I am getting confused on what should be a fairly simple concept.
A flu test gives true positives 90% of the time, and true negatives 85% of the time. 5% of people have the flu.
We want to find the probability that a healthy person tests positive two times.
To me I feel we should look at the false positives, people that test positive but are healthy. Which calculates at 100-90 = 10. Therefore the probability of testing positive twice and being healthy is 10% x 10% = 1%.
The results I'm seeing from another source is that we should be calculating the false positive rate from the true negative, i.e 100 - 85 = 15. And then 15% x 15% = 2.25%
This seems back to front to me, could I get clarification on which is actually right?
In theory if we have 100 people who tested positive, then to me it seems 90 of them actually have the virus (90% true positive), and 10 tested positive but don't (10% false positive).
As for the numbers involved... $90\%$ is the probability that a person tests positive given that they are truly sick. That is, $\Pr(\text{positive}\mid\text{is sick})$
On the other hand, $100\%-90\% = 10\%$ is the probability that a person tests negative given that they are truly sick... that is $$1-\Pr(\text{positive}\mid\text{is sick}) = \Pr(\neg \text{positive}\mid\text{is sick})=\Pr(\text{negative}\mid\text{is sick})$$ Again, this $10\%$ value is the probability of a false negative. It is not, as you thought, the probability of testing positive given not sick.
Reworded... $\Pr(A\mid B)+\Pr(\neg A\mid B)=1$ is a valid identity. You tried to say that $\Pr(A\mid B)+\Pr(A\mid\neg B)=1$ which is false and invalid.