Bayes' Theorem/General Result

Theorem
Let $\Pr$ be a probability measure on a probability space $\struct {\Omega, \Sigma, \Pr}$. Let $\set {B_1, B_2, \ldots}$ be a partition of the event space $\Sigma$.

Then, for any $B_i$ in the partition:


 * $\map \Pr {B_i \mid A} = \dfrac {\map \Pr {A \mid B_i} \, \map \Pr {B_i} } {\map \Pr A} = \dfrac {\map \Pr {A \mid B_i} \, \map \Pr {B_i} } {\sum_j \map \Pr {A \mid B_j} \map \Pr {B_j} }$

where $\displaystyle \sum_j$ denotes the sum over $j$.

Proof
Follows directly from the Total Probability Theorem:
 * $\displaystyle \map \Pr A = \sum_i \map \Pr {A \mid B_i} \, \map \Pr {B_i}$

and Bayes' Theorem:
 * $\map \Pr {B_i \mid A} = \dfrac {\map \Pr {A \mid B_i} \, \map \Pr {B_i} } {\map \Pr A}$