Is there a non-circular explanation for computing the probability of the intersection of two dependent events?

Question

The explanation with which I'm familiar goes like this. Define the conditional probability P(A|B) as the probability of their intersections P(A and B) divided by the probability P(B). To figure out the probability of the intersection P(A and B) then, simply rearrange the terms of equation algebraically to obtain P(A and B) = P(A|B) X P(B).

The problem: What I'm struggling to understand is how to apply this equation in practice. It seems like whenever we want to figure out P(A and B), we need to figure out the values of P(A|B) and P(B).But to figure out the value of P(A|B), we already must figure out the value of P(A and B). So how is the equation P(A and B) = P(A|B) X P(B) helpful for figuring out the value of P(A and B)? In other words, to solve for it, we must figure out the value of P(A|B), which we can't do without knowing the value of P(A and B).

Answer 1

In practice, a lot of time you can calculate $P(A|B)$ by assuming $B$ happened and see what is the probability of $A$. For example, let's say you draw 2 cards face down, then randomly choose 1 to flip up, and let $B$ is the event that exactly 1 red card and 1 blue card was drawn, and $A$ is the event that the flipped up card is blue. Then you know $P(A|B)$ without knowing $P(B)$ nor $P(A\bigcap B)$, because assuming you have 1 red card and 1 blue card faced down, randomly picking 1 of them to flip up give you blue card with probability $\frac{1}{2}$.

Answer 2

If the occurrence of an event B is affected by the occurrence of another event A then we say that A and B are dependent events. So when the experiment is performed, it is known that event A has occurred. Does this affect the probability of B? This probability is the conditional probability. In principle you need to know both P(A and B) and P(B) to figure of P(A|B) (or P(A|B) and P(B) to find P(A and B)).

However, if the probability of B is unaffected by the prior occurrences of A, then we say that A and B are independent or that P(B|A)=P(B), which implies that P(A and B)= P(A)P(B).

Answer 3

Or, … just compute $P(A\cap B)$ by whatever methods you may have at hand. Say, we are rolling a fair die and $A$ is "number is even" and $B$ is "number is prime", both having probability $\frac12$ - what could be more straightforward than to observe that "$A$ and $B$" means "number is $2$" and conclude that $P(A\cap B)=\frac16$?

There is nothing "inherent" that can avoid what you call circularity. In a way, working with $P(A\mid B)$ (or $P(B\mid A)$) is just another way of expressing the fact that $P(A\cap B)$ need not be the product of $P(A)$ and $P(B)$.