a-and-pab-pa-b-pb-which-equation-correctly-shows-the-derivation-of-bayes-formula

Question

Algebraically derive Bayes' theorem starting from the conditional probability identity P(B∣A)=P(A∩B)/P(A)P(B|A) = P(A ∩ B) / P(A) and P(A∣B)=P(A∩B)/P(B)P(A|B) = P(A ∩ B) / P(B). Which equation correctly shows the derivation of Bayes' formula?

Accepted Answer

P(B∣A)=P(A∣B)P(B)/[P(A∣B)P(B)+P(A∣Bc)P(Bc)]P(B|A) = P(A|B)P(B) / [P(A|B)P(B) + P(A|B^c)P(B^c)]

Answer

P(B∣A)=P(A∣Bc)P(Bc)/P(A)P(B|A) = P(A|B^c)P(B^c) / P(A)

Answer

P(B∣A)=P(A∪B)/P(A)P(B|A) = P(A ∪ B) / P(A)

Answer

P(B∣A)=P(B)/P(A)P(B|A) = P(B) / P(A)