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.

6 0

3 years ago

What information is needed in order to solve problems using the Pythagorean theorem?

gladu [14]

You would need to know the length of two sides either a2, b2, c2. you would also need to know what sides you are solving for, either one of the legs or hypotenuse. hope this helps, if you know how to solve a Pythagorean theorem problem this should be easy to answer

8 0

3 years ago

Can someone please help me on 20 and 21

Feliz [49]

For number 20 x = 4

3 0

3 years ago