# Conditional probability and bayes theorem problem involving a medical test

I have a test that checks if a patient is sick (E = {patient is sick}) and gives either a positive (A={result is positive}) or a negative result.

Given that $P(A|E) = 0.95 = P(A^c | E^c)$ and $P(E) = 0.005$, calculate $P(E|A)$.

Since $P(E|A)=\frac{P(A|E)P(E)}{P(A)}$ I only need to find $P(A)$.

I thought of 2 ways of doing this but I'm not sure they work,

$$P(A)= P(A|E)P(E) + P(A | E^C) P (E^C)$$

Then I need to find $P(A|E^c)$, and this I'm unsure about. I imagine that $P(A|E^c) = 1 - P(A^c | E^c)$ since if I assume $E^c$ happens, then either $A$ or $A^c$ has to happen. If I do it this way, I get $P(A)=0.0545$.

The other way would be using

$$P(A^c|E^c)=\frac{P(A^c \cap E^c)}{P(E^c)} = \frac{P((A\cup E)^c)}{P(E^c)} = \frac{1 - P(A\cup E)}{P(E^c)} = \frac{1-P(A)-P(E)+P(A\cap E)}{P(E^c)}$$

And if I understand things correctly, $P(A \cap E) = 0$ in this sense, so I can get $P(A)$ out of that. This methos gives me $P(A)=0.04975$

So, I'm doing something wrong but the difference between the results is small enough that I don't get something like a probability over 1.

Any help would be greatly appreciated.

Your first calculation is correct. $P(A)=.0545$. For your second calculation, in fact, $\Pr(A\cap E)\ne 0$ (it equals $(.005)(.95)=.00475$.