Supplementary exercise 4.115 of IPS7e
-------------------------------------

Positive predictive value of a diagnostic test for an abnormal gene 
related to cystic fibrosis. Note that the calculations here will follow the 
scheme from slide 3L-10 closely. The information given to us is:

  test sensitivity = P(T+|D+) = 0.90
  test specificity = 1-P(T+|D-) = 1-0 = 1
  prevalence (European ancestry) = 1/25 = 0.04

We calculate the predictive value of a positive test result as follows:

  P(D+|T+) = P(T+|D+)*P(D+) / [P(T+|D+)*P(D+) + P(T+|D-)*P(D-)]
           = 0.90*0.04 / [0.90*0.04 + 0*0.96] = 0.90*0.04 / 0.90*0.04 = 1

The positive predictive value is 1! This should actually not come as a
surprise because the text of the question says explicitly that the test
is never positive for people who are not carriers. In other words, the
test never produces a false positive. Therefore, if a subject (Jason) tests
positive, we can be sure the person is indeed a carrier.

Let us for illustration also compute the negative predictive value. The
formula for this is similar to the formula for the positive predictive
value when switching D+ to D- and T+ to T-,

  P(D-|T-) = P(T-|D-)*P(D-) / [P(T-|D-)*P(D-) + P(T-|D+)*P(D+)]
           = 1*0.96 / [1*0.96 + 0.10*0.04] = 0.96 / 0.964 = 0.996

Also here we can be pretty sure the test gives the correct answer.