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:
![P(E_{i}|A)=\frac{P(A|E_{i})P(E_{i})}{\sum\liits^{n}_{i=1}{P(A|E_{i})P(E_{i})}}](https://tex.z-dn.net/?f=P%28E_%7Bi%7D%7CA%29%3D%5Cfrac%7BP%28A%7CE_%7Bi%7D%29P%28E_%7Bi%7D%29%7D%7B%5Csum%5Cliits%5E%7Bn%7D_%7Bi%3D1%7D%7BP%28A%7CE_%7Bi%7D%29P%28E_%7Bi%7D%29%7D%7D)
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>,
![P(A)=\sum\limits^{n}_{i=1}{P(A|B_{i})P(B_{i})}](https://tex.z-dn.net/?f=P%28A%29%3D%5Csum%5Climits%5E%7Bn%7D_%7Bi%3D1%7D%7BP%28A%7CB_%7Bi%7D%29P%28B_%7Bi%7D%29%7D)
Denote the events as follows:
<em>X</em> = fetus have a chromosome abnormality.
<em>Y</em> = the test is positive
The information provided is:
![P(X)=0.04\\P(Y|X)=0.90\\P(Y^{c}|X^{c})=0.85](https://tex.z-dn.net/?f=P%28X%29%3D0.04%5C%5CP%28Y%7CX%29%3D0.90%5C%5CP%28Y%5E%7Bc%7D%7CX%5E%7Bc%7D%29%3D0.85)
Using the above the probabilities compute the remaining values as follows:
![P(X^{c})=1-P(X)=1-0.04=0.96](https://tex.z-dn.net/?f=P%28X%5E%7Bc%7D%29%3D1-P%28X%29%3D1-0.04%3D0.96)
![P(Y^{c}|X)=1-P(Y|X)=1-0.90=0.10](https://tex.z-dn.net/?f=P%28Y%5E%7Bc%7D%7CX%29%3D1-P%28Y%7CX%29%3D1-0.90%3D0.10)
![P(Y|X^{c})=1-P(Y^{c}|X^{c})=1-0.85=0.15](https://tex.z-dn.net/?f=P%28Y%7CX%5E%7Bc%7D%29%3D1-P%28Y%5E%7Bc%7D%7CX%5E%7Bc%7D%29%3D1-0.85%3D0.15)
(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})](https://tex.z-dn.net/?f=P%28Y%5E%7Bc%7D%29%3DP%28Y%5E%7Bc%7D%7CX%29P%28X%29%2BP%28Y%5E%7Bc%7D%7CX%5E%7Bc%7D%29P%28X%5E%7Bc%7D%29)
![=(0.10\times 0.04)+(0.85\times 0.96)\\=0.004+0.816\\=0.82](https://tex.z-dn.net/?f=%3D%280.10%5Ctimes%200.04%29%2B%280.85%5Ctimes%200.96%29%5C%5C%3D0.004%2B0.816%5C%5C%3D0.82)
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})}](https://tex.z-dn.net/?f=P%28X%7CY%29%3D%5Cfrac%7BP%28Y%7CX%29P%28X%29%7D%7BP%28Y%7CX%29P%28X%29%2BP%28Y%7CX%5E%7Bc%7D%29P%28X%5E%7Bc%7D%29%7D)
![=\frac{(0.90\times 0.04)}{(0.90\times 0.04)+(0.15\times 0.96)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%280.90%5Ctimes%200.04%29%7D%7B%280.90%5Ctimes%200.04%29%2B%280.15%5Ctimes%200.96%29%7D)
![=0.20](https://tex.z-dn.net/?f=%3D0.20)
Thus, the proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.