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Independence of Union and Intersection of Two Events

Suppose 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.

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