Question

A home pregnancy test is not always accurate. suppose the probability that the test indicates that a woman is pregnant when she actually is not is 0.02, and the probability the test indicates that a woman is not pregnant when she really is, is 0.03. assume that the probability that a woman who takes the test is actually pregnant is 0.7. what is the probability that a woman is pregnant if the test yields a not pregnant result?

Answer 1

Set Events:
T=tests positive~T=tests negativeP=subject is pregnant~P=subject is not pregnant
We are givenP(T n ~P)=0.02P(~T n P)=0.03P(P)=0.7
recall by definition of conditional probabilityP(A|B)=P(A n B)/P(B)

Need to find P(P|~T)
First step: make a contingency diagram of probabilities (intersection, n)
          P       ~P       sum 
T       0.67   0.02     0.69=P(T) 
~T     0.03   0.28     0.31=P(~T) 
sum   0.70  0.30     1.00
      =P(P)  =P(~P)

therefore
P(P|~T)=P(P n ~T)/P(~T)=0.03/0.31   [ both read off the contingency table ]
=0.0968