Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The problem:

If we have

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



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


share|cite|improve this question
up vote 1 down vote accepted

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)$

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.