Conditional Probability

[jpanknin]

In my textbook (Devore, 9E), the section on conditional probability states that $P(A \cap B) = P(B | A) * P(A)$. But if $P(A \cap B) = P(B \cap A)$ and the intersection property is commutative, how do we know with certainty whether $P(A \cap B) = P(B | A) * P(A)$ or $P(A \cap B) = P(A | B) * P(B)$?

Seems there should be a rule or guideline for interpretation here given that $P(A | B) \neq P(B | A)$ (or not necessarily).

 

[blamocur]

In my textbook (Devore, 9E), the section on conditional probability states that $P(A \cap B) = P(B | A) * P(A)$. But if $P(A \cap B) = P(B \cap A)$ and the intersection property is commutative, how do we know with certainty whether $P(A \cap B) = P(B | A) * P(A)$ or $P(A \cap B) = P(A | B) * P(B)$?

Seems there should be a rule or guideline for interpretation here given that $P(A | B) \neq P(B | A)$ (or not necessarily).

What exactly is your question here?

 

[Dr.Peterson]

In my textbook (Devore, 9E), the section on conditional probability states that $P(A \cap B) = P(B | A) * P(A)$. But if $P(A \cap B) = P(B \cap A)$ and the intersection property is commutative, how do we know with certainty whether $P(A \cap B) = P(B | A) * P(A)$ or $P(A \cap B) = P(A | B) * P(B)$?

I think you're demonstrating that they will be equal! $P(B | A) P(A)=P(A \cap B) = P(A | B) P(B)$

Both are true, and you can use whichever is useful for some particular purpose.

Seems there should be a rule or guideline for interpretation here given that $P(A | B) \neq P(B | A)$ (or not necessarily).

Why does that matter?

 

[jpanknin]

I think you're demonstrating that they will be equal! $P(B | A) P(A)=P(A \cap B) = P(A | B) P(B)$

Both are true, and you can use whichever is useful for some particular purpose.

Why does that matter?

Let me think about this a bit. I started working through your response and I managed to figure part of it out and confuse myself on a few other parts.