Are these events independent?

[Lucdej]

Hello everyone.
Let us consider 3 events A,B,C such that: $$P((A \cap B )\cup C)=P(A)*P(B)*P(C)$$ Notice that the second term is a union and not an intersection. Are they independent? And what if the assumption was: $$P(A \cap( B \cup C))=P(A)*P(B)*P(C)$$? I know that the independence condition requires us to check whether the probability of the intersection of each pair factorizes plus the probability of the intersection of all of them factorizes as well. But I do not know how to prove that they are/they are not independent.
Thank you.

 

[Deleted member 4993]

Hello everyone.
Let us consider 3 events A,B,C such that: $$P((A \cap B )\cup C)=P(A)*P(B)*P(C)$$ Notice that the second term is a union and not an intersection. Are they independent? And what if the assumption was: $$P(A \cap( B \cup C))=P(A)*P(B)*P(C)$$? I know that the independence condition requires us to check whether the probability of the intersection of each pair factorizes plus the probability of the intersection of all of them factorizes as well. But I do not know how to prove that they are/they are not independent.Thank you.

Please explain the word "factorizes", with example, in your statement:

I know that the independence condition requires us to check whether the probability of the intersection of each pair factorizes plus the probability of the intersection of all of them factorizes as well.

 

[pka]

Let us consider 3 events A,B,C such that: \(\displaystyle P((A \cap B )\cup C)=P(A)*P(B)*P(C)\)Notice that the second term is a union and not an intersection. Are they independent? And what if the assumption was: \(\displaystyle P(A \cap( B \cup C))=P(A)*P(B)*P(C)\)? I know that the independence condition requires us to check whether the probability of the intersection of each pair factorizes plus the probability of the intersection of all of them factorizes as well. But I do not know how to prove that they are/they are not independent.

Lucdej, please review what you posted. In particular, note that \(\displaystyle P((A \cap B )\cup C)\not=P(A \cap (B \cup C))\)