Answer
a)\[P({{A}^{C}}\cup {{B}^{C}})=1-P(A\cap B)\]
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 know that,
\[\begin{align}
& P(A\cup B)=P(A\cap (A\cup B)+P({{A}^{C}}\cap (A\cup B)) \\
& P(A\cup B)=P(A)+P({{A}^{C}}\cap (A\cup B)) \\
\end{align}\]
Therefore,
\[\begin{align}
& P({{A}^{C}}\cap (A\cup B))=P(A\cup B)-P(A) \\
& P({{A}^{C}}\cap (A\cup B))=P(A)+P(B)-P(A\cap B)-P(A) \\
& P({{A}^{C}}\cap (A\cup B))=P(B)-P(A\cap B) \\
\end{align}\]
We express \[P({{A}^{C}}\cap (A\cup B))\]as: \[P(B)-P(A\cap B)\].
