Answer
a)\[(F\cap R\cap V)=\{E\}\]
b)\[({{F}^{C}}\cap R\cap {{V}^{C}})=\{T,S\}\]
c)\[(F\cap {{R}^{C}}\cap V)=\{A,I\}\]
Work Step by Step
(a)
Suppose that each of the 12 letters in the word "TESSELLATION" is written on a chip.
Let F be the letters in the first half of alphabet that means letters between alphabets A to M and then the outcomes of event F are:
$F=\{A,E,I,L\}$
Let R be the letters that are repeated and then the outcomes of event R are:
\[R=\{T,E,S,L\}\]
Let V be the letters that are vowels and then the outcomes of event V are:
\[V=\{A,E,I,O\}\]
We have find \[(F\cap R\cap V)\]as:
\[\begin{align}
& (F\cap R\cap V)=\{(A,E,I,L)\cap (T,E,S,L)\cap (A,E,I,O)\} \\
& (F\cap R\cap V)=\{E\} \\
\end{align}\]
Therefore, there is only one letter ‘E’ in the event \[(F\cap R\cap V)\].
(b)
We are given that,
$F=\{A,E,I,L\}$
\[R=\{T,E,S,L\}\]
\[V=\{A,E,I,O\}\]
We have find \[({{F}^{C}}\cap R\cap {{V}^{C}})\] so we have find first \[{{F}^{C}}\] and \[{{V}^{C}}\],
\[{{F}^{C}}=\{T,S,O,N\}\]
\[{{V}^{C}}=\{T,S,L,N\}\]
So, we get \[({{F}^{C}}\cap R\cap {{V}^{C}})\]:
\[\begin{align}
& ({{F}^{C}}\cap R\cap {{V}^{C}})=\{(T,S,O,N)\cap (T,E,S,L)\cap (T,S,L,N)\} \\
& ({{F}^{C}}\cap R\cap {{V}^{C}})=\{T,S\} \\
\end{align}\]
Therefore, the event \[({{F}^{C}}\cap R\cap {{V}^{C}})=\{T,S\}\].
(c)
We are given that,
$F=\{A,E,I,L\}$
\[R=\{T,E,S,L\}\]
\[V=\{A,E,I,O\}\]
We have find \[(F\cap {{R}^{C}}\cap V)\] so we have find first \[{{R}^{C}}\],
\[{{R}^{C}}=\{A,I,O,N\}\]
So, we get the event \[(F\cap {{R}^{C}}\cap V)\]:
\[\begin{align}
& (F\cap {{R}^{C}}\cap V)=\{(A,E,I,L)\cap (A,I,O,N)\cap (A,E,I,O)\} \\
& (F\cap {{R}^{C}}\cap V)=\{A,I\} \\
\end{align}\]
Therefore, the event \[(F\cap {{R}^{C}}\cap V)=\{A,I\}\].
