We can extend Bayes' theorem taking into consideration more probability events. Suppose that the events B₁,...,B_n are conditionally independent given A. Let ~A denote the complement of A. Then:

P(A|B₁,...,B_n) = P(B₁,...,B_n|A) * P(A) / P(B₁,...,B_n)

= [P(B₁|A) * ... * P(B_n|A) * P(A)] / [P(B₁|A) * ... * P(B_n|A) * P(A) + P(B₁|~A) * ... * P(B_n|~A) * P(~A)]

Proof:

Since the events B₁,...,B_n are conditionally independent given A (and also given ~A), we have the following:

P(B₁,...,B_n|A)=P(B₁|A) * ... * P(B_n|A)

Applying the simple form of Bayes' theorem and this fact, we thus have the following:

P(A|B₁,...,B_n) = P(B₁,...,B_n|A) * P(A) / P(B₁,...,B_n)

= P(B₁|A) * ... * P(B_n|A) * P(A) / [P(B₁,...,B_n|A)*P(A)+P(B₁,...,B_n|~A)*P(~A)]

= [P(B₁|A) * ... * P(B_n|A) * P(A)] / [P(B₁|A) * ... * P(B_n|A) * P(A) + P(B₁|~A) * ... * P(B_n|~A) * P(~A)]

This completes the proof as required.