Question

isn't Bayes' theorem all about conditional probability and isn't actually the conditional probability?

Answer 1

@satellite73 @TuringTest any help please?

Answer 2

baye's theorem is just some reworking the algebra for \(P(A|B)=\frac{P(A\cap B)}{P(B)}\)at least in its most basic form

Answer 3

sorry I didn't get it

Answer 4

since \(P(A\cap B)=P(B|A)P(A)\) and \(P(B)=P(A\cap B) +P(A^c\cap B)=P(B|A)P(A)+P(B|A^c)P(A^c)\) you get bayes by substitution

Answer 5

for two event bayes says
\[P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^c)P(A^c)}\]

Answer 6

it is just an algebraic reworking of the definition, but now the conditions have switched

Answer 7

so yes, it is in fact the conditional probability as you said

Answer 8

@satellite73 so Baye's thm is the basics/fundamental of what we study today in the modern books of probability about conditional probability?

Answer 9

It's easy to understand if you use venn diagrams

Answer 10

I do understand what is Baye's theoram in general (I mean how it works) I believe that is sort of the fundamental of modern conditional probability but I want to get suggestions from others so that I can see myself if I am wrong or not