Answer
\[[({{A}_{1}}\cup {{A}_{2}}\cup ........\cup {{A}_{k}})\cup (A_{1}^{C}\cap A_{2}^{C}\cap ........\cap A_{k}^{C})]=S\]
Work Step by Step
Let \[{{A}_{1}},{{A}_{2}},........,{{A}_{k}}\] be any set of events defined on a sample space S and the complement of sets \[{{A}_{1}},{{A}_{2}},........,{{A}_{k}}\]are \[A_{1}^{C},A_{2}^{C},........,A_{k}^{C}\].
We have to find the value of event \[({{A}_{1}}\cup {{A}_{2}}\cup ........\cup {{A}_{k}})\cup (A_{1}^{C}\cap A_{2}^{C}\cap ........\cap A_{k}^{C})\].
According to DeMorgan’s law, the complement of an union of \[{{A}_{1}},{{A}_{2}},........,{{A}_{k}}\] is the intersection of the complements \[A_{1}^{C},A_{2}^{C},........,A_{k}^{C}\]that is,
\[(A_{1}^{C}\cap A_{2}^{C}\cap ........\cap A_{k}^{C})={{({{A}_{1}}\cup {{A}_{2}}\cup ........\cup {{A}_{k}})}^{C}}\]
Therefore, we can write the event \[({{A}_{1}}\cup {{A}_{2}}\cup ........\cup {{A}_{k}})\cup (A_{1}^{C}\cap A_{2}^{C}\cap ........\cap A_{k}^{C})\] as
\[\begin{align}
& ({{A}_{1}}\cup {{A}_{2}}\cup ........\cup {{A}_{k}})\cup (A_{1}^{C}\cap A_{2}^{C}\cap ........\cap A_{k}^{C}) \\
& =({{A}_{1}}\cup {{A}_{2}}\cup ........\cup {{A}_{k}})\cup {{({{A}_{1}}\cup {{A}_{2}}\cup ........\cup {{A}_{k}})}^{C}} \\
\end{align}\]
By using the definition, for any event A,
\[(A\cup {{A}^{C}})=S\]
Therefore,
\[\begin{align}
& =({{A}_{1}}\cup {{A}_{2}}\cup ........\cup {{A}_{k}})\cup {{({{A}_{1}}\cup {{A}_{2}}\cup ........\cup {{A}_{k}})}^{C}} \\
& =S \\
\end{align}\]
So, all possible outcomes in the sample space S are the outcomes of the event \[({{A}_{1}}\cup {{A}_{2}}\cup ........\cup {{A}_{k}})\cup (A_{1}^{C}\cap A_{2}^{C}\cap ........\cap A_{k}^{C})\].
