Is there a way to proof to prove this for independence using the method of contradiction ?
Let $A_1, A_2,$ and $A_3$ be events, and let $B_i$ represent either $A_i$ or its complement $A^c _i$. Then there are eight possible choices for the triple $(B_1,B_2,B_3)$. Prove that the events $A_1, A_2, A_3$ are independent if and only if $P(B_1 \cap B_2 \cap B_3) = P(B_1)P(B_2)P(B_3) $, for all eight of the possible choices for the triple $(B_1, B_2, B_3)$.
The first thing I thought of is contradiction but I guess writing out $8 $ equations and showing that if they are independent, then $A_1, A_2, A_3$ must also be independent is also a good way. Is there any other way to prove this ?
Thanks in advance.