Answer
a)We proved that,
\[A\cup (B\cup C)=(A\cup B)\cup C)=(A\cup B\cup C)\]
b)We proved that,
\[A\cap (B\cap C)=(A\cap B)\cap C)=(A\cap B\cap C)\]
Work Step by Step
(a)
Let A, B and C be any three events defined on a sample space S.
Let x be a member of \[A\cup (B\cup C)\]. Then x belongs to either A or \[(B\cup C)\] (or both). If x belongs to A, it also belongs to \[(A\cup B)\cup C)\].
If x belongs to \[(B\cup C)\], it belongs to B or C or both, so it must belong to \[(A\cup B)\cup C\] and also belongs to \[(A\cup B\cup C)\]. Now, suppose x belongs to \[(A\cup B)\cup C)\]. Then it belongs to either \[(A\cup B)\]or C or both. If it belongs to C, it must belong to \[A\cup (B\cup C)\]. If it belongs to \[(A\cup B)\], it must belong to either A or B or both, so it must belong to \[A\cup (B\cup C)\] and also belongs to \[(A\cup B\cup C)\].
(b)
Let A, B and C be any three events defined on a sample space S.
If ‘x’ is a member or outcome of \[A\cap (B\cap C)\], then x belongs to A and to\[(B\cap C)\]. If it is a member of A and of \[(B\cap C)\], then it belongs to \[(A\cap B)\]and to \[C\]. Thus, it is a member of \[(A\cap B)\cap C\] and also belongs to \[(A\cap B\cap C)\].
Conversely, choose ‘x’ in\[(A\cap B)\cap C\]. If it belongs to \[(A\cap B)\]or to \[(A\cap C)\]. Thus it also is a member of \[A\cap (B\cup C)\] and also belongs to \[(A\cap B\cap C)\].
Therefore, we proved that \[A\cap (B\cap C)=(A\cap B)\cap C)=(A\cap B\cap C)\].
