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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

pts = 2415 % Patients Admitted

Months = 5 / (pts * pr)

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 —

.

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