Probability : Do relationships $P(A \cap (A \cup B)) = P(A \cap B)$ and $P(A \mid A) = 1$ hold?

Question

For this, it may be important to mention that $A \text{ and } B$ are NOT mutually exclusive events. I am unsure if it is possible to have multiple events of A or B though, as I am only given the $P(A)$, $P(B)$, and $P(A \cap B)$. That being said, here are my questions:

Can you assume $P(A \cap (A \cup B)) = P(A \cap B)$?
Is it also reasonable to assume that $P(A \mid A) = 1$?

I feel like logically these both make sense, but I can't find anything online about something like this. Thanks in advance.

Answer 1

The first question is purely a set theory question, and not really related to probability. Since $A \subseteq A \cup B$ the intersection $A \cap (A \cup B)$ is just $A$. So $P(A \cap (A \cup B)) = P(A)$.

The second follows from the definition of conditional probability. For two events $A$ and $B$ (with $P(A) \ne 0$), we have $P(B \mid A) = P(A \cap B) / P(A)$. When $A=B$, this becomes $P(A \mid A) = P(A \cap A) / P(A) = P(A)/P(A)=1$.

Answer 2

angryavian deal with both parts formally, but here is an intuitive way to think about the second part: if you know that $A$ happened, what is the chance that $A$ happened?