Why is the P(B) in the denominator of Bayes theorem not equal to the joint probability of all evidences?

Question

I have three evidences: B1, B2, B3 (this means they are already known)
And I want to predict A given these evidences, so according to the bayes theorem, it should be:

P(A|B) = P(B1 | A) * P(B2 | A) * P(B3 | A) * P(A) / P(B)

Is this correct?

If yes, my question is that is the denominator P(B) in this formula equal to P(B1) * P(B2) * P(B3)? My teacher already told me this is wrong, but I just don't understand why.

Could experts please explain why as simple as possible because I am very new to probability and statistics.

Thank you very much for your help.

thanks for your reply, but according to wiki, your formula doesn't look like the Bayes theorem formula:
http://en.wikipedia.org/wiki/Bayes'_theorem

Answer 1

If I have understood (!!) : I think that the formula is :

P(A|B) = [P(A|B1) * P(B1) + P(A|B2) * P(B2) + P(A|B3) * P(B3)] / P(B)

P(B) = P(B1) * P(B2) * P(B3) only if B1, B2, B3 are independant. Perhaps we are not in this case here (!?)