## 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.