Conditional probability, and Bayes rule (and Bayes rule with background knowledge)

Question

So, I've got a few questions:

$1)$ is $P(a,b) = P(b,a)$ ?

$2)$ How do I get some intuition for Bayes rule? I know don't really understand what is happening. $P(h|d) = (P(d | h) P(h) / P(d))$. I get the top half of the equation, we're finding the probability that d is occurring given h, but what does the bottom "do"? I think of multiplication as two events occurring in sequence, but what does a divide do?

$3)$ I don't really understand what a Bayes rule with background knowledge ($b$) is derived from. It seems to be a pretty esoteric equation that my professor gave me:

$$\mathbb P(h|d, b)=\frac{\mathbb P(d|h,b)\mathbb P(h|b)}{\mathbb P(d|b)}.$$ None of this is for homework.

Answer 1

For 2 and 3:

Consider probability space $(\Omega,\mathcal F,\mathbb P)$ and a finite or at least countable partition of $\Omega$. This will be $A_1,A_2,\ldots$, such that $A_i\cap A_j=\emptyset\,\forall i\neq j$ and $\bigcup\limits_{i}A_i=\Omega$ and $\mathbb P(A_i)>0 \,\forall A_i\in\mathcal F$. Then, for every $B\in\mathcal F$: $\mathbb P(B)=\sum\limits_i \mathbb P(B|A_i)\mathbb P(A_j)$.

This result is true and is classically proved like this:

$B=B\cap \Omega=B\cap \bigcup\limits_iA_i=\bigcup\limits_i(B\cap A_i)\Longrightarrow \mathbb P(B)=\sum\limits_i\mathbb P(B\cap A_i)=\sum\limits_i\mathbb P(B|A_i)\mathbb P(A_i)$.

Now, we can deduce Bayes rule as follows:

$$\mathbb P(A_i|B)=\frac{\mathbb P(A_i\cap B)}{\mathbb P(B)}=\frac{\mathbb P(A_i)\mathbb P(B|A_i)}{\mathbb P(B)}=\frac{\mathbb P(A_i)\mathbb P(B|A_i)}{\sum\limits_j\mathbb P(B|A_j)\mathbb P(A_j)}.$$

We can interpret this by telling that if $A_1,A_2,\ldots$ are possible causes of $B$, then this rule will help us know which one is the most probable cause.

Also, please do clarify what you mean by $\mathbb P(a,b)$.