Question

Janelle is taking a pregnancy test. She estimates that 0.5% of women at any given time are pregnant. Janelle takes an over-the-counter pregnancy test. The test tests positive (says your pregnant) for pregnant women 90% of the time. It gives a false positive 10% of the time.

Janelle tests positive (the test says she is pregnant). What is the probability that Janelle is actually pregnant? Use a tree diagram and Bayes theorem to show if Janelle is actually pregnant.

Answer 1

Tree Diagram:

0.90 Test Positive 0.005 Pregnant 0.10 Test negative Women 0.10 Test Positive 0.995 Not Pregnant 0.90 Test Negative

X : Event of a women pregnant

P(X) = 0.05/100 =0.005

$\overline{}$$\overline{X}$: Event of a women not pregnant

P($\overline{X}$) = 1-0.005 =0.995

A : Test Positive

$\overline{A}$ : Test Negative

The test tests positive (says your pregnant) for pregnant women 90% of the time

Probability that test is positive, given that women is pregnant = P(A|X) =0.90

False Positive 10% of the time,

Probability that the test is Positive, given that the women is not pregnant = P(A|$\overline{X}$) = 0.10

Given Janelle tests positive,  probability that Janelle is actually pregnant = P(X|A)

By Bayes theorem,

P(XA) = PO P(X)PAX) P(X)P(AX) + P(X)P(AX)

P(X) P(A|X) = 0.005 x 0.90 = 0.0045

P($\small \overline{X}$)P(A|$\overline{X}$) = 0.995 x 0.10 = 0.0995

P(XA) = P P(XP(AX) P(X)P(AX) + P(X)P(AX) 0.0045 0.0045 0.0045 + 0.0995.0.1040 = 100 = 0.043269231

Janelle tests positive,  probability that Janelle is actually pregnant = P(X|A) = 0.043269231