Given:
A is the number of people who have a disease is 0.01
B is when a drug tests positive
$\mathsf P(B\mid A)=0.9$
$\mathsf P(B\mid A')= 0.05$
I solved for $\mathsf P(B)=0.0585$ and $\mathsf P(A \cap B) = 0.009$
The question asks what is the probability of a person having the disease given that two independent tests that were done on him turned up positive?
Essentially the questions asks for $\mathsf P(A\mid C)$, where $C$ is both tests turning up positive or $B^2$.
I found $\mathsf P(C)$ from multiplying $0.0585$ by itself, but how do I find the probability that: A occurs and B occurs on both tests?
I'm leaning towards $\mathsf P(A \cap B) \times \mathsf P(A \cap B)$ since both tests are independent. However, when I divide by $\mathsf P(C)$ I got $0.02367$ which is smaller than $\mathsf P(A\cap B)$.
Shouldn't two positive independent tests increase the probability that the person indeed has the disease? I think my difficulty is in the syllogism of defining what A occurring and B occurring on both tests is.
Given the person has the disease, the probability of $2$ positives is $(0.9)^2$.
Given the person does not have the disease, the probability of $2$ consecutive positives is $(0.05)^2$.
Thus the probability of what you call $C$ is $(0.1)(0.9)^2+(0.99)(0.05)^2$.
The required conditional probability is therefore $$\frac{(0.1)(0.9)^2}{(0.1)(0.9)^2+(0.99)(0.05)^2}.$$
Remark: When you do the calculation, you will find that the answer accords better with the intuition. Yes, two tests are (a lot) better than one. The assumption of independence is, however, not entirely reasonable.
The answer has been brief. If detail is needed, please leave a comment.