Probability of having a disease - Bayes' Theorem problem 3% of the country has a disorder. However, the health institute recently developed a test for the disorder that has a 97% "true positive" rate (the probability that a person will test positive given that they have the disorder) and a 2% false positive rate (the probability that a person will test positive given that they do NOT have the disorder). If you take the test (independently) twice simultaneously and it shows that you got positive on one result and negative on another, what's the probability you actually have the disorder?

I've tried to solve this question for a couple of hours now, but I've had no luck. It seems like a Bayes' Theorem problem, but it seems like there are three "given" events to condition on. I'm not completely sure how to approach this problem. Can someone please help me? I'm not entirely sure how to utilize the fact that 3% of the country has a disorder.
I thought it might be the following:
$$\frac{0.03(0.97 \cdot 0.03 + 0.03 + 0.97)}{0.03 \cdot (0.97 \cdot 0.03 + 0.03 \cdot 0.97) + 0.97 \cdot (0.02 \cdot 0.98 + 0.98 \cdot 0.02)} \approx 0.043902,$$
but I'm really not sure.
 A: In order to find the posterior probability, you need two updates on the prior probability of any citizen having this disorder ($3\%$).
In general,
$P(T+| D) \cdot P(D) = P(D | T+) \cdot P(T+)$,
where $D$ means "the citizen has the disorder" and $T+$ means "the citizen tests positive".
In this case, $P(D)=3\%$ and $P(T+|D)=97\%$.
$P(T+)$ can be calculated as a partition of a domain:
$P(T+) = P(T+|D)\cdot P(D) + P(T+|\bar{D})\cdot P(\bar{D})$,
where $\bar{D}$ means "the citizen does not have the disorder", and $P(\bar{D})=1-0.03=97\%$. $P(T+|\bar{D})$ is the false positive rate (the probability of testing positive given that one doesn't have the disorder). In this case, $P(T+|\bar{D})=2\%$. Then,
$\begin{align}
P(T+) &= 0.97\cdot 0.03 + 0.02\cdot0.97 \\
&= 0.0485
\end{align}$
The probability that a citizen has the disorder after testing positive, then, is
$\begin{align}
P(D|T+) &= \dfrac{0.97}{0.0485}\cdot 0.03 \\
&= 60\%
\end{align}$
However, they test themselves again, and this time it comes back negative. You just have to apply this procedure again:
$P(T-|D)\cdot P(D) = P(D|T-)\cdot P(T-)$,
except you use $P(D)=60\%$, the updated probability that a citizen who tested positive has the disorder. In this case,
$\begin{align}
P(T-|D) &= 1-0.97 = 3\% \\
P(T-|\bar{D}) &= 1-0.02 = 98\%
\end{align}$
In my calculations, it came out to $P(D|T-)=4.4\%$
A: Assume the second test is independent of the first test (this means that the second test result is not affected by the first test). The probability that a person who has the disorder gets one + and one - testing results is 2 x 0.03 x 0.97 = 0.0582, while the probability a healthy person gets one + and one - testing results is 2 x 0.02 x 0.98 = 0.0392. Then applying Bayes theorem, we will get
P = (0.03 x 0.0582)/(0.03 x 0.0582 + 0.97 x 0.0392) = 4.39 %
This is the same as LmnICE's result, though the methods are somehow different.
