(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?
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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?
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.
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 $\frac{P(A,B,C)}{P(A,B)} = \frac{\text{red}}{\text{red+brown}}$.
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.)