Answer to Question #327873 in Statistics and Probability for Keke

Question

A new pregnancy test was given to 100 pregnant women and 100 non pregnant women.The test indicated pregnancy of 92 of 100 pregnant and to 12 of the 100 non pregnant women.If a randomly selected woman takes this test and the test indicates that she is pregnant.What is the probability that she is not pregnant?

Answer 1

Let the event A is that the test indicates that the woman is pregnant,

B₁ - the woman is pregnant,

B₂ - the woman is not pregnant.

There are 100 pregnant and 100 non pregnant women,

$P(B_1) =P(B_2)=0.5,\\ P(A|B_1)=0.92,P(A|B_2)=0.12.$

The sought probability we'll find using Bayes theorem:

$P(B_2|A)=\\ =\cfrac{P(A|B_2)\cdot P(B_2)}{P(A|B_1)\cdot P(B_1)+P(A|B_2)\cdot P(B_2)}=\\ =\cfrac{0.12\cdot0.5} {0.92\cdot0.5+0.12\cdot0.5} =\cfrac{3}{26}.$