I'm not sure how to approach these 2 problems.

Question

1: A certain disease has an incidence rate of 0.5%. The false negative rate on a test for the disease is 5%; the false positive rate is 5%. Compute the probability that a person who tests positive actually has the disease. (You may find it useful to construct a probability contingency table.)

The probability a person who tests positive actually has the disease is __________

Give your answer accurate to at least 3 decimal places.

2:

P(A)=0.4, P(B)=0.65 and P(A and B)=0.2.

P(B∣A)= __________

P(notB∣A)= __________ 

P(notB∣notA)= __________

Answer 1

Use Bayes Theorem

Basic formula is

P(x)P(y/x) = P(y)(P(x/y)

replace x and y by A and B and positive and negative test results

P(A) = .4, P(B)=.65, P(A^B) = Probability of events A and B happening together given as = .2 which should = P(A)P(B) if independent events, but instead = .4(.65) = .26 .

P(A)P(B/A)= P(B)P(A/B)

.4P(B/A) = .65P(A/B)

P(B/A) = P(A^B)/P(A) = .2/.4 = .5

P(A/B) = P(A^B)/P(B) = .2/.65 = 2/65 = about .031

P(~B/A) = 1-P(B/A) = 1-.5= .5

P(