Bayes' Theorem: Dividing P(B|A)*P(A) by P(B) for Accurate Probability

[xeon123]

in the Bayes' theorem, why P(B|A)*P(A) is divided by P(B)? What we want no achieve with the fraction?

 

[D H]

Start with the definition of conditional probability:
$$P(B|A) \equiv \frac{P(A\cap B)}{P(A)}$$
and similarly,
$$P(A|B) = \frac{P(A\cap B)}{P(B)}$$
Solving for $P(A\cap B)$ yields
$$P(A\cap B) = P(B|A)P(A) = P(A|B)P(B)$$
Bayes' theorem derives directly from this.