Mutual Independence of union of events

Question

How to prove that the $P(A_{1} \cup A_{2}\cap A_{3} \cup A_{4} \cap A_{5} \cup A_{6}) = P(A_{1} \cup A_{2}) *P(A_{3} \cup A_{4}) * P (A_{5} \cup A_{6}) $ provided $P(A_{1} \cap A_{2} \cap A_{3} \cap A_{4} \cap A_{5} \cap A_{6}) = P(A_{1})*P (A_{2}) *P(A_{3}) *P(A_{4}) * P(A_{5}) *P(A_{6})$ mutually independent

Answer 1

Parentheses would be helpful: I think you mean $P((A_1 \cup A_2) \cap (A_3 \cup A_4) \cap (A_5 \cup A_6))$. The hypothesis must be not just $P(A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5 \cap A_6) = P(A_1) P(A_2) \ldots P(A_6)$, but rather that the events $A_1, \ldots, A_6$ are all mutually independent. Then $A_1 \cap A_2$, $A_3 \cap A_4$, $A_5 \cap A_6$ are mutually independent, and the result follows.

To see that $P(A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5 \cap A_6) = P(A_1) P(A_2) \ldots P(A_6)$ is not enough, consider a case where $P(A_6) = 0$.