Answer:
(a) The proportion of women who are tested, get a negative test result is 0.82.
(b) The proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.
Step-by-step explanation:
The Bayes' theorem states that the conditional probability of an event <em>E</em>
, of the sample space <em>S,</em> given that another event <em>A</em> has already occurred is:

The law of total probability states that, if events <em>E</em>₁, <em>E</em>₂, <em>E</em>₃... are parts of a sample space then for any event <em>A</em>,

Denote the events as follows:
<em>X</em> = fetus have a chromosome abnormality.
<em>Y</em> = the test is positive
The information provided is:

Using the above the probabilities compute the remaining values as follows:



(a)
Compute the probability of women who are tested negative as follows:
Use the law of total probability:


Thus, the proportion of women who are tested, get a negative test result is 0.82.
(b)
Compute the value of P (X|Y) as follows:
Use the Bayes' theorem:



Thus, the proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.