There are 3 events: A = "Person is positive", B = "Person is negative", C = "A test says the Person is negative".
Given $P(B) \in [0,\infty] $, $P(C|A)$ is well defined and $P(C|B) > 0$.
Show that P(B|C) is well defined and describe P(B|C) as a function of $P(B)$.
Has anybody some Ideas how to start, I just can't see anything here?
Bayes' Theorem states that $P(A|B)\cdot P(B) = P(B|A)\cdot P(A)$
So $$P(B|C) = \frac{P(C|B)\cdot P(B)}{P(C)}$$
$$P(C|A) = \frac{P(C\cap A)}{P(A)}$$ $$P(C|B) = \frac{P(C\cap B)}{P(B)}$$ $A$ and $B$ are mutually exclusive and collectively exhaustive: there are only two possibilities and exactly one of them is true. As a result, $$P(C) = P(C\cap A) + P(C\cap B)$$ and rearranging the two prior formulas, $$P(C) = P(C|A)\cdot P(A) + P(C|B)\cdot P(B)$$ so for the first formula we get $$P(B|C) = \frac{P(C|B) \cdot P(B)}{P(C|A)\cdot P(A) + P(C|B)\cdot P(B)}$$ Now if we divide top and bottom by $P(C|B)$, we get $$P(B|C) = \frac{P(B)}{\frac{P(C|A)}{P(C|B)}\cdot P(A) + P(B)}$$ and we can get rid of $P(A)$ because $P(A) = 1 - P(B)$. Substituting that gives you the result.