What is it $\frac{P(A\cap B\cap C)}{P(B\cap C)}$? (My question is prompted by this post.)  
I'm not understanding why $\frac{P(A\cap B\cap C)}{P(B\cap C)}$is not equal to $\frac{P(A)P(B)P(C)}{P(B)P(C)} = P(A)$. I know it's wrong but I cannot understand why. Can you explain it please?
 A: If $A$, $B$, and $C$ are mutually independent you can say $P(A\cap B\cap C)= P(A)P(B)P(C)$. But, in general case, they are not independent, and you can't always say that.
A: It could be wrong, and it could be right, based on the independency/dependency of the events. Of course it's not generally true.
An example when it's not true:
Lets say
\begin{align*}
  \Omega &:= \left\{ x \in \mathbb{Z} \; : \; 1 \le x \le 31 \right\}, \\
  A &:= \left\{ x \in \Omega \; : \; 2|x \right\}, \\
  B &:= \left\{ x \in \Omega \; : \; 3|x \right\}, \\
  C &:= \left\{ x \in \Omega \; : \; 5|x \right\}
\end{align*}
In this case:
\begin{align*}
  A \cap B \cap C &:= \left\{ x \in \Omega \; : \; 30|x \right\}, \\
  B \cap C &:= \left\{ x \in \Omega \; : \; 15|x \right\}
\end{align*}
And the corresponding probabilities:
\begin{align*}
  P\left(A\right) &= \frac{15}{31} \\
  P\left(B\right) &= \frac{10}{31} \\
  P\left(C\right) &= \frac{6}{31} \\
  P\left(A \cap B \cap C\right) &= \frac{1}{31} \\
  P\left(B \cap C\right) &= \frac{2}{31}
\end{align*}
Hence:
\begin{align*}
  \frac{P\left(A \cap B \cap C\right)}{P\left(B \cap C\right)} &= \frac{\frac{1}{31}}{\frac{2}{31}} = \frac{1}{2}\\
  \frac{P\left(A\right) P\left(B\right) P\left(C\right)}{P\left(B\right) P\left(C\right)} &= P\left(A\right) = \frac{15}{31}
\end{align*}
And they are not equal. (And also $P\left(A \cap B \cap C\right) \ne P\left(A\right) P\left(B\right) P\left(C\right)$, so $A, B, C$ are not independent.)
(Some homework (as an example when it's true, remark what a small difference is enough): check for $\Omega := \left\{ x \in \mathbb{Z} \; : \; 1 \le x \le 30 \right\}$. It will be true, and $A, B, C$ will be independent. So it can be true, but not always, just in the case of independence.)
A: Suppose the three different domains A, B, C as shown below. A∩B∩C is the red colored surface. What you ask is similar to the Bayes' rule:
Given that A and B occur, what is the probability that C also occurs?

This is basically the probability $P(C|A,B)$ or if you prefer $P(C|A∩B)$, and this is given by Bayes' rule:
$$P(C|A,B) = \frac{P(A,B,C)}{P(A,B)}$$
If I use the colors from the picture above $\frac{P(A,B,C)}{P(A,B)} = \frac{\text{red}}{\text{red+brown}}$.
