I do not understand the green part of this deriviation in a book about probability theory $\color{blue}{(1)}$: $$ P(A \mid B) = \color{green}{\frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)} \frac{P( A \cap B)}{P(A)}} = \frac{P(A)}{P(B)}P(B \mid A)$$
To me it looks like if we substituted these probabilities with normal numbers we could assign it randomly like this: $$ P(A \cap B) = 5, P(B) = 7, P(A) = 10) $$ And get$\color{blue}{(2)}$:
$$ \frac{5}{7} = \frac{10}{7} \frac{5}{10} $$
Which is true and I can tell that it gives us the so called backwards conditional probability as a final result.
My teacher calls this the conditional probability "reversal" formula, but this is actually what is know as Bayes Theorem. Other people just write it as$\color{blue}{(3)}$: $$ P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B)} $$
- Is it correct that this is actually Bayes Theorem that he is trying to derive just without using the name for it in $\color{blue}{(1)}$ ?
- What is the best way to explain my back of the envelope deductions in $\color{blue}{(2)} $ ? Because I guess it is just some algebra trick being used here but I would really appreciate if anyone could put mathematical terms on what is going on exactly.
EDIT: Yeah okay, I understand my confusion now. So they use a trick, they multiply with $\frac{P(A)}{P(A)}$ So they get: $$\frac{P(A)}{P(A)}\frac{P(A \cap B)}{P(B)} $$ This is equivalent to: $$ P(A)\frac{1}{P(A)} P(A \cap B)\frac{1}{P(B)} $$ And because multiplication is commutative we can just assemble it differently: $$\frac{P(A)}{P(B)}\frac{P(A \cap B)}{P(A)} $$
By definition : $$p(A | B) = \dfrac{p(A \cap B)}{p(B)} \text{ and } p(B | A) = \dfrac{p(A \cap B)}{p(A)}$$ then : $$p(A | B) p(B) = p(A \cap B) = p(B | A) p(A)$$ And Finally : $$p(A | B) = \dfrac{p(A)}{p(B)} p(B | A)$$