Answer
a)\[P(B)=0.2\]
b)\[P({{A}^{C}}\cap (A\cup B))=P(B)-P(A\cap B)\]
Work Step by Step
(a)
Let A and B be any two events defined on S.
By DeMorgan’s law,
\[\begin{align}
& P({{A}^{C}}\cup {{B}^{C}})=P{{(A\cap B)}^{C}} \\
& P({{A}^{C}}\cup {{B}^{C}})=1-P(A\cap B) \\
\end{align}\]
We express \[P({{A}^{C}}\cup {{B}^{C}})\]as: \[1-P(A\cap B)\].
(b)
Let A and B be any two events defined on S.
We are given that,
\[\begin{align}
& P(\text{at least one of them occurs})=P(A\cup B)=0.3 \\
& P(A\text{ occurs and }B\text{ does not occur})=P(A\cap {{B}^{C}})=0.1 \\
\end{align}\]
Therefore,
\[\begin{align}
& P(B)=P(A\cup B)-P(A\cap {{B}^{C}}) \\
& P(B)=0.3-0.1 \\
& P(B)=0.2 \\
\end{align}\]
