Likelihood that a sample is from one of two known distributions I start with exactly two, mutually-exclusive populations, each with a distribution along a single parameter: Let's say the height of males and female persons.  Now a new person shows up. By measuring the new person's height, I'd like to estimate the probability that person is either male or female. In my case, the two populations each appear to have a log-normal distribution and the distribution parameters (mean and standard deviation) have been estimated from a large subsample of each population.
How do I think about this and make such a calculation?
Assume I do not have the raw data for the populations, only the distribution parameters for the two log-normal distributions.
 A: Is it not simply the ratio of the PDFs at the newly sampled value?  Or more specifically..
P(sex=male|height) = PDF_male(height)/[PDF_male(height)+PDF_female(height)]
A: There are a ton of ways to do this, and much of machine learning focuses on such a probability prediction.
The easiest way to do this would be to fit a logistic regression to your data with height as the predictor of gender. Then you input the new height and get a probability of being male vs female.
Depending on how gross your distributions are, you may want to use a more complex model. For instance, if your male and feline heights have the same mean but different variances (let’s assume normality), then you might want to include a quadratic term in your logistic regression. If the distributions are all kinds of funky, you may have no idea what the function should be, so you might want to let a neural network run wild and find some nonlinear features (which is more or less what’s going on behind the scenes of neural nets).
A: You know the probability of height given gender. You seek the probability of gender given height. This is a job for Bayes rule.
Let $g$ denote gender and let $h$ denote height where $g \in \{M,F\}$ and $h \in (0,\infty)$. Bayes rule says
\begin{equation}
p(g|h) = \frac{p(h|g)\,p(g)}{p(h)} ,
\end{equation}
where
\begin{equation}
p(h) = \sum_g p(h|g)\,p(g) . 
\end{equation}
Of course $p(g=M) + p(g=F) = 1$ and consequently $p(g=F) = 1-p(g=M)$. In addition you have assumed $p(h|g=M)$ and $p(h|g=F)$ are both log-normal. (Nothing wrong with that; nothing special about it either.)
Given the height $h$, the odds in favor of $g=M$ relative to $g=F$ are
\begin{equation}
\frac{p(g=M|h)}{p(g=F|h)} = \frac{p(h|g=M)}{p(h|g=F)} \times \frac{p(g=M)}{p(g=F)} ,
\end{equation}
which says that the posterior odds ratio equals the likelihood ratio times the prior odds ratio.
It should be clear that what's missing from your setup is $p(g=M)$, which is the probability that the next person to be measured is male. (If that probability were 1/2, then the prior odds ratio would equal one.)
