Why is $p(A) \times p(B|A) = p(B) \times p(A|B)$? Is there an easy way or simple example to get why it must be that
$$
p(A) \times p(B|A) = p(B) \times p(A|B)
$$?
I get that $p(A) \times p(B|A)$ can be seen as probability of $B$ occurring, weighted by the probability of $A$ occurring on which it is being conditioned on.
What's the background for this equation and is it generally valid?

Edited Question
Is there an easy way or simple example to get why then it must be that
$$
p(A \cap B) = p(A) p(B|A)
$$?
If I think of $A$ and $B$ as overlapping circles in a Venn-Diagramm, then it seems to me that $B|A$ already would describe the overlap.
 A: First, recall 
$$P(A|B)=\frac{P(A∩B)}{P(B)}$$
and consequently
$$P(A∩B)=P(A|B)P(B)$$
The trick here is to realize the very simple fact that $P(A∩B)= P(B∩A)$. This fact is quite intuitive; the probability that UNC wins and Duke loses is the same as the probability that Duke loses and UNC wins. So, in reality, we have two options:
$$P(A∩B)=P(A|B)P(B)$$
and
$$P(B∩A)=P(B|A)P(A)$$
and so
$$P(B|A)P(A)=P(A|B)P(B)$$
A: Short answer in words:  they're both equal to the probability of (A and B) occurring.
(Probability of A) times (Probability of B, given that A has happened) equals (Probability that both A and B happen).
Similarly, (Probability of B) times (Probability of A, given that B has happened) equals (Probability that both A and B happen).
A: If $A$ and $B$ are events such that $P(A), P(B) > 0$, how can I compute the probability that both events occurred? That is, how can I determine $P(A\cap B)$?
Well, if $A$ and $B$ are independent events, then I know that
$$P(A\cap B) = P(A)P(B)\tag{1}$$
(that is just the definition of independence) and so $P(A\cap B)$ is straightforward
to compute.  Wouldn't it be wonderful if I could write $P(A\cap B)$ in general
as $P(A)\alpha(B;A)$ where $\alpha(B;A)$ is some quantity
(obviously dependent on $A$ and $B$) with the magical property that
that $P(A)\alpha(B;A)$ has value $P(A \cap B)$? Now for all this to
happen, we must have that $\alpha(B;A) = P(B)$ when
$A$ and $B$ are independent (so that $(1)$ holds). We can say a bit
more about this magical function.  Since $(A\cap B) \subset A$, and so
$P(A\cap B) \leq P(A)$,
$\alpha(B;A)$ has maximum value $1$, and since it is possible that
$P(A\cap B) = 0$, $\alpha(B;A)$ can be as small as $0$. So, since
$\alpha(B;A)$ always has value in $[0,1]$, that is, it looks like a
probability and quacks (acts?) like a probability, we could even call it
a probability if we like, except we have not really said
 what $\alpha(B;A)$ is the probability of.  But whatever this
thingy is or what it means, we always have that
$$\begin{align}
P(A\cap B) &= P(A)\alpha(B;A) \tag{2}\\
\alpha(B;A) &= \frac{P(A\cap B)}{P(A)}\tag{3}
\end{align}$$
Similarly, if we interchange the roles of $A$ and $B$ in the above,
we can write
$$\begin{align}
P(B\cap A) &= P(B)\alpha(A;B) \tag{4}\\
\alpha(A;B) &= \frac{P(B\cap A)}{P(B)}\tag{5}
\end{align}$$
and since $A\cap B = B \cap A$, we can use $(2)$ and $(4)$ to 
deduce that
$$P(A)\alpha(B;A) = P(B)\alpha(A;B)\tag{6}$$
which looks  almost the same as what the OP is asking about.
So, what are these quantities $\alpha(B;A)$ and $\alpha(A;B)$ that 
are "defined" by $(3)$ and $(5)$?  Well, one way to think about this
is to consider that in order for both $A$ and $B$ to have occurred, 
obviously $A$ must have occurred (which has probability $P(A)$), and
given that we have already assumed that $A$ has occurred, we should
think of $\alpha(B;A)$ as the conditional probability of $B$ given
that we have already assumed that $A$ has occurred. Thus, we call
$\alpha(B;A)$ as the

conditional probability of $B$ given that $A$ has occurred, denoted 
  $P(B\mid A)$ and defined as
  $$ P(B\mid A) 
= \frac{P(A\cap B)}{P(A)} ~ \text{provided that} ~ P(A) > 0.\tag{7}$$

