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?
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
