A pharmaceutical company manufacturing pregnancy test kits wants to determine th

Question

A pharmaceutical company manufacturing pregnancy test kits wants to determine the probability of a woman not being pregnant when the test results indicate pregnancy. It is estimated that the probability of pregnancy among potential users of the kit is 10%. According to the company laboratory test results 1 out of 100 non - pregnant women tested pregnant (false positive). On the other hand, 1 out of 200 pregnant women tested non - pregnant (false negative)

a)A woman has just used the pregnancy test kit manufactured by the company and the results showed pregnancy. What is the probability that she is not pregnant?

b)A woman has just used the pregnancy test kit manufactured by the company and the results

showed that she is not pregnant. What is the probability that she is pregnant

Explanation / Answer

First some notation:

AC is the compliment of A P(AC) = 1 - P(A)

P(A intersect B) is the probability of A AND B happening.

P(A|B) is read probability of A given B and means the probability of A happening given that B has ALREADY happened. The mathematical definition is:

P(A|B) = P(A intersect B) / P(B)

Bayes theorem says

P(A|B) = P(B|A) P(A) / (P(B|A) P(A) + P(B|AC) P(AC))

Now onto the problem.

First, define events

A = User is pregnant
B = User is not pregnant
C = Positive result on test (pregnant)
D = Negative result on test (not pregant)

P(A) = .1 P(C|B) = .01 P(D|A) = .005

Notice that C and D are compliments, and So are A and B. Therefore.

P(B) = 1 - P(A) = .9

P(D|B) = 1 - P(C|B) = .99

We are interested in finding P(A|D)

by Bayes theorem

P(A|D) = P(D|A) P(A)/ (P(D|A) P(A) + P(D|B) P(B))

= .005*.1/(.005*.1 + .99*.9) = .00056

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