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

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

Answer 1

[Image Analyst]

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.

Answer 2

[Walter Roberson]

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

Answer 3

[Star Strider]

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 —

.