Why is P(A,B|C)/P(B|C) = P(A|B,C)?

Question

I understand $P(A\cap B)/P(B) = P(A|B)$. The conditional is the intersection of A and B divided by the whole area of B.

But why is $P(A\cap B|C)/P(B|C) = P(A|B \cap C)$?

Can you give some intuition?

Shouldn't it be: $P(A\cap B \cap C)/P(B,C) = P(A|B \cap C)$?

Answer 1

Any probability result that is true for unconditional probability remains true if everything is conditioned on some event.

You know that by definition, $$P(A\mid B) = \frac{P(A\cap B)}{P(B)}\tag{1}$$ and so if we condition everything on $C$ having occurred, we get that $$P(A\mid (B \cap C)) = \frac{P((A\cap B)\mid C)}{P(B\mid C)}\tag{2}$$ which is the result that puzzles and surprises you; you think it should be $$P(A\mid (B \cap C)) = \frac{P(A\cap B \cap C)}{P(B\cap C)}.$$

So, let's start by setting $D = B\cap C$ write $P(A\mid (B \cap C)) = P(A\mid D)$ as in $(1)$ to get \begin{align} P(A\mid (B \cap C)) &= P(A\mid D)\\ &= \frac{P(A\cap D)}{P(D)}\\ &= \frac{P(A\cap (B \cap C))}{P(B\cap C)}\\ &= \frac{P(A\cap B \cap C)}{P(B\cap C)}\tag{3}\end{align} which is what you think the result should be. But observe that if you multiply and divide the right side of $(3)$ by $P(C))$, you can get \begin{align} P(A\mid (B \cap C)) &= \frac{P(A\cap B \cap C)}{P(B\cap C)}\times \frac{P(C)}{P(C)}\\ &= \dfrac{\dfrac{P(A\cap B \cap C)}{P(C)}}{\dfrac{P(B\cap C)}{P(C)}}\\ &= \dfrac{P(A\cap B \mid C)}{P(B\mid C)} \end{align} which is just $(2)$. In short, the intuition about $(2)$ is that it is just $(3)$ (which you agree with) re-written in terms of conditional probabilities conditioned on the same event $C$.

Answer 2

Just draw the Venn diagram. We then have $$\Pr[A \cap B \mid C] = \frac{\text{"1"}}{\text{"C"}}, \quad \Pr[B \mid C] = \frac{\text{"1"} + \text{"2"}}{\text{"C"}}, \quad \Pr[A \mid B \cap C] = \frac{\text{"1"}}{\text{"1"} + \text{"2"}},$$ and the relationship follows by dividing the first expression by the second.

Answer 3

\begin{align*} \frac{P(A,B|C)}{P(B|C)} &= \frac{P(A,B,C)}{P(C)}\frac{P(C)}{P(B,C)} \\ &= \frac{P(A,B,C)}{P(B,C)} \\ &= P(A|B,C) \end{align*}

Answer 4

My intuition is the following ...

Conditioning on $C$ means that we are considering only the cases when $C$ is given. Now, suppose that I live in a world where $C$ is always given.

My pepole know nothing about and cannot imagine a world without $C$. For some reason, our mathematicians denote probability of $X$ by $\hat{P}(X)$. They have also already discovered the rule

$$\hat P(A|B) = \frac{\hat P(A\cap B)}{\hat P(B)}\text{.}$$

Now, you as an Earthling, know a world where $C$ is not part of the assumptions in everyday life. So, when you come to our planet you can immediately notice, that every our probability $\hat P(X)$ actually correspond to your $P(X|C)$.

You are immediately able to rewrite the RHS, following the upper discovery:

$$\frac{P(A\cap B\mid C)}{P(B \mid C)}\text{.}$$

But ... What is the LHS? Well, what is the probability of $A$ when $B$ is given when $C$ is (also) given? Precisely $$P(A\mid B\cap C)\text{,}$$ hence the formula.