# Is there a way to find $P(H_h)$, $P(E_\epsilon|H_\eta) (1 \leq \epsilon \leq e)$ and $P(E_\epsilon) (1 \leq \epsilon \leq e)$?

The problem:

If we have

$P(H_\eta|E_1,E_2,...,E_e)(1 \leq \eta \leq \mathbb{H})$

and

$P(E_1,E_2,...,E_e)$

for all True and False values of $E_\epsilon(1 \leq \epsilon \leq e)$ and $H_\eta(1 \leq \eta \leq \mathbb{H})$.

Can we find

$P(H_h)$, $P(E_\epsilon|H_h) (1 \leq \epsilon \leq e)$ and $P(E_\epsilon) (1 \leq \epsilon \leq e)$

??

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Yes. $P(H_{\eta}) = \sum P(H_{\eta} | E_1, E_2, \ldots, E_e) P(E_1, E_2, \ldots, E_e)$ where the sum is over all possible values of $E_1, E_2, \ldots, E_e$.

$P(E_{\epsilon}) = \sum_{ { E_1, \ldots , E_{\epsilon -1}, E_{\epsilon +1}, \ldots , E_e } } P(E_1, \ldots, E_e)$

The last one you need to use Bayes' Law: $P(E_{\epsilon} | H_{\eta}) = P(E_{\epsilon}, H_{\eta}) / P(H_{\eta})$. We've determined $P(H_{\eta}$ already, so we just have to get $P(E_{\epsilon}, H_{\eta})$.

$P(H_{\eta}, E_{\epsilon}) = \sum_{ \{ E_1, \ldots ,E_{\epsilon - 1}, E_{\epsilon + 1}, \ldots ,E_e \} } P(H_{\eta}, E_1 , \ldots E_e ) = \sum P(H_{\eta} | E_1 ,\ldots , E_e) P(E_1 ,\ldots E_e)$

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By definition of conditional probability and the fact that $P(H_h) = \sum P(H_h, E_1, E_2, \ldots, E_e)$. – Max Oct 25 '11 at 19:13
@Craig I needed modify a little this question. Is there a way to find this new probabilities? – GarouDan Oct 26 '11 at 15:41
Very interesting formulas. I will take a look and return. Thx. – GarouDan Oct 26 '11 at 19:09

Yes, the theorem which allows you to calculate $P(H_{h})$ from the given probabilities is called [Bayes' Theorem]. The extended form listed on the Wikipedia entry linked should cover this situation nicely.

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