Are $A,B,C$ independent given that $P((A \cap B )\cup C)=P(A)\cdot P(B)\cdot P(C)$

Question

Can someone help me shading light on this question about independence?The answers look conflicting. Are these 3 events independent? Not Solved yet. Can anyone help?

Answer 1

Assume that

$$P((A\cap B)\cup C) = P(A)P(B)P(C) \tag{*}$$

holds. Then we claim the following.

Claim. $\text{(*)}$ holds if and only if exactly one of the followings is true:

1. $P(C) = 0$ and $P(A\cap B) = 0$.
2. $P(A) = P(B) = P(C) = 1$.

Proof. Since 'if' part is easy to check, we only prove 'only if' part. If $P(C) = 0$, then

$$ P(A \cap B) = P((A \cap B) \cup C) \stackrel{\text{(*)}}= P(A)P(B)P(C) = 0. $$

Next, assume that $P(C) > 0$. Using $P((A\cap B) \cup C) \geq P(C)$, we get

$$ 0 \stackrel{\text{(*)}}= P((A \cap B) \cup C) - P(A)P(B)P(C) \geq (1 - P(A)P(B))P(C). $$

Then dividing both sides by $P(C)$ shows that $P(A)P(B) \geq 1$, which forces $P(A) = P(B) = 1$. From this,

$$ P(C) = P(A)P(B)P(C) \stackrel{\text{(*)}}= P((A\cap B)\cup C) \geq P(A\cap B) = 1, $$

which proves the desired claim. $\square$