In continuation to this question
$p(h|D) = \frac{p(D|h)p(h)}{p(D)}$
$p(h) = $prior probability of hypothesis $h$
$p(D)$ = prior probability of training data $D$
$p(h|D)$ = probability of $h$ given $D$
$p(D|h)$ = probability of $D$ given $h$
I am inputting to a machine a training data $D$ in form of a table with $n$ training instances.
The machine read only $m$ training examples among $n$ training examples in order to get a hypothesis as output.
I have the following understanding: .
$p(h) = $prior probability of hypothesis $h$ = $\frac{\left\vert{Version- space}\right\vert}{\left\vert{Hypothesis-space}\right\vert}$
$p(D)$ = prior probability of training data $D$ = $\frac{\left\vert{Observed-training-data}\right\vert}{\left\vert{Training-data}\right\vert}$ = $\frac{{m}}{{n}}$
$p(D|h)$ = probability of $D$ given $h$ = $\frac{\left\vert{Training-examples-satisfied-by-hypothesis-h}\right\vert}{\left\vert{Training data}\right\vert}$ = $\frac{\left\vert{Training-examples-satisfied-by-hypothesis-h}\right\vert}{{n}}$
I do not know how to interpret
$p(h|D)$ = probability of $h$ given $D$
in the above forms.
Is my understanding about formulas for corresponding probabilities is correct and what is formula for $p(h|D)$?