Continuous conditional distribution This is about continuous conditional probability distributions. Why is it that they are allowed to take on single values, while this is a no-no with non conditional continuous distributions(to my knowledge) . 
What makes these the exception?
For an example, let X and Y be continuous random variables, it seems to be allowed to say P[ Y | X = 4] can be displayed as:
$f(x,y)/f_{x}(4) $
Thanks 
 A: Conditional probability is classically defined as:
$$\mathsf P(Y\in \mathcal B\mid X\in\mathcal A) ~=~ \dfrac{\mathsf P(X\in\mathcal A, Y\in\mathcal B)}{\mathsf P(X\in\mathcal A)}$$
This is okay as long as the event $\{X\in\mathcal A\}$ has non-zero probability mass.   In particular when the interval $\mathcal A$ contains discrete value(s) but the random variable is continuous, this definition is a "no-no", as you say ; eg: when $\mathsf P(X=x)=0$ it can not be done.
However that is probability mass.   We can work analogously with probability densities; if we take care.   If $X,Y$ are both continuous then conditional probability density is the ratio of the joint density to the marginal density (as long as that is non-zero at the point in question).
$$f_{Y\mid X}(y\mid x) ~=~ \dfrac{f_{X,Y}(x,y)}{f_X(x)}$$
And if $\mathcal B$ is a continuous interval.
$$\mathsf P(Y\in\mathcal B\mid X=x) ~=~ \dfrac{\int_\mathcal B f_{X,Y}(x,y)\operatorname d y}{f_X(x)}$$
We can also work with mixtures.   If $Y$ is a discrete random variable and $X$ a continuous one then we can say:
$$\mathsf P(Y=y\mid X=x) = \dfrac{f_{X\mid Y}(x\mid y)~\mathsf P(Y=y)}{f_X(x)}$$

Well, basically.   The modern approach is to use Measure Theory to tie this all together under a more abstract but rigorous umbrella.
