Question 1206680
i) This is simply 0.05, as we are given that the probability of a person having the disease is 0.05.
ii) There are two possibilities, the person has the disease and the test is positive, or the person doesn't have the disease and the test is positive. The probability of the first one is {{{0.05*0.9}}}, and the probability of the second one is {{{0.95*0.2}}}. Adding these together gives us 0.235.
iii) This is a conditional probability, so we use Bayes' Theorem. It tells us that P(~A|B)=(P(B|~A)P(~A))/P(B). P(B|~A) is the probability that the test result is positive given a person does not have the disease. Since there is a 0.8 chance that the result is negative given a person doesn't have the disease, there is a 0.2 chance that the result is positive given that a person doesn't have the disease. We calculated P(B) in the previous parts, and P(~A)=1-P(A)=0.95, so P(~A|B)=0.2*0.95/0.235, which is approximately 0.80851.
iv) We use Bayes' Theorem again, giving us P(A|~B)=(P(~B|A)P(A))/P(~B). P(~B|A) is the probability that the test result is negative given the person has the disease, which is 1-0.9=0.1. We know P(A), and P(~B)=1-P(B)=0.765. So plugging these in gives us P(A|~B)=0.1*0.05/0.765, which is approximately 0.006536.