Probability: Prove that events are independent

Question

I'm self-learning probability and struggle on the following task:

If $A$ and $B$ are independent events, prove that $A \cup B$ and $A \cap B$ are also independent.

This is one of those cases where one event contains the other so they are always independent but I never came across to prove it.

Any clues? Please avoid answers in comments, because I cannot accept this way. Thanks!

Answer 1

Let $A=$ heads on first coin toss and $B=$ heads on second toss. They're independent. Then $$ \Pr(A\cup B) = \frac 3 4 \ne 1 = \Pr(A\cup B\mid A\cap B). $$ So they're not independent.

If event $E$ "contains" event $F$, that means $\Pr(E\mid F)=1$. If $\Pr(E)< 1$, then that conflicts with independence. So the statement in the question that if one event contains the other then they're independent is very far from true.

Answer 2

$A\cap B$ is never independent from $A\cup B$, unless either $A\cap B$ is negligible or $A\cup B$ is almost certain.

In fact, you should have $P(A\cap B)=P((A\cup B)\cap A\cap B)=P(A\cup B)P(A\cap B)$, hence $(P(A\cup B)-1)P(A\cap B)=0$.

Answer 3

$$P(A \cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)=P(A)+P(A')P(B)$$ $$P(\,(A \cup B)\cap (A \cap B)\,)=P(A\cap B)=P(A)P(B)$$ $$P(A \cup B)P(A\cap B)=P(A)^2P(B)+P(A')P(A)P(B)^2=P(A)P(B)(\,P(A)+P(A')P(B)\,)$$ therefore if $P(A)\ne 0$ , $P(A)\ne1$, $P(B)\ne 0$, and $P(B)\ne1$, then $$P(\,(A \cup B)\cap (A \cap B)\,)\ne P(A \cup B)P(A\cap B)$$