A common blood test performed on pregnant women to screen for chromosome abnormalities in the fetus measures the human chorionic gonadotropin (hCG) hormone. Suppose that in a given population, 4% of fetuses have a chromosome abnormality. The test correctly produces a positive result for a fetus with a chromosome abnormality 90% of the time and correctly produces a negative result for a fetus without an abnormality 85% of the time.

What proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality?

Let's call A the event that a fetus have a chromosome abnormality, A' the a fetus doesn't have a chromosome abnormality, T that the test said that the fetus have a chromosome abnormality or it is positive and T' that the test said that the fetus doesn't have a chromosome abnormality or it is negative

Now, the proportion of women who get a positive test result that are actually carrying a fetus with a chromosome abnormality is equal to the probability P(A/T) that a fetus have a chromosome abnormality given that the test is positive.Therefore it is calculated as:

P(A/T)=P(A∩T)/P(T)

Where P(T)=P(A∩T)+P(A'∩T)

So, the probability P(A∩T) that a that a fetus have a chromosome abnormality and the test is positive is calculated as:

P(A∩T)=0.04*0.9=0.036

Because 0.04 is the probability that a fetus have a chromosome abnormality and 0.9 is the probability that the test is positive given that the fetus have a chromosome abnormality.

At the same way, P(A'∩T) is equal to:

P(A'∩T)=(1-0.04)*(1-0.85)=0.96*0.15=0.144

Because 0.96 is the probability that a fetus doesn't have a chromosome abnormality and 0.15 is the probability that the test is negative given that the fetus doesn't have a chromosome abnormality.

So, P(T) and P(A/T) are equal to:

P(T)=0.036+0.144=0.18

P(A/T)=0.036/0.18=1/5