Question

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.

(a) What proportion of women who are tested get a negative test result? Write a probability statement and find the answer.
(b) What proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality? Write a probability statement and find the answer.

Answer 1

(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.

Explanation:

The Bayes' theorem states that the conditional probability of an event E
`_(i)`, of the sample space S, given that another event A has already occurred is:

`P(E_(i)|A)=\fracP(A{\sum\liits^(n)_(i=1)P(A}`

The law of total probability states that, if events E₁, E₂, E₃... are parts of a sample space then for any event A,

`P(A)=\sum\limits^(n)_(i=1)P(A`

Denote the events as follows:

X = fetus have a chromosome abnormality.

Y = the test is positive

The information provided is:

`P(X)=0.04\\P(Y|X)=0.90\\P(Y^(c)|X^(c))=0.85`

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

`P(X^(c))=1-P(X)=1-0.04=0.96`

`P(Y^(c)|X)=1-P(Y|X)=1-0.90=0.10`

`P(Y|X^(c))=1-P(Y^(c)|X^(c))=1-0.85=0.15`

(a)

Compute the probability of women who are tested negative as follows:

Use the law of total probability:

`P(Y^(c))=P(Y^(c)|X)P(X)+P(Y^(c)|X^(c))P(X^(c))`

`=(0.10* 0.04)+(0.85* 0.96)\\=0.004+0.816\\=0.82`

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:

`P(X|Y)=(P(Y|X)P(X))/(P(Y|X)P(X)+P(Y|X^(c))P(X^(c)))`

`=((0.90* 0.04))/((0.90* 0.04)+(0.15* 0.96))`

`=0.20`

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