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

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 E$_{i}$, of the sample space S, given that another event A has already occurred is:

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

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

$P(A)=\sum\limits^{n}_{i=1}{P(A|B_{i})P(B_{i})}$

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:

$P(Y^{c})=P(Y^{c}|X)P(X)+P(Y^{c}|X^{c})P(X^{c})$

          $=(0.10\times 0.04)+(0.85\times 0.96)\\=0.004+0.816\\=0.82$

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:

$P(X|Y)=\frac{P(Y|X)P(X)}{P(Y|X)P(X)+P(Y|X^{c})P(X^{c})}$

             $=\frac{(0.90\times 0.04)}{(0.90\times 0.04)+(0.15\times 0.96)}$

             $=0.20$

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