Understanding the definition of $P(Y = y \mid X = x)$ Let $X: \Omega \rightarrow E_X$ and $Y: \Omega \rightarrow E_Y$ be random variables. By definition, we have that $P(Y = y \mid X = x)$ is defined as follows:
$$
P(Y = y \mid X = x) = \frac{P(X = x \cap Y = y)}{P(X = x)}
$$
Question: Isn't it possible that $X$ and $Y$ have distinct probability distributions (say, $P_1$ and $P_2$)? If so, which probability distribution does this definition refer to when it uses $P(\cdots)$?
 A: I'll give a concrete example.  Say you throw two dice and get a pair of numbers:
$$
\begin{array}{cccccccccc}
(1,1) & (1,2) & (1,3) & \cdots & (1,6) \\
(2,1) & (2,2) & (2,3) & \cdots & (2,6) \\
(3,1) & (3,2) & (3,3) & \cdots & (3,6) \\
\vdots & \vdots & \vdots & & \vdots \\
(6,1) & (6,2) & (6,3) & \cdots & (6,6)
\end{array}
$$
Suppose they're not equally likely, but rather, but dice have various biases, and furthermore because of magnets embedded in them the two tosses are not independent.  So there's some family of $36$ positive numbers that add up to $1$ assigned to these outcomes.
Now let
$$
\begin{align}
X & = \text{the sum of the numbers given by the two dice,} \\
Y & = \text{the maximum of the numbers given by the two dice.}
\end{align}
$$
So $X$ and $Y$ have different distributions.  (In particular, $Y$ cannot be more than $6$ whereas $X$ could be as high as $12$.
Now suppose we talk about $P(X=7)$ and $P(Y=4)$.
Q: Which probability distribution does "$P$" refer to?A: It refers to the $36$ numbers assigning probabilities to the $36$ pairs above.  It is the distribution of the random pair.
One often uses the Greek letter $\Omega$ to refer to the probability space that is the domain of the functions $X$ and $Y$ and whose members are (in this case) those $36$ pairs.  The distribution $P$ is a measure assigning probabilities to subsets of that space.
The notation $$P(X=7)$$ means
$$
P\{\omega\in\Omega : X(\omega)=7\} = P\{ (6,1),(5,2),(4,3),(3,4),(2,5),(1,6) \}.
$$
A: The  distribution of $Y$ given $X$  is neither the distribution of $Y$ nor the distribution of $X$. The distribution of $Y$ given $X$ can be  interpreted as the best guess you would give for $Y$ given the outcome of $X$. So it refers to values of $Y$ but it has available information from $X$.
A: Since $X,Y$ are random variables in a probility space $(\Omega,\sigma,P)$, the sets $\{\omega:X(\omega)=x\}$, $\{\omega:Y(\omega)=y\}$ are measurable and so it makes sense to talk about the probability of their intersection. The probability distribution of $X,Y$ play a role here, but it doesn't seem to be what you're thinking of. The probability $P$ is one for every event $E\in\Omega$. What varies is the induced probability on $\Bbb R$, $P_X(B)=P(X\in B)$ for $B$ a borel set, but this is mildly irrelevant here.
To understand this conceptually, consider the following: a train arrives to town with uniform distribution over the interval of time $(0,15)$ (that is, at some fixed hour say $1000$, the train arrives uniformly from $1000$ to $1015$, call this random variable $X$). The train departs uniformly from the moment it arrives to $1030$, call this random variable $Y$. 
Then $P(X\leqslant t)=\dfrac{t}{15}$ for $t\in (0,15)$. What is the probability that the train leaves before time $s$, given it arrived at time $t$? This is the probability that a random variable uniformly distributed over $(t,30)$ takes values at most $s$: 
$$P(Y\leqslant s\mid X=t)=\frac{s}{30-t} , s\in (t,30)$$
Differentiating this gives the probability density $$P(Y=s\mid X=t)=\frac{1}{30-t}$$
Integrating the former over the variable $t$ against the probability density of $X$ gives the probability distribution of $Y$, which is something not that simple the obtain at first sight. 
