Proof

(related to Theorem: Bayes' Theorem)

By hypothesis, \(p(B) > 0\). It follows from the definition of conditional probability. \[p(A_i|B)=\frac{p(A_i\cap B)}{p(B)}=\frac{p(B\cap A_i)}{p(B)},\quad\quad i=1,2,\ldots,n.\quad\quad( * )\]

Applying the probability of joint events to the nominator of \(( * )\) , we get \[p(A_i|B)=\frac{p(B|A_i)p(A_i)}{p(B)},\quad\quad i=1,2,\ldots,n.\quad\quad( * * )\]

Because by hypothesis all events \(A_1,A_2,\ldots,A_n\) are mutually exclusive and collectively exhaustive with \(p(A_i) > 0\) for \(i=1,2,\ldots,n\), we can apply the law of total probability to the denominator of \( ( * * ) \) and get the required result:

\[p(A_i|B)=\frac{p(B|A_i)p(A_i)}{\sum_{i=1}^np(B|A_i)p(A_i)},\quad\quad i=1,2,\ldots,n.\]


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References

Bibliography

  1. Bosch, Karl: "Elementare Einf├╝hrung in die Wahrscheinlichkeitsrechnung", vieweg Studium, 1995, 6th Edition