# Conditional Probability - Cancelling the numerator with denominator...

If this question has been answered somewhere else, I apologize, and welcome the redirection.

I've been given the formula for conditional probability (the probability of event A happening, given event B) as: $$P(A | B) = \frac{P(A\text{ and }B)}{P(B)}$$

My question is if $P(A\text{ and }B) = P(A) * P(B)$ then wouldn't the probability of event B just cancel each other out in the numerator and denominator?

In my head, I know this can't be right, so I think I must have made a mistake when finding the probability of (A and B) but I'm not sure where I went wrong...or what formula I should be using instead.

Thanks in advance for any help/guidance!

$P(A \text{ and }B)=P(A)P(B)$ if and only if the events $A$ and $B$ are independent. For instance, consider flipping a coin, and let $A$ be the event that it lands on heads, and let $B$ be the event that it lands on tails. Clearly $P(A \text{ and }B)=0$, but $P(A)P(B)=1/4$.

if P(A and B)=P(A)∗P(B) then wouldn't the probability of event B just cancel each other out in the numerator and denominator?

Yes, but only if that were the case.   It isn't always so.   The product rule only applies in the case of independent events.   Not all events are independent; that's why we have conditional probabilities.

or what formula I should be using instead.

That entirely depends on what you have been given to work with.   What are your events?   What is the experiment/model?   Do you know $\mathsf P(B\mid A)$?   And so forth.

If you have the former two you should be able to evaluate $\mathsf P(A\cap B)$ from first principles.   If you have the latter, you can use: $\;\mathsf P(A\cap B) = \mathsf P(A)\;\mathsf P(B\mid A)\;$.

You may already have encountered Bayes' Rule, or maybe you will soon be introduced to it.

$$\mathsf P(A\mid B) = \dfrac{\mathsf P(A)\;\mathsf P(B\mid A)}{\mathsf P(B)}$$