# 6.2 - A Generalization

6.2 - A Generalization
Bayes' Theorem

Let the $$m$$ events $$B_1, B_2, \ldots, B_m$$ constitute a partition of the sample space $$\mathbf{S}$$. That is, the $$B_i$$ are mutually exclusive:

$$B_i\cap B_j=\emptyset$$ for $$i\ne j$$

and exhaustive:

$$\mathbf{S}=B_1\cup B_2\cup \ldots B_m$$

Also, suppose the prior probability of the event $$B_i$$is positive, that is, $$P(B_i)>0$$ for $$i=1, \ldots, m$$. Now, if $$A$$ is an event, then $$A$$ can be written as the union of $$m$$ mutually exclusive events, namely:

$$A=(A\cap B_1)\cup(A\cap B_2)\cup\ldots\cup (A\cap B_m)$$

Therefore:

\begin{align} P(A) &= P(A\cap B_1)+P(A\cap B_2)+\ldots +P(A\cap B_m)\\ &= \sum\limits_{i=1}^m P(A\cap B_i)\\ &= \sum\limits_{i=1}^m P(B_i) \times P(A|B_i)\\ \end{align}

And so, as long as $$P(A)>0$$, the posterior probability of event $$B_k$$ given event $$A$$ has occurred is:

$$P(B_k|A)=\dfrac{P(B_k \cap A)}{P(A)}=\dfrac{P(B_k)\times P(A|B_k)}{\sum\limits_{i=1}^m P(B_i)\times P(A|B_i)}$$

Now, even though I've presented the formal Bayes' Theorem to you, as I should have, the reality is that I still find "reverse conditional probabilities" using the brute force method I presented in the example on the last page. That is, I effectively re-create Bayes' Theorem every time I solve such a problem.

 [1] Link ↥ Has Tooltip/Popover Toggleable Visibility