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....
