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: $P(A\cap B \cap C)/P(B,C) = P(A|B \cap C)$?

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    $\begingroup$ Maybe it's easier to comprehend in the multiplicative form: $P(A,B\mid C)=P(A\mid B,C)P(B\mid C)$? $\endgroup$ – Hagen von Eitzen Jan 28 '17 at 13:56

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 $$P(A\mid (B \cap C)) = \frac{P((A\cap B)\mid C)}{P(B\mid C)}\tag{2}$$ exactly as your intuition tells you. But, you can set $D = B\cap C$ and start with the definition of $P(A\mid (B \cap C)) = P(A\mid D)$ as in $(1)$ $$P(A\mid (B \cap C)) = P(A\mid D) = \frac{P(A\cap D)}{P(D)} = \frac{P(A\cap (B \cap C))}{P(B\cap C)} = \frac{P(A\cap B \cap C)}{P(B\cap C)}\tag{3}$$ and then multiply and divide by $P(C))$ on the right of $(3)$ to write the final result as $(2)$ as in Taylor's answer.


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.

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\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*}

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    $\begingroup$ -1 Although quite correct, the question asked for some intuition, this does not contain any. $\endgroup$ – Jack Aidley Jan 27 '17 at 11:48
  • $\begingroup$ What is the meaning of $P(A,B)$? $\endgroup$ – Xi'an Jan 27 '17 at 12:48
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    $\begingroup$ it means P(A and B) :: the joint probability, $\endgroup$ – nyxee Jan 27 '17 at 13:18
  • $\begingroup$ @Xi'an I think it was the original notation $\endgroup$ – Taylor Jan 27 '17 at 15:03

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.


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