Do These Probability Conditions Imply Independence of Events A, B, and C?

[Patt]

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.

 

[mitochan]

Hi.
If all A,B and C are independent
P((A\cap B)\cup C)=1-(1-P(A)P(B))(1-P(C))=P(A)P(B)+P(C)-P(A)P(B)P(C),
P(A\cap (B\cup C))=P(A)[1-(1-P(B))(1-P(C))]=P(A)[P(B)+P(C)-P(B)P(C)] and of course
P(A\cap B \cap C)=P(A)P(B)P(C).

I have no idea how you made assumptions.

 

[PeroK]

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.

My first thoughts are:

If A, B, C are independent, then you can get an equation for P(C) in terms of P(A) and P(B). So, under certain circumstances your equation would hold.

If A, B, C are not independent, then the equation may hold by an arithmetic coincidence and it should be possible to construct an example.

I don't see that either equation you quote demands that A, B, C be independent or not independent.

 

[PeroK]

PS My second thought is that if A, B, C are independent, you can't get your equation to hold. It's impossible.

 

[WWGD]

You could start expanding using $P(A \cup B)=P(A)+P(B)-P(A \cap B)$ and then see what happens if the three are independent.

 

[Stephen Tashi]

PS My second thought is that if A, B, C are independent, you can't get your equation to hold. It's impossible.

What about P(A)=P(B)=P(C) = 0 ?

 

[WWGD]

What about P(A)=P(B)=P(C) = 0 ?

But this brings up the old discussion of whether we accept/consider events of probability 0. Would they even be in the ( standard, afaik) sample space?

 

[Stephen Tashi]

But this brings up the old discussion of whether we accept/consider events of probability 0. Would they even be in the ( standard, afaik) sample space?

One of the axioms of probability is that $\emptyset$ is a set in the probability space and the measure of the empty set is zero.

We can also consider the case A = B = C with P(A) = 1.