If $A$ and $B$ are independent, are $A$ and $A\cup B$ also independent?

Question

I have somewhat of a hard time visualizing what independent means. I can understand that it would mean that the two events do not impact each other, such as flipping 2 coins. I also understand that it would mean that $P(A\cap B)=P(A)\times P(B)$.

So for this example I know that $P(A\cap(A\cup B))=P(A)\times P(A\cup B)$ if they were in fact independent. However, I do not know how to proceed from here. The only additional information given is the following: $P(A)=0.7$ and $P(B|A)=0.5$. However I do not know if this information is relevant.

Answer 1

No. For one thing, $A$ and $\emptyset$ are always independent, but $A$ and $A=A\cup\emptyset$ aren't unless $P(A)=0$ or $P(A)=1$.

As for the more mundane instance of a dice roll, i.e. $\{1,\cdots, 6\}$ with the uniform probability, $A=\{2,4,6\}$ and $B=\{3,6\}$ are independent, but $$P(A\cap (A\cup B))=\frac12\ne \frac12\cdot\frac 23=P(A)\cdot P(A\cup B).$$

Answer 2

No, $A$ and $A \cup B$ need not be independent. In fact $A \cap (A \cup B)$ is same as $A$. Independence of $A$ and $A \cup B$ would mean $P(A)=P(A)P(A \cup B)$ which means $P(A \cup B)=1$ which need not be true. For example, in two independent tosses of a fair coin let $A$ be the event that the first toss results in heads and $B$ be the event that the second toss results in heads. Then $P(A\cup B)= \frac 3 4$.

Answer 3

$A\cap(A\cup B)=A$ so $A\cup B$ and $B$ are independent iff $P(A)=P(A)\times P(A\cup B)$ or equivalently: $$P(A)=0\text{ or }P(A\cup B)=1$$This however is in general not a consequence of $P(A\cap B)=P(A)\times P(B)$.