This is just a hypothetical question I made up.
Assume that the coronavirus infects $1$ in $10$ persons. So, $P(D)=0.1,P(D^C)=0.9$. The RAPID testing kits developed to detect the presence of antibodies in the blood are not highly accurate. Their accuracy is $0.3$. Thus, $P(T|D)=0.3=P(T^C|D^C)$.
(i) What is the chance that a person has the disease, given he tests positive? $P(D|T)$
(ii) What is the chance that a person has the disease, given he tests positive twice? $P(D|TT)$
Solution.
(i) $P(D|T)=\frac{P(DT)}{P(T)}=\frac{P(T|D)P(D)}{P(T|D)P(D)+P(T|D^C)P(D^C)}=\frac{(0.3)(0.1)}{(0.3)(0.1)+(0.7)(0.9)}=0.04545=4.5 \%$
(ii) $P(D|TT)=\frac{P(DTT)}{P(TT)}=\frac{P(T|TD)P(TD)}{P(TT)}=\frac{P(T|TD)P(T|D)P(D)}{P(TT|D)P(D)+P(TT|D^C)P(D^C)}=\frac{P(T|TD)P(T|D)P(D)}{P(T|TD)P(T|D)P(D)+P(T|TD^C)P(T|D^C)P(D^C)}=0.02$
Are the above probabilities correct? And why does the chance of being infected fall, after two rounds, when in fact, the test should confirm, that you have the disease intuitively?
I tried switching the numbers and taking another testing kit with $P(T|D)=0.7$. In this case, the more you test, the more certain you are that you have the disease given additional information.
In the hypothetical posed above, as $P(T|D) < 0.5$, I believe that you become more and more doubtful, if you contracted the disease.