Notice that when $A$ and $B$ are independent, $P(B\mid A) = P(B)$,
that is, knowing that $A$ has occurred does not in the least change
your estimate of the probability of $B$.
All of which is fine and dandy, but what is the intuition behind
all this? Well, many people make sense of probabilities as long-term
relative frequencies and so let us consider a sequence of $N$ trials
of the experiment, $N$ large. If the event $A$ occurred $N_A$ times on
these $N$ trials and the event $B$ occurred on $N_B$ trials, then
$$P(A) \approx \frac{N_A}{N}, \quad P(B) \approx \frac{N_B}{N}.$$
Similarly, $$P(A\cap B) \approx \frac{N_{A\cap B}}{N}$$ where we
note that the trials on which $A\cap B$ occurred are necessarily
a subset of the $N_A$ trials on which $A$ occurred (as well as a
subset of the $N_B$ trials on which $B$ occurred).
But, $P(B\mid A)$ is the (conditional) probability of $B$ given that
event $A$ has occurred, and so let us consider the $N_A$ trials on
which $A$ has occurred.  On this subsequence of $N_A$ trials, what is
the relative frequency of $B$? Well, $B$ occurred on exactly $N_{A\cap B}$
of these $N_A$ trials, and so
$$P(B\mid A) \approx \frac{N_{A\cap B}}{N_A}
= \frac{\frac{N_{A\cap B}}{N}}{\frac{N_{A}}{N}}\approx \frac{P(A\cap B)}{P(A)}$$
which matches the definition in $(7)$.

OK, so now let the down-voting begin....
A: Just trying a graphical intuition... Hope it flies...
A: $p(B\mid A)p(A) = p(B\cap A)=p(A\mid B)p(B)$
A: A simple example is like this. Suppose we have A containing two events, "A = 1" or "A = 2", and B containing three events, "B = 1", "B = 2", or "B = 3". The (arbitrary) cell counts for the occurrence of the events are listed in the following table.

Now we can do a simple example:
\begin{align*}
P(A=1) & = \frac{6}{21} \\
P(B=2) &= \frac{7}{21} \\
P(A=1 \mid B=2) &= \frac{2}{7} \\
P(B=2 \mid A=1) &= \frac{2}{6} \\
\end{align*}
which yields to
\begin{align*}
P(A=1) \times P(B=2 \mid A=1) &= \frac{6}{21} \times \frac{2}{6} = \frac{2}{21} \\
P(B=2) \times P(A=1 \mid B=2) &= \frac{7}{21} \times \frac{2}{7} = \frac{2}{21}
\end{align*}
It's easy to verify that the equation does hold for other occurrence of event A and B. 
The equation is generally valid. That's no magic of course. 
If A and B are independent, then $P(B|A) = P(B)$, $P(A|B) = P(A)$. The two sides of the equation reduce to $P(A)P(B)$. The equation holds. 
If A and B are not independent, then it's easy to be proved by the definition of conditional probability (or Bayes' theorem).
A: Let us motivate the formula for conditional probability. We keep things are simple as possible. $S$, our sample space, will be finite. Let $E$ and $F$ be subsets of it (events).
When we write, $P(E|F)$ we are saying, "probability of $E$ given $F$". Thus, instead of the possibilities being in $S$, they are now all in $F$ since we are told that event $F$ has occurred. The new sample space is $F$. To find the probability of $E$ given $F$, it remains to count the number of possibilities which are in $E$. It would be wrong to write $|E|$ since this is counting the possibilities outside of $F$ (the new sample space). Therefore, there are $|E\cap F|$ possibilities of $E$ which happen within $F$. 
It follows by the finite probability formula (event size divided by sample size), 
$$ P(E|F) = \frac{|E\cap F|}{|F|} $$
But let us rewrite this formula to have probabilities on the right side. We use, 
$$ P(E|F) = \frac{|E\cap F|}{|S|} \cdot \frac{|S|}{|F|} = \left( \frac{|E\cap F|}{|S|} \right) \bigg/\left( \frac{|F|}{|S|} \right) = \frac{P(E\cap F)}{P(F)} $$
