Question involving Bayes' Rule The question is as follows: > Suppose a screening test for AIDS has the following features: (i) If a blood sample comes from someone with AIDS then the test will be positive 95% of the time. (ii) If the blood sample comes from someone without AIDS then the test will be negative 95% of the time. Suppose that 5% of the population has AIDS. If a blood sample tests positive, what is the probability that the person whose blood was tested has AIDS? Now I think the approach to solve this comes from Bayes' Rule. The partition $H_i$ would be if the person has AIDS or not, and the event $A$ would be if the test is positive or not. Would I be correct in assuming this? Bayes' Rule: $$\mathbb{P}(H_i\mid A) = \frac{\mathbb{P}(H_i) \mathbb{P}(A\mid H_i)}{\sum_{j=0}^\infty\mathbb{P}(H_j)\mathbb{P}(A\mid H_j)}$$ My guess is : $$\mathbb{P}(H_i\mid A) = \frac{(0.05)(0.95)}{(0.05)(0.95) + (0.95)(*)(0.95)^2}$$ The star is where I'm not 100% sure.

The chance of someone testing positive while having AIDS (=0.95*0.05) is the same as the chance of someone testing positive who does not have AIDS (=0.05*0.95)!

So, if you test positive, you are equally likely to have AIDS as to not have AIDS. So, if a person under these conditions tests positive, they have a 50% chance of having AIDS.