Value of $P(A|A \cup B)$ in terms of $P(A)$ and $P(B)$ I did a few calculations (using Bayes' theorem) that got me to
$$ \frac{P(A)}{P(A \cup B)}$$
And it is not possible to write $P(A \cup B)$ in terms of the individual probabilities unless $A$ and $B$ are disjoint, which is information I do not have.
Someone pointed out that this was trivial, that since $A$ and $B$ occurred then the probability of $A$ occurring is $1$. Is this correct? 
 A: That someone is confusing union (at least one between $A$ and $B$ occurs) with intersection (both occur).
The formula to remember in order to express $P(A\mid A\cup B)$ in terms of $P(A)$, $P(B)$ and $P(A\mid B)$ - or, better, $P(B\mid A)$ -, is $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
A: 
And it is not possible to write $P(A∪B)$ in terms of the individual probabilities unless $A$ and $B$ are disjoint, which is information I do not have.

$\def\P{\mathsf P} $Not quite.   You cannot express the probability for the union in terms of probabilities for each event, unless the events are disjoint or independent.
$$\P(A\mid A\cup B) = \dfrac{\P(A)}{\P(A\cup B)} = \dfrac{\P(A)}{\P(A)+\P(B)-\P(A\cap B)}$$
So, if disjoint then $\P(A\mid A\cup B) =\dfrac{\P(A)}{\P(A)+\P(B)}$
Otherwise, if independent then $\P(A\mid A\cup B)=\dfrac{\P(A)}{P(A)+\P(B)(1-\P(A))}$


Someone pointed out that this was trivial, that since $A$ and $B$  occurred then the probability of $A$ occurring is $1$. Is this correct? 

Very much not so.   $A$ being a subset of the union does not guarantee that $A$ will happen whenever an outcome of $A\cup B$ occurs.
edit:  However, should $B$ be a subset of $A$, then; $\P(A\mid A\cup B) = 1$.  But is $B\subseteq A$?
A: It is true that $P(A|A \cap B)=1$, but that is not what you have.
