Flipping Variables for a Poisson Distribution

Question

Not sure exactly how to word this, but I'm stuck on a current question about poisson distributions.

So far, the questions have gone along the lines of "What is the probability that x SUCCESSES will occur in a two-day period?" and "What is the probability that at least x SUCCESSES will occur in a two-day period? I've found the respective probabilities for these on python, but how would I approach a question worded like this:

"What is the probability that at least one of these two-day periods sees at least 2 SUCCESSES?"

For reference, my basic code looks like this for a question about at least 2 successes in a 2-day period:

mu = 0.11

k = 2

P = stats.poisson.cdf(k-1, mu)

P1 = 1-P

How can I figure out how to instead find the probability about two-day periods with at least two SUCCESSES?

Answer 1

You already know the answer to: "What is the probability that at least k SUCCESSES will occur in a two-day period?" In particular, you know this for $k=2$.

Let's call this probability $p_1$ (you called it P1, so I changed it slightly).

Your question: "What is the probability that at least one of these two-day periods sees at least 2 SUCCESSES?" For this to be meaningful, you need a number of those periods, let's call this $m$.

And, also, let's call this probability you are looking for $p_m$.

If your data is created by a proper Poisson process (and I will presume this here in my answer), then the events at different times are independent (given the Poisson process). That means there are two cases:

First Case: The two-day periods are disjoint:

If all your two-day periods are disjoint, then the counts of successes in each of those two-day periods are independent, and, since they are of the same length, their distributions are identical. So, first, you note that the probability $p_1$, that you have already figured out, is the same for all of those two-day periods. Next, you realize that the probability that you have at least one two-day period with at least two SUCCESSES is the same as one minus the probability that none has at least two SUCCESSES. Since the probability of "there are less than two SUCCESSES in a two-day period" is $1-p_1$, and because, as stated above, the counts in all those periods are independent, we get: $$ p_m = 1 - (1-p_1)^m. $$ (Recall, that the probability of a collection of independent events to all occur together is equal to the product of their individual probabilities.)

Second Case: The two-day periods are not disjoint: In this case, the counts of SUCCESSES in those two-day periods are not anymore independent. Think of e.g. having the same period twice. Then, of course, the number of SUCCESSES in those "two" periods is not independent. So the analysis would be far more complicated in this situation, so I hope you can content yourself with having only a solution to the first case.