Prob: Observation coming from a specific continuous distribution I have a two-staged random process: First, I draw a type, which can be either $\tilde f$ or $\tilde g$. The (unconditional) probability of drawing the former type is $P_F$. 
Then, these two types have another characteristic, which is continuously distributed with CDFs $F(x), G(x)$ and PDFs $f(x)$, $g(x)$. 
I observe a realization $\tilde z$ on the second characteristic, and would like to compute the probability whether the realization $\tilde z$ belonged to $\tilde f$ or $\tilde g$.
For a better visualization, imagine that there are winning and loosing (paper) lottery tickets. A share $P_F$ of tickets is winner ($\tilde f$). These paper tickets have random weights, which are differently distributed for winner and looser tickets. You observe that a ticket weights $0.05g$ - what is the likelihood that it's a winner?
$$Prob(\tilde z \sim F(g) | \tilde z = z) = \frac{Prob( z \sim F(z) \wedge f(z))}{Prob( z \sim F(z) \wedge f(z)) + Prob( z \sim G(z) \wedge g(z))}\\
= \frac{P_F f(z)}{P_F f(z) + (1 - P_F)g(z)}$$
This is my approach - but it comes from discrete logic. I'm not sure whether it applies - something about continuously distributed variables having probability zero attached to any discrete value, and even if it does, the notation must be horribly wrong.
Could someone shed some light on this?
 A: You are correct. You are also correct that the probability of any discrete value from a continuous distribution is zero. In this case you can handle the problem by only looking at the probability of belonging to a small interval around $\bar{z}$, which is nonzero, and can be calculated as follows.
$$P(x<X<x+\epsilon) \approx \epsilon f(x)$$
Let $\hat{f}$ and $\hat{g}$ be the events that you draw either type $\bar{f}$ or $\bar{g}$, and $X$, $Y$ be random variables coming from $f(x)$ and $g(x)$.
Let's assume we've drawn something from the small interval $z_0 = [z, z+\epsilon]$.
$$P(Z\in z_0) = P(Z\in z_0|\hat{f}) P(\hat{f})+ P(Z\in z_0|\hat{g})P(\hat{g}) = 
P(X\in z_0) P(\hat{f})+ P(Y\in z_0)P(\hat{g})$$
$$P(\hat{f}|Z\in z_0) = \frac{P(Z\in z_0|\hat{f})P(\hat{f})}{P(Z\in z_0)} \approx
\frac{\epsilon f(z) P_F}{\epsilon f(z) P_F + \epsilon g(z) (1-P_F)} = 
\frac{f(z) P_F}{ f(z) P_F +  g(z) (1-P_F)} $$
This is the same result as you got, and since it does not depend on the interval size, you can easily take the limit as $\epsilon \to 0$, which will make the approximation into an equality.
Hope this helps you.
