Question

section 5.4 homework score: 5.47/14 answered: 6/14 question 7 a home pregnancy test was given to women, then pregnancy was verified through blood tests. the following table shows the home pregnancy test results.

	pregnant	not pregnant	total
negative	4	69	73
total	84	76	160

find the following. round answers to 4 decimal places.
a. p(positive | pregnant) =
b. what is the probability that the woman is pregnant given that the test is positive?
c. given that a woman is pregnant, what is the probability that the test is negative?
d. p(not pregnant | negative) =

Explanation:

Step1: Recall conditional - probability formula

The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}$. In terms of frequency from a contingency table, if $A$ and $B$ are events, $P(A|B)=\frac{n(A\cap B)}{n(B)}$, where $n(A\cap B)$ is the number of elements in the intersection of $A$ and $B$, and $n(B)$ is the number of elements in $B$.

Step2: Calculate P(Positive | Pregnant)

We want to find the probability that the test is positive given that the woman is pregnant. From the table, the number of pregnant women is $n(\text{Pregnant}) = 84$, and the number of pregnant women with a positive test is $n(\text{Pregnant}\cap\text{Positive})=80$. So $P(\text{Positive}|\text{Pregnant})=\frac{n(\text{Pregnant}\cap\text{Positive})}{n(\text{Pregnant})}=\frac{80}{84}\approx0.9524$.

Step3: Calculate P(Pregnant | Negative)

We want to find the probability that the woman is pregnant given that the test is negative. The number of women with a negative test is $n(\text{Negative}) = 73$, and the number of pregnant women with a negative test is $n(\text{Pregnant}\cap\text{Negative}) = 4$. So $P(\text{Pregnant}|\text{Negative})=\frac{n(\text{Pregnant}\cap\text{Negative})}{n(\text{Negative})}=\frac{4}{73}\approx0.0548$.

Answer:

$P(\text{Positive}|\text{Pregnant})\approx0.9524$
$P(\text{Pregnant}|\text{Negative})\approx0.0548$