Are these 3 events independent? Not Solved yet. Can anyone help?

Question

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.

Last Edit: Lozano and antkam's andwers seem conflicting. Can anyone help?

Thank you

Answer 1

The first condition, $P((A \cap B )\cup C)=P(A)*P(B)*P(C)$, is actually very, very restrictive...

• For any $2$ events, $P(X \cup Y) \ge P(X)$

• For any event, $P(Z) \le 1$

So together we have $P((A \cap B) \cup C) \ge P(C) \ge P(A)P(B)P(C)$ for any $3$ events. If you really have a situation where the first term and the last term are equal, then all three are equal.

$$P(C) = P(A)P(B)P(C) \implies P(A)=P(B)=1 \,\,\,\,\,\,\text{or} \,\,\,\,\,\, P(C)=0$$

• Case 1: $P(A)=P(B)=1$, in this case, $P((A \cap B) \cup C) = 1 = P(C)$ and yes $A,B,C$ are independent.

• Case 2: $P(C) = 0$, in this case, $P((A \cap B) \cup C) = 0$ which further implies $P(A \cap B) = 0$, so you can conclude $A,B$ are exclusive. (All three can still be independent if $P(A)=P(B)=P(C)=0$.)

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The second condition, $P(A \cap( B \cup C))=P(A)*P(B)*P(C)$ is much less restrictive than the first, and I won't analyze it in full. However there are examples either way...

• Example where $A,B,C$ are independent: Just take $P(A)=P(B)=P(C)=0$, or, $P(B)=P(C)=1$.

• Example where $A,B,C$ are dependent: Roll an $8$-sided die with these faces: $A,A,A,ABC,BC,BC,BC,blank$. Then $LHS=RHS=1/8$.

Also, any example where $A,B,C$ are independent must involve some "degenerate" event ($P=0$ or $1$), because:

$$P(A)P(B)P(C) = P(A \cap (B \cup C)) = P((A\cap B) \cup (A \cap C)) \ge P(A\cap B)$$

so if they are independent we have $P(A)P(B)P(C) \ge P(A)P(B)$ which can happen only if $P(C)=1$ or $P(A)P(B)=0$, i.e. some event must be degenerate.

Answer 2

In boolean logic it'd be $A*B*(C-1)$ if they're independent. Thus here it is true, only if $P(C)=.5$

Answer 3

We know that $P((A \cap B) \cup C) = P(A \cap B) + P(C) - P(A \cap B \cap C) $

Substituting in what we know from the problem we get $P(A) \cdot P(B) \cdot P( C) = P(A \cap B) + P(C) - P(A \cap B \cap C) $

Which implies

$- P(A) \cdot P(B) \cdot P( C) + P(A \cap B) + P(C) = P(A \cap B \cap C) $

Which shows there can't be three-way independence.