Question

a home pregnancy test was given to women, then pregnancy was verified through blood tests. the following table shows the home pregnancy test results.

	positive	negative	total
not pregnant	5	60	65
total	67	64	131

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 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. $0.9375$