a home pregnancy test was given to women, then pregnancy was verified through blood tests. the following…
a home pregnancy test was given to women, then pregnancy was verified through blood tests. the following table shows the home pregnancy test results.\n| | positive | negative | total |\n|--|--|--|--| \n| pregnant | 62 | 4 | 66 |\n| not pregnant | 5 | 60 | 65 |\n| total | 67 | 64 | 131 |\nfind the following. round answers to 4 decimal places.\na. p(positive | pregnant) =\nb. what is the probability that the woman is pregnant given that the test is positive?\nc. given that a woman is pregnant, what is the probability that the test is negative?\nd. p(not pregnant | negative) =
Answer
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 the table values, 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 $P(\text{Positive}|\text{Pregnant})$
We know that $n(\text{Positive}\cap\text{Pregnant}) = 62$ and $n(\text{Pregnant})=66$. So $P(\text{Positive}|\text{Pregnant})=\frac{62}{66}\approx0.9394$.
Step3: Calculate $P(\text{Pregnant}|\text{Positive})$
$n(\text{Pregnant}\cap\text{Positive}) = 62$ and $n(\text{Positive}) = 67$. So $P(\text{Pregnant}|\text{Positive})=\frac{62}{67}\approx0.9254$.
Step4: Calculate $P(\text{Negative}|\text{Pregnant})$
$n(\text{Negative}\cap\text{Pregnant}) = 4$ and $n(\text{Pregnant})=66$. So $P(\text{Negative}|\text{Pregnant})=\frac{4}{66}\approx0.0606$.
Step5: Calculate $P(\text{Not Pregnant}|\text{Negative})$
$n(\text{Not Pregnant}\cap\text{Negative}) = 60$ and $n(\text{Negative}) = 64$. So $P(\text{Not Pregnant}|\text{Negative})=\frac{60}{64}= 0.9375$.
Answer:
a. $0.9394$ b. $0.9254$ c. $0.0606$ d. $0.9375$