Probability an event happens at least once in an interval given the event can happen a maximum of 5 times (truncated Poisson distribution)

Question

I am having a hard time in understanding how to factor in the maximum number of times a event can happen while calculating this probability. I understand the probability of an event happening in a time interval follows a Poisson distribution. But I am not sure how to proceed after that!

Answer 1

You have $X \sim \mathsf{Pois}(\lambda),$ and you seek $$P(X = k\,|\,X \le 5) = \frac{P(X = k,\, X \le 5)}{P(X \le 5)} = \frac{P(X = k)}{P(X \le 5)},$$ for $k = 0, 1, \dots, 5.$

The denominator $P(X \le 5)$ 'boosts' the six relevant values so that they add to $1.$ The general idea is sometimes discussed as 'truncation'. (For a general discussion of truncation, mainly from the point of view of continuous distributions, see this Wikipedia article.)

For the specific case $\lambda = 10$ the following computation in R statistical software compares the original and truncated Poisson PDFs. (Ignore brackets [ ].)

lam = 10;  k = 0:5;  pdf = dpois(k, lam)
t.pdf = pdf/ppois(5, lam)
round(cbind(k, pdf, t.pdf), 5)
     k     pdf   t.pdf
[1,] 0 0.00005 0.00068
[2,] 1 0.00045 0.00677
[3,] 2 0.00227 0.03384
[4,] 3 0.00757 0.11279
[5,] 4 0.01892 0.28198
[6,] 5 0.03783 0.56395

In the figure below, original Poisson probabilities are shown as centers of small red circles and truncated probabilities are shown as vertical bars.