Independence of Union and Intersection of Two Events
\[A\]
and \[B\]
are such that \[A \cap B\]
and \[A \cup B\]
are independent.Then
\[P(A \cap B) \times P(A \cup B)=P((A \cap B ) \cap (A \cup B))\]
(1)But
\[(A \cap B) \subseteq P(A \cup B)\]
so (1) becomes \[P(A \cap B) \times P(A \cup B)=P((A \cap B )\]
.Now divide both sides by
\[P((A \cap B )\]
to give \[P(A \cup B)=1\]
.This implies that
\[A\]
and \[B\]
are mutually exhaustive. 
