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Disease $G$ infects $1$ out of every $1000$ people. There's a test for Disease $G$ which is guaranteed to test positive for someone with the disease. For those without the disease, they will test positive $1\text{%}$ of the time.

You tested positive for Disease $G$, and your doctor wants to place you on an expensive emergency treatment. What's the probability you actually have it?

$P(Z) =1/1000$, $P(+|Z) =1/1000$, $P(Z') =999/1000$. $P(Z|+) = ???$ Also, how to calculate $P(+|Z')$?

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    "$P(+\mid Z')$" means "The probability of a positive test if you are not infected". Do you really need to _calculate_ that? – 2017-01-14
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    TCSHS, you appear to be confusing *conditional probability* with *joint probability*. The (conditional)probability of testing positive when given that the subject has the disease is said to be certain ($\mathsf P(Z\mid +)=1$). Thus the (joint)probability of testing positive *and* having the disease, $\mathsf P(Z\cap +)$, is $1/1000$. – 2017-01-14

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Actually, you are given that: $~\mathsf P(Z)=1/1000, \mathsf P(+\mid Z)= 1, \mathsf P(+\mid Z')=1/100$

That is, respectively, the probability of: a subject having the disease, testing positive given the subject is with disease (certain), testing positive given the subject is without disease ($1\%$).

You want to find the probability of having the disease given a positive test, $\mathsf P(Z\mid +)$, so use Bayes' Rule (and the Law of Total Probability):$$\mathsf P(A\mid B)~=~\dfrac{\mathsf P(A)\,\mathsf P(B\mid A)}{~\mathsf P(A)\,\mathsf P(B\mid A)+\mathsf P(A')\,\mathsf P(B\mid A')~}$$