Chapter 7 - Section 7.6 - Bayes' Theorem and Applications - Exercises - Page 519: 18

$0.36$

According to Bayes' theorem: $P(C'|T')=\dfrac{P(T'|C')P(C')}{P(T'|C')P(C')+P(T'|C)+P(C)}~~~~~~~~(1)$ Here, we have $P(T|C')=0 \\ P(C)=75 \% =0.75 \\ P(T'|C)=60 \%=0.6$ Now, we will now use formula (1) and the given data to obtain: $P(F'|T')=\dfrac{P(T|F')P(F')}{P(T|F')P(F')+P(T|F)+P(F')}=\dfrac{(1-0)(1-0.75)}{(1-0)(1-0.75)+(0.6) (0.75)}$ or, $ \approx 0.36$