Solution Consider a person chosen at random from among those screened. Let A be the
event that the person tests positive for the disease, B1 the event that the person actually
has the disease, and B2 the event that the person does not have the disease. Then
P(A | B1) = 0.99, P(Ac | B1) = 0.01, P(Ac | B2) = 0.97, and P(A | B2) = 0.03.
Also, because 5 people in 1,000 have the disease,
P(B1) = 0.005 and P(B2) = 0.995.