Question

Suppose that 55% of all women that use a pregnancy test really are pregnant. Additionally, suppose that a pregnancy test accurately indicates that a woman was pregnant (+) 99% of the time and accurately indicates that a woman wasn’t pregnant (-) 99.2% of the time. What is the probability that the test gives a positive (+) reading? a. 0.4519 b. 0.5481 c. 0.55 d. 0.992 e. 0.99

Answer 1

P(P) = Probability of a women who uses a pregnancy test really is pregnant = 55/100=0.55

P(NP) = Probability of a women who uses a pregnancy test really is not pregnant = 1-0.55=0.45

+ : Pregnant test is +

- : Pregnant test is -

Probability that test is + given that the woman was pregnant = P(+|P) = 99/100 =0.99

Probability that test is - given that the woman was not pregnant = P(-|NP) = 99.2/100 =0.992

Probability that test is + given that the woman was not pregnant = P(+|NP) =

1- Probability that test is - given that the woman was not pregnant 99/100 = 1- P(-|NP) = 1- 0.992 = 0.008

Probability that test is + given that the woman was not pregnant = P(+|NP) = 0.008

probability that the test gives a positive (+) reading = P(+)

Event of Test is positive :(+)

= (Women is pregnant and test +) OR (Women is not Pregnant and test positive)

=(P and +) OR (NP and +)

P(+) = P(P and +) + P(NP and + )

P(+) = P(P) x P(+|P) + P(NP)+P(+|NP)

P(P) x P(+|P) = 0.55 x 0.99= 0.5445

P(NP) x P(+|NP) = 0.45 x 0.008 =0.0036

P(+) = P(P) x P(+|P) + P(NP)+P(+|NP) = 0.5445+ 0.0036 = 0.5481

probability that the test gives a positive (+) reading = 0.5481

Answer :

b. 0.5481