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

Question

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 $P(E_\epsilon) (1 \leq \epsilon \leq e)$

??

Answer 1

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)$

Answer 2

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.