a. By the law of total probability,
[tex]P(E)=P(A_1\cap E)+P(A_2\cap E)+P(A_3\cap E)[/tex]
and using the definition of conditional probability we can expand the probabilities of intersection as
[tex]P(E)=P(E\mid A_1)P(A_1)+P(E\mid A_2)P(A_2)+P(E\mid A_3)P(A_3)[/tex]
[tex]P(E)=0.1\cdot0.3+0.6\cdot0.5+0.8\cdot0.2=0.49[/tex]
b. Using Bayes' theorem (or just the definition of conditional probability), we have
[tex]P(A_1\mid E)=\dfrac{P(A_1\cap E)}{P(E)}=\dfrac{P(E\mid A_1)P(A_1)}{P(E)}[/tex]
[tex]P(A_1\mid E)=\dfrac{0.1\cdot0.3}{0.49}\approx0.0612[/tex]
c. Same reasoning as in (b):
[tex]P(A_2\mid E)=\dfrac{P(E\mid A_2)P(A_2)}{P(E)}\approx0.612[/tex]
d. Same as before:
[tex]P(A_3\mid E)=\dfrac{P(E\mid A_3)P(A_3)}{P(E)}\approx0.327[/tex]
(Notice how the probabilities conditioned on [tex]E[/tex] add up to 1)
