I'm refreshing on bayes theorem and conditional probability and I ran across these practice problems. I was trucking along until problem 9, which states:
HIV The New York State Health Department reports a 10% rate of the HIV virus for the “at-risk” population. Under certain conditions, a preliminary screening test for the HIV virus is correct 95% of the time. (Subjects are not told that they are HIV infected until additional tests verify the results.) If someone is randomly selected from the at-risk population, what is the probability that they have the HIV virus if it is known that they have tested positive in the initial screening?
Seems pretty straight forward. Let P(H) denote the probability of having HIV.
P(H) = .1 P(~H) = .9 P(+ | H) = .95
and we need to find
P(H | +) = P(H)P(+ | H)/[(P(H)P(+ | H)) + (P(~H)P(+ | ~H))
After running the numbers, they somehow conclude that
P(+ | ~H) = 1 - P(+ | H) = .05
I figured this out by working backwards from their answer.
The problem is I have no idea how they reached this conclusion. Doing what they did above they should be including both P(- | H) and P(+ | ~H) together since these are the cases that are the opposite of P(+ | H).
The reason I'm confused is this doesn't seem like the proper way to refer to the probability that they tested positive given that they dont have HIV, strictly because what I suppose they did above also includes the case where they tested negative even though they have HIV.
Can someone walk me through their logic here? Thanks!