I have solved one of those standard bayes rule application exercises a la:

Given a prevalence value of a disease, the sensitivity and the specificity of a test, calculate the probability that the patient has the disease given one positive test.

Now I am asked to determine how many (statistically independent) positive tests in a row are needed to obtain a probability of 99% that the patient has the disease.

How can I adapt the formula used for the first part of the exercise: \begin{equation} P(Dis|+) = \frac{P(+|Dis) \cdot P(Dis)}{P(+)} \end{equation} to incorporate several positive tests and to find out the needed number for the 99% threshold value?


1 Answer 1


Now the propability of disease given N positive tests (and only those tests) is

$ \begin{equation} P(Dis|N+) = \frac{P(N+|Dis) \cdot P(Dis)}{P(N+)} \end{equation} $


$ P(N+) = P(+)^N $


$ P(N+|Dis) = P(+|Dis)^N$

Then you just solve the smallest N that satisfies

$P(Dis|N+) \geq 0.99$.


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