Question: When P(A) is small, P(A|B) P(A) * (P(B|A) /P(B|A c ) Why is this true? Expand Bayes' formula for P(A|B) in a Taylor series in

When P(A) is small,

P(A|B) P(A) * (P(B|A) /P(B|Ac)

Why is this true? Expand Bayes' formula for P(A|B) in a Taylor series in P(A).

P(A) stands for the probability of an event A

and Bayes' formula is given by P(A|B) = P(B|A)*P(A) /( P(B|A)*P(A) + P(B|Ac)*P(Ac))

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