# I need to find the probability of an event happening given the population mean, a sample size and how many times I need it to happen

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Colin Leonelli on 24 Sep 2021 at 18:37
Answered: Star Strider on 25 Sep 2021 at 14:47
Cases happen at a rate of 0.9 per 9,000 patients per month. If I have 2,415 patients admitted, what is the probability of 5 cases

Image Analyst on 25 Sep 2021 at 14:16
rate = 0.9 / 9000
cases = rate * 2415
rate =
0.0001
cases =
0.2415
So, assuming a one month time period, since it's not 5, the probability is zero. Are you willing to allow some spread in the 0.9? If so, then how is it distributed? Is it a Normal distribution? If so, then there will be some probability of 5 cases, otherwise if it's exactly 0.9 (unlikely) then the probability is 0 and you won't have exactly 5 cases until you get 50000 patients seen.

Walter Roberson on 25 Sep 2021 at 14:29
p = 0.9/9000
n=5
N=2415
P = p^n*(1-p)^(N-n) * C(N, n)
where C(a, b) is a!/(b! (a-b)!) except you need to implement it better than using factorial
This would be for the situation of exactly 5 cases, not 5 or more cases

Star Strider on 25 Sep 2021 at 14:47
The question as stated actually doesn’t make much sense, because ‘cases’ should have some sort of probability distribution.
That aside, to find out the time required for a given number of patients to be diagnosed with that particular condition is relatively straightforward —
pr = 0.9 / 9000 % Monthly Probability
pr = 1.0000e-04
pts = 2415 % Patients Admitted
pts = 2415
Months = 5 / (pts * pr)
Months = 20.7039
So it would take a bit less than 21 months for that patient population to be diagnosed with 5 iincidents of whatever that condition is.
The calculation is — rearranging — or —  .