 
Bayes theorem:
 Let us assume that, for some event A there are n possible causes, B_{1} to B_{n}.
For example, event A could be you finding a stray 100 note.
Event B_{i} is that note slipping from Bundle 'i'. 
 You wish to know P(B_{i}A), the probability that one of these causes, the ith, resulted in your observed event, A.
(Where P(XY) is the probability of randomly selecting X, given Y.) 
 Assuming:
 Each possible cause is mutuallyexclusive and nonoverlapping.
 For each of these causes, the probability of observing A is known.
In other words, if the kth explanation were correct, P(AB_{k}) is the probability of observing A. 
 The probability of each causal event, P(B_{i}), is also known.
If these causes are equally possible then, given n possible causes, P(B_{i})= ^{1}/n, but this is of academic interest, because they cancel out of the equation. 
 Then the probability that event B_{i}, resulted in your observed event, A, is:
P(B_{i}A) = 
P(B_{i})P(AB_{i}) 

P(A) 
 P(A), is the probability of observing A from any of the bundles, or Σ[P(B_{k})P(AB_{k})]
