I'm finishing my guideline of exercises from a course I'm doing of Probability and Statistics and I'm presented with the next problem:
Suppose A,B and C are events and you know that:
i. A is independent of B∩C and B∪C
ii. B is independent of A∩C
iii. C is independent of A∩B
iv. P(A) P(B) P(C) > 0 (P meaning probability)
then the objective is to prove that A, B y C are independent.
I tried rewritting P(A∩B∩C) with the expectations to get it to be equal to P(A)*P(B)*P(C) but i couldn't finde the way.
Since $A$ is independent of $B\cup C,$ $$P(A)[P(B)+P(C)-P(B\cap C)]=P(A)P(B\cup C)=P(A\cap (B\cup C))=P((A\cap B)\cup (A\cap C))=P(A\cap B)+P(B\cap C)-P(A\cap B \cap C). $$ Using the independence of $A$ from $B\cap C$ and adding to both sides, $$ P(A)[P(B)+P(C)]=P(A\cap B)+P(A \cap C). $$ Using the independence of $A\cap B$ from $C$ and of $A\cap C$ from $B,$ $$ P(A)[P(B)+P(C)]=P(A\cap B)+P(A\cap C)=\frac{P(A\cap B \cap C)}{P(C)}+\frac{P(A\cap B \cap C)}{P(B)}=\frac{P(A\cap B \cap C)[P(B)+P(C)]}{P(B)P(C)}. $$ Cancelling terms and multiplying, $$P(A)P(B)P(C)=P(A\cap B \cap C).$$