Is there something wrong with Bayes' Theorem

Question

Bayes' theorem (simplified): $P(A|B) = \frac{P(A) \times P(B|A)}{P(B)}$

Per the books/teachers I've had the honor to meet, the probability of an event $P(E)$ is such that $0 \leq P(E) \leq 1$ i.e. $P(E) = 0$ is possible.

But if $P(B) = 0$ then, $P(A|B) = \frac{P(A) \times P(B|A)}{P(B)} = P(A|B) = \frac{P(A) \times P(B|A)}{0} =$ undefined.

How do I resolve this problem, if it is a problem in the first place?

Answer 1

$P(B) = 0$ means that $B$ itself is impossible, or at least impossible to have happen "randomly". Which means among other things that $P(B|A)$ is also $0$, so really you have $0/0$, which is indeterminate and could take on any limiting value, which makes sense, you're asking about impossible situations here.

Answer 2

$P(A|B)$ yields the probability of $A$ given $B$ occurs. If $P(B) = 0$, then $P(A|B)$ being undefined makes sense because how can you find the probability of $A$ given $B$ occurs if $B$ never occurs? It's impossible and therefore undefined; so it's not necessarily a problem, but there's no point in doing $P(A|B)$ if you know that $P(B) \text{ is } 0$, though it is logically consistent.