a. [tex]P(A')=1-P(A)=\boxed{0.7}[/tex]
b. [tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)=\boxed{0.4}[/tex]
c. By the law of total probability,
[tex]P(B)=P(A\cap B)+P(A'\cap B)[/tex]
so that
[tex]P(A'\cap B)=P(B)-P(A\cap B)=\boxed{0.1}[/tex]
d. Similar to (c); we have
[tex]P(A)=P(A\cap B)+P(A\cap B')[/tex]
[tex]\implies P(A\cap B')=P(A)-P(A\cap B)=\boxed{0.2}[/tex]
e. [tex]P(A\cup B)'=1-P(A\cup B)=\boxed{0.6}[/tex]
f. By DeMorgan's law,
[tex]P(A'\cup B)=P(A\cap B')'=1-P(A\cap B')=\boxed{0.8}[/tex]
