I do not quite understand how the probability, when I know $C$ happened, that $A$ occured is $\frac{P(A∩C)}{P(A∩C)+P(B∩C)}$. I am not sure how to write those probabilities, but I can show you what I mean with a numbers example. Here is the tree diagram:
$C$ and $D$ are the same events each time, but they have different probabilities. Let me make up an example:
Please correct me if I am using syntax wrong, I am a learner and this is why I add comments to each statement.
$P(A)=.3$, the probability that $A$ occurs
$P(B)=.7$, respectively that $B$ occurs
For $A$:
$P_A(C)=.4$
$P_A(D) =.6$
I think that $P_A(...)$ means "when $A$ happened probability of (...)"
For $B$:
$P_B(C)=.8$
$P_B(D) =.2$
At the end:
$$ P(A∩C) = (.3)(.4) = .12 $$ $$ P(A∩D) = (.3)(. 6) = .18 $$ $$ P(B∩C) = (.7)(.8) = .56 $$ $$ P(B∩D) = (.7)(.2) = .14 $$ With this numbers example the probability that, when $C$ happened, that $A$ happened would be $\frac{.12}{.12+.56}$.
I would like to understand why this works and also thoughts on notation. The exercise where I found myself wondering about this uses this notation: $P(A|C)$
We have the equation
$$ P(A \cap C)= P(C)P(A|C)\,. $$
In order for $A$ and $C$ to happen simultaneously, $C$ must happen (which takes place with probability $P(C)$) and $A$ must also happen (which takes place with probability $P(A|C)$, since we know that $C$ has already happened).
We can rearrange this into the following. $$ P(A|C)=\frac{P(A\cap C)}{P(C)} $$
You get your fraction by observing that $P(C)=P(A\cap C)+P(B\cap C)$. Why is this true? Well, you know that either $A$ happens or $B$ happens (but not both). So, in order for $C$ to happen, either $A$ and $C$ must happen together, or $B$ and $C$ must happen together.