Chapter 7 - Section 7.3 - Bayes' Theorem - Exercises - Page 476: 14

$\frac{7}{15}$

By using Baye's theorem, we get $P(F_{2}|E)=\frac{P(F_{2})P(E|F_{2})}{P(F_{1})P(E|F_{1})+P(F_{2})P(E|F_{2})+P(F_{3})P(E|F_{3})}=\frac{\frac{1}{2}\times\frac{3}{8}}{\frac{1}{6}\times\frac{2}{7}+\frac{1}{2}\times\frac{3}{8}+\frac{1}{3}\times\frac{1}{2}}=\frac{7}{15}$