Probability - A Conceptual Doubt

We have

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$

If $P(B)=0$ then the RHS is undefined.

Also, if we're given that $B$ happens then it cannot be the case that $P(B)=0$. That's a contradiction.

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Here is an example of when we can still calculate $P(A|B)$ when $P(B)=0$

Let $X\sim N(0,1)$ and $Y\sim N(0,1)$

Let $A$ be the event that $X\lt1$ and let $B$ be the event that $X+Y=2$

Then $P(A)=\Phi(1) \approx 0.8413$ and $P(B)=0$ since the normal distribution is continuous but still $P(A|B)=0.5$

It is to avoid the divide-by-zero error that occurs when you try to divide by zero.

The definition of a conditional probability mass function, that $\mathsf P(A\mid B):=\mathsf P(A\cap B)\div\mathsf P(B)$ is only viable when $B$ has an non-zero measure.

Still, however, when $\mathsf P(B)=0$ there are other compatible definitions for conditional probability measures that can be used; although they are not necessarily probability mass functions.