$A,B, C$ are independent $\implies (A \cup B)$ and $(B \cup C)$ are independent? Given that $A,B, C$ are independent events, I am trying to prove that $A \cup B$ and $B \cup C$ are independent.
$$P((A \cup B) \cap (B \cup C)) \\= P(B \cup (A \cap C)) \\= P(B) + P(A \cap C) - P(A \cap B \cap C) \\= P(B) + P(A)P(C) - P(A)P(B)P(C)\\ = P(B) + P(A)P(C) \left(1 - P(B) \right).$$
This clearly does not equal $P(A \cup B)P(B \cup C).$ How do I find an explicit counterexample to the claim that $A \cup B$ and $B \cup C$ are independent events? I am struggling because of the condition that $A,B,C$ must be independent.
 A: Consider an experiment of tossing two coins where each outcome is uniform. We have $\Omega =$ {H,T}$^2$. Label events:
$A =$ Event that first coin toss is heads. $P(A) = \frac{1}{2}$
$B =$ Event that second coin toss is heads $P(B) = \frac{1}{2}$
$C = \emptyset$ $P(C) = 0$
Check that $A,B,C$ are independent.
$P(A \cap B) = P({HH}) = \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2} $
$P(A \cap C) = P(\emptyset) = 0 = \frac{1}{2} \cdot 0 $
$P(B \cap C) = P(\emptyset) = 0 = \frac{1}{2} \cdot 0 $
$P(A \cap B \cap C) = P(\emptyset) = 0 = \frac{1}{2} \cdot \frac{1}{2} \cdot 0 $
Now, check whether $A \cup B$ and $B \cup C$ are independent.
$P((A \cup B) \cap (B \cup C)) = P({HH,TH}) = \frac{1}{2}$. However, this is not equal to $P(A \cup B) \cdot P(B \cup C) = \frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8}$, so not independent.
A: Let us assume that with a few inependent events A,B,C the equality is satisfied:
$${P(B)+P(A)P(C)(1-P(B))=P(A\cup B)P(B\cup C) (I)}$$
From set theory knowing that:
$${A \cup B = A \cup (B \setminus (A \cap B))}$$
$${P(A \cup B)=P(A \cup (B \setminus (A \cap B)))}$$
Since sets {${A}$} and {${B \setminus (A \cap B)}$} do not have any common elements, then we can use that ${P(X \cup Y)=P(X)+P(Y)}$, resulting in:
$${P(A \cup B)=P(A \cup (B \setminus (A \cap B)))=P(A)+P(B)-P(A \cap B)}$$ while ${P(A \setminus B)=P(A)-P(B)}$ is also true while ${A \supset B}$.
Then rewriting the right side of the (I) equation we get:
$${P(A\cup B)P(B\cup C)=(P(A)+P(B)-P(A \cap B))(P(B)+P(C)-P(B \cap C))}$$
$${=(P(A)+P(B)-P(A)P(B))(P(B)+P(C)-P(B)P(C))}$$
$${=P(A)P(C)[(1-2P(B)+P^2(B))]+P(B)[(P(A)+P(B)+P(C)-P(B)P(C)-P(A)P(B))] (II)}$$
Now we can find the equality between (I) equation's left side and (II):
$${P(B): 1=P(A)+P(B)+P(C)-P(B)P(C)-P(A)P(B)}$$
$${P(A)P(C): 1-P(B)=1-2P(B)+P^2(B) (III)}$$
From the (III) equation we have:
$${0=-P(B)+P^2(B)=P(B)(-1+P(B))}$$
Which means it is forced that P(B) is either 0 or 1 and it obviuosly satisfies (I) equation.
So the condition for your statement to be true, ${P(B)=0 \lor 1}$ have to be true.
A: Distilling the OP's counterexample in the self-answer: consider a single toss of a fair coin, and let $A,B$ and $C$ be the events of obtaining 2 Tails, a Tail, and 3 Tails, respectively.
Then $A \cup B=\{\text{tail}\}=B \cup C,$
and $P(A \cup B)P(B \cup C)=\frac14\ne\frac12=P\big((A \cup B)\cap(B \cup C)\big).$
