Show that $\mathbb{P}(B|C)$ is fully defined, i.e. the denominator is never $0$ or $\infty$.

Question

We have $3$ events, $A,B,C$ and $\mathbb{P}(B)=p \in ]0,1[$. We know that $\mathbb{P}(C|B) > 0$ and $r := \frac{\mathbb{P}(C|A)}{\mathbb{P}(C|B)}$. Show that $\mathbb{P}(B|C)$ is fully defined, i.e. the denominator is never $0$ or $\infty$.

I tried using the Bayes' Theorem and the conditional probability formula to somehow simplify the expression and use $\mathbb{P}(B)=p$ but I have get $\mathbb{P}(C)$ in the denominator, which is unknown. So now I am stuck on how to solve this.

Answer 1

$P(B|C)=\frac{P(B\cap C)}{P(C)}$. Since $P(C|B)\gt 0$, then $P(C)\gt 0$. In general all probabilities are $\le 1$.