Conditional probability and indeterminate form

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Consider a finite and discrete setup. It is well known that, given two events $A$ and $B$, the probability of $A$ knowing $B$ is:

$$P(A | B) = \frac{P(A \wedge B)}{P(B)}.$$

Suppose now that $B$ is impossible, i.e. $P(B) = 0$. Similarly, $P(A \wedge B) = 0$. Moreover, suppose that $A \neq B$. Therefore, I would get an indeterminate form:

$$P(A | B) = \frac{0}{0}.$$

Anyway, it seems more plausible to me that:

$$P(A | B) = \begin{cases} \displaystyle\frac{P (A \wedge B)}{P(B)} & \text{if}~P(B) > 0\\ 0 & \text{otherwise} \end{cases}.$$

Is this formalism right?