# Bayes, three tests, probability patient has disease if exactly two come out positive

I know how to apply Bayes Theorem to determine the likelihood a patient has a given disease if multiple trials return a positive result. In this case, if two trials both return positive:

$$P(\text{User}\mid++) = \frac{P(\text{User})P(+\mid\text{User})^2}{P(\text{User})P(+\mid\text{User})^2+(1 − P(\text{User})) P(+\mid\neg\text{User})^2}$$

However, how could I modify the above formula if one conducts three trials, and exactly two tests come out positive? What would be the likelihood that the patient has the disease then?

I would assume that $_3C_2$ comes into play as well, but I'm lost as to how to incorporate the negative result into the probability calculation.

• So "User" means a patient who has a given disease? Sep 18, 2018 at 0:28
• Draw the tree diagram. It should have 4 trials (User/Non-User ; +/- ; +/- ; +/-) and 16 outcomes. 6 of the outcomes pertain to the event that exactly two tests of three return positive results; 3 of these 6 outcomes also pertain to the patient being a User/Unhealthy/Diseased. The required probability P(User | exactly two tests of three return positive results) is then [sum of the probabilities of the latter 3 outcomes]/[sum of the probabilities of the former 6 outcomes], exactly as given by Graham. For reference, a similar tree diagram is here. Feb 21, 2022 at 16:32

I would assume that $^3\mathrm C_2$ comes into play as well,
To shorten typesetting, let $U$ be the event of the subject being a user (???), $S$ be the count of positive test results among three tests (which are presumed to be conditionally iid for a given subject, each with conditional probabilities $P(T\mid U)$ and $P(T\mid \neg U)$ ).
\begin{align}\mathsf P(U\mid S=2) &=\dfrac{\mathsf P(S=2\mid U)~\mathsf P(U)}{\mathsf P(S=2\mid U)~\mathsf P(U)+\mathsf P(S=2\mid \neg U)~\mathsf P(\neg U)}\\[1ex]&= \dfrac{^3\mathrm C_2\mathsf P(T\mid U)^2~\mathsf P(\neg T\mid U)~\mathsf P(U)}{^3\mathrm C_2\mathsf P(T\mid U)^2~\mathsf P(\neg T\mid U)~\mathsf P(U)+^3\mathrm C_2\mathsf P(T\mid \neg U)^2~\mathsf P(\neg T\mid\neg U)~\mathsf P(\neg U)}\\[1ex]&= \dfrac{\mathsf P(T\mid U)^2(1-\mathsf P(T\mid U))~\mathsf P(U)}{\mathsf P(T\mid U)^2(1-\mathsf P(T\mid U))~\mathsf P(U)+\mathsf P(T\mid \neg U)^2(1-\mathsf P(T\mid\neg U))(1-\mathsf P(\neg U))}\end{align}