FWIW, there’s a nice proof of Bayes’ theorem in Russel and Norvig’s textbook, which I haven’t seen posted here yet.
Is this the one you meant?
P(A & B) = P(B | A) P(A) = P(A | B) P(B) Hold the second two statements equal and divide by P(A): P(B | A) = P(A | B) * P(B) / P(A)
P(A & B) = P(B | A) P(A) = P(A | B) P(B)
Hold the second two statements equal and divide by P(A):
P(B | A) = P(A | B) * P(B) / P(A)
Yes—though the accompanying text helped quite a bit too.
FWIW, there’s a nice proof of Bayes’ theorem in Russel and Norvig’s textbook, which I haven’t seen posted here yet.
Is this the one you meant?
Yes—though the accompanying text helped quite a bit too.