So, I understand what they are but not perfectly and I'd like to.
Take this example. We flip a fair coin. If it shows Heads, then we flip a fair 6-sided die, else, flip a 3-sided die.
Let A be the event an even number is rolled. Find $P(A|H)$
When I approach these questions, I never know whether to convert it into $P(A ∩ H)/P(H)$ or just answer it using the conditional. I'm not even sure you can without converting it but this is precisely why I'm confused. How do you approach conditionals? Even more complex ones
Edit: what's with the down votes? Did I say something wrong
To summarize the discussion in the comments:
There are $9$ possible outcomes to this game, though they are not equally probable.
They split into two groups, according to the coin toss. Within each group, the outcomes are equally probable.
The first group consists of six outcomes, namely: $$(H,1),\,(H,2),\,(H,3),\,(H,4), \,(H,5), \,(H,6)$$
using what I hope is obvious notation.
Now, this group collectively has probability $\frac 12$ since these are all the outcomes attached to tossing $H$ and that is a probability $\frac 12$ event. And these outcomes must be equally probable as the die is assumed to be fair. Thus each of these six outcomes has probability $\frac 1{12}$.
Of course, the other group, $(T,1),\,(T,2),\,(T,3)$ also has, collectively, probability $\frac 12$ so each of those three outcomes must have probability $\frac 16$, but we do not need this to address your question.
There are three outcomes that comprise the event $A\cap H$, namely: $$(H,2),\,(H,4),\,(H,6)$$
As each of these outcomes has probability $\frac 1{12}$, we see that $$P(A\cap H)=3\times \frac 1{12}=\frac 14$$
It follows that $$P(A\,|\,H)=\frac {P(A\cap H)}{P(H)}=\frac {1/4}{1/2}=\frac 12$$ and we are done.