haven't checked the computations for your second example; the calculation logic is looks correct, but the conclusion is incorrect: you are making an assumption about the distribution (Poisson) which is plausible but not implied by the problem statement.
If you take again my example, which is consistent with your problem description, the probability to see 0 cars in 10 minutes is 0.9 x 0.9 = 0.81, which gives you 19% of seeing one car or more. We could arbitrarily change my example to give you a wide variety of probabilities.
From your problem statement, the only thing you can say is that "in the long run, you'll see 0.5 cars every 5 minutes". Beyond that you can't make a statement on what should be expected within 10 minutes, unless you make some assumptions about the distribution of the cars arrivals.