I'm studying Bayes' Rule and came across the diseases problem. And wonder if it also works in other circumstances.
Let us say $D$ is the event having the diseases, $T$ is the event testing positive for the diseases and $N$ is the event testing negative . Usually, we will be given with the probability of $P(D)$, $P(T|D)$ and $P(N|\overline{D})$. So we can figure out $P(D|T) = \frac{P(T|D)P(D)}{P(T|D)P(D)+P(T|\overline{D})P(\overline{D})}$.
But What if I know $P(D)$, $P(D|T)$ and $P(\overline{D}|N)$, and trying to figure out $P(T|D)$. Is that possible? So far I can only get $$P(T|D) = \frac{P(\overline{D}|T)[P(\overline{D}|T)P(T))+P(D|T)P(T)]}{P(D)} = \frac{P(\overline{D}|T)[P(T|\overline{D})P(\overline{D})+P(T|D)P(D)]}{P(D)}$$ It seems like if I don't know $P(T)$ or $P(T|D)$ itself I can' find the answer. I wonder if there is a way to figure this out knowing the Bayes' Rule. Thank you.
IF $T$ and $D$ are independent with $P(D)=P(D\mid T)= 1-P(\overline{D}\mid N)$ then you are stuck. But assuming the test is not completely useless then you can say:
$P(D) = P(D\mid T) P(T) +P(D\mid N)P(N) = P(D\mid T) P(T) +(1-P(\overline D\mid N))(1-P(T))$
so $P(T)=\dfrac{P(D) -1+P(\overline D\mid N)}{ P(D\mid T) -1+P(\overline D\mid N)}$
and $P(T\mid D) = \dfrac{P(D\mid T) P(T)}{P(D)} = \dfrac{P(D\mid T)\,(P(D) -1+P(\overline D\mid N))}{P(D)\, (P(D\mid T) -1+P(\overline D\mid N))}$