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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.

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  • $\begingroup$ So "User" means a patient who has a given disease? $\endgroup$ – Michael Hardy Sep 18 '18 at 0:28
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I would assume that $^3\mathrm C_2$ comes into play as well,

Yes and no. They cancel.

but I'm lost as to how to incorporate the negative result into the probability calculation.

Just include them.

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 $P(T\mid \neg U)$ ).

$$\begin{align}\mathsf P(U\mid S=2) &=\dfrac{\mathsf P(S=2\mid U)~\mathsf P(U)}{\mathsf P(S=2\mid U)~\mathsf P(U)+\mathsf P(S=2\mid \neg U)~\mathsf P(\neg U)}\\[1ex]&= \dfrac{^3\mathrm C_2\mathsf P(T\mid U)^2~\mathsf P(\neg T\mid U)~\mathsf P(U)}{^3\mathrm C_2\mathsf P(T\mid U)^2~\mathsf P(\neg T\mid U)~\mathsf P(U)+^3\mathrm C_2\mathsf P(T\mid \neg U)^2~\mathsf P(\neg T\mid\neg U)~\mathsf P(\neg U)}\\[1ex]&= \dfrac{\mathsf P(T\mid U)^2(1-\mathsf P(T\mid U))~\mathsf P(U)}{\mathsf P(T\mid U)^2(1-\mathsf P(T\mid U))~\mathsf P(U)+\mathsf P(T\mid \neg U)^2(1-\mathsf P(T\mid\neg U))(1-\mathsf P(\neg U))}\end{align}$$

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