# Bayes Theorem - Multiple independent tests

Assume that 1% of the population suffers from an illness. There are two tests that diagnose the disease which are independent. A citizen decides to take both of the tests. Each of the tests makes a correct diagnosis of the disease with a probability of 0.95.

A)Find the probability of a person being sick given the fact that they tested positive in both tests.

B)Find the probability of a person being sick given the fact that they tested positive in at least one test.

If S is "person is sick", T1 is "test 1 is positive" and T2 is "test 2 is positive" then

$$P(S/T1T2)=\frac {P(T1T2/S)*P(S)}{P(T1T2)}$$

$$P(S/T1T2)=\frac {P(T1T2/S)*P(S)}{P(T1) * P(T2)}$$

I can calculate all the values except P(T1T2/S). In many similar examples on this site, the solution to the problem is P(T1T2/S)= P(T1/S)*P(T2/S). However, I do not understand how the former is accurate since if im not mistaken, P(AB/C) = P(A/B)*P(A/C) is valid only if A and B are conditionally independent given C. In this problem I thought that T1 and T2 were independent, are they conditionally independent given S as well?

Consider the following table. It represents the situation taking a population of 1,000,000 Citizens

As you can see,

A)

$$\mathbb{P}[D|T^{++}]=\frac{9025}{11500}\approx 78.478\%$$

B)

...$$\approx 9.366\%$$

• Aren't you doing the exact same thing as the other answers that I mentioned? For example, P(T1' S) = P(T2' S) = 0.0005, but how do you conclude that P(T1'T2'S) = P(T1'S)P(T2'S)? Commented Aug 23, 2021 at 9:50
• @I1corice: my table solution is to visualize the problem. I think it is useful for you. Commented Aug 23, 2021 at 9:55