Question: The probability of getting sick is 5%. The probability of correct detection after getting sick is 80%. The probability of detection error in healthy people is 1%. Three consecutive tests are all getting sick. What is the actual probability of getting sick?
So this problem is confusing me because it seems to be a Bayes' law rule (as follows), but I don't know if the question itself is a typo. Why are they asking us of the "actual" probability of getting sick, when they give us the probability of getting sick (5%)? Here's what I'm thinking:
$$\Pr(\text{getting sick | 3 test results are positive}) = \tfrac{\Pr(\text{getting sick AND 3 are sick})}{\Pr(\text{3 test results are positive})}$$
$$\Pr(\text{3 test results are "sick"}) = \Pr(\text{3 are actually sick; 3 tests are correct}) + \Pr(\text{2 are actually sick, 1 isn't sick; 2 tests are correct, 1 is incorrect}) + \Pr(\text{1 is actually sick, 2 aren't sick; 1 test is correct, 2 are incorrect}) + \Pr(\text{0 are actually sick, 3 aren't sick; 3 tests are incorrect})$$
So I know how to find this probability, but I'm still confused as to how to find the "actual" probability. Is the actual probability no longer 5%, now that we are given this info regarding the 3 people?
I think that the question is about the same person. What the question asks is: without prior knowledge, the probability of someone being sick is 5%. A person is chosen at random and three tests made to him are positive, what is the probability of him being sick? (All test results are independent given the person is sick or healthy)
Answer:
$$\Pr(sick\vert 3 tests + )=\tfrac{\Pr(3 tests + \vert sick ) \Pr (sick)}{\Pr (3 tests +)}=\tfrac{0.8^3 \cdot 0.05}{0.8^3 \cdot 0.05 + 0.01^3 \cdot 0.95}=0.99996$$