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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 $p(h|D)$?

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The training data is already defined as the subset the machine is using for learning.

So I would redefine your parameters as follow :

Let $E$ be the dataset, and $D$ the subset of data used for training, with $card(E)=n$ and $card(D)=m$

then you have:

$p(h) = \frac{\vert{Instances-of-E-for-wich-h-is-satisfied}\vert}{card(E)}$

$p(D) = \frac{card(D)}{card(E)} = \frac{m}{n}$

$p(h/D) = \frac{\vert{Instances-of-D-for-wich-h-is-satisfied}\vert}{card(D)}$

$p(D/h) = \frac{\vert{Instances-of-D-for-wich-h-is-satisfied}\vert}{\vert{Instances-of-E-for-wich-h-is-satisfied}\vert}$

You can verify that you have $p(h/D) = \frac{p(D/h)p(h)}{p(D)}$

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