I have seen many similar questions on this forum; however, this question has two diagnostic tests, which appear to complicate the problem a bit.
Two tests (Test A and Test B) for a disease exist. Suppose that 1% of the population has the disease. Among those with the disease, 10% (and 5%) will incorrectly test negatively for the presence of the disease with Test A (and Test B). Among those without the disease, 5% (and 6%) will test incorrectly positive with Test A (and Test B). It is important to note that the results for the tests are independent.
(a) Given that both tests are positive when administered to a person selected randomly for the population, what is the probability that this person has the disease?
(b) Given that Test A is positive when administered to a person randomly selected from this population, what is the probability that Test B will also be positive?
(c) Given that a person selected randomly from this population actually has the disease, what is the probability that at least one of the two difference diagnostic tests given to this person will be positive?
(a)P(disease | Test A and Test B are positive) = P(Test A and Test B are positive|disease) * P(disease) / (P(Test A and Test B are positive)). Since we have conditional independence (the test results are independent from each other), the RHS becomes P(Test A is positive|disease) * P(Test B is positive|disease) * P(disease) / (P(Test A and Test B are positive)) = 0.00855/P(Test A and Test B are positive). However, I cannot find the denominator for this problem. How might I go about that?
(b) The formula for this answer follows from Bayes' Theorem: P(A and B are both positive) * P(B is positive) / P(A is positive), where I need to make use of the same argument that is in the denominator in (a).
(c) This is equivalent to saying 1 - P(A and B are both negative | disease) = $1-0.05 \cdot 0.10,$ due to the conditional independence. Therefore, this answer is $0.995$, right?