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. find the following. give answers as a reduced fraction or decimal to at least 4 decimal places. a. ( p(\text{positive } | \text{ 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(\text{not pregnant } | \text{ negative}) = )

Explanation:

Step1: Recall the formula for conditional probability

The formula for conditional probability is \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). In terms of the table, \(P(A|B)=\frac{\text{Number of elements in }A\cap B}{\text{Number of elements in }B}\).

Step2: Solve part (a)

For \(P(\text{Positive}|\text{Pregnant})\), \(A = \text{Positive}\), \(B=\text{Pregnant}\).
\(P(\text{Positive}|\text{Pregnant})=\frac{\text{Number of pregnant and positive}}{\text{Number of pregnant}}=\frac{73}{78}\approx0.9359\)

Step3: Solve part (b)

For the probability that the woman is pregnant given that the test is positive. Let \(A=\text{Pregnant}\), \(B = \text{Positive}\).
\(P(\text{Pregnant}|\text{Positive})=\frac{\text{Number of pregnant and positive}}{\text{Number of positive}}=\frac{73}{80}=0.9125\)

Step4: Solve part (c)

For the probability that the test is negative given that a woman is pregnant. Let \(A=\text{Negative}\), \(B=\text{Pregnant}\).
\(P(\text{Negative}|\text{Pregnant})=\frac{\text{Number of pregnant and negative}}{\text{Number of pregnant}}=\frac{5}{78}\approx0.0641\)

Step5: Solve part (d)

For \(P(\text{Not Pregnant}|\text{Negative})\). Let \(A=\text{Not Pregnant}\), \(B=\text{Negative}\).
\(P(\text{Not Pregnant}|\text{Negative})=\frac{\text{Number of not - pregnant and negative}}{\text{Number of negative}}=\frac{78}{83}\approx0.9398\)

Answer:

a. \(0.9359\)
b. \(0.9125\)
c. \(0.0641\)
d. \(0.9398\)