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