Question

score on last try: 0.75 of 1 pts. see details for more. at least one scored part is incorrect. jump to first changable incorrect part. > next question get a similar question you can retry this question below a home pregnancy test was given to women, then pregnancy was verified through blood t following table shows the home pregnancy test results.

	positive	negative	total
not pregnant	7	69	76
total	87	73	160

find the following. round answers to 4 decimal places. a. p(positive | pregnant) = 0.9524
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) = 0.0548

Explanation:

Step1: Recall conditional - probability formula

The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{n(A\cap B)}{n(B)}$, 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 $P(\text{Pregnant}|\text{Positive})$

We know that $n(\text{Pregnant}\cap\text{Positive}) = 80$ and $n(\text{Positive})=87$. So $P(\text{Pregnant}|\text{Positive})=\frac{80}{87}\approx0.9195$.

Step3: Calculate $P(\text{Negative}|\text{Pregnant})$

We know that $n(\text{Negative}\cap\text{Pregnant}) = 4$ and $n(\text{Pregnant}) = 84$. So $P(\text{Negative}|\text{Pregnant})=\frac{4}{84}\approx0.0476$.

Step4: Calculate $P(\text{Not Pregnant}|\text{Negative})$

We know that $n(\text{Not Pregnant}\cap\text{Negative}) = 69$ and $n(\text{Negative})=73$. So $P(\text{Not Pregnant}|\text{Negative})=\frac{69}{73}\approx0.9452$.

Answer:

a. $0.9524$
b. $0.9195$
c. $0.0476$
d. $0.9452$