Is the next true? If so, why?
P(A∩B|C) = P(A|B∩C) x P(B|C)
I saw as part of solution of an exercise, but I can't prove that is true, how is it possible to arrive to that?
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1 Answer
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Via a few steps using conditional probability definition:
$$\begin{align}P(A,B|C)&=\frac{P(A,B,C)}{P(C)}\\ &=\frac{P(A|B,C)P(B,C)}{P(C)}\\ &=\frac{P(A|B,C)P(B|C)P(C)}{P(C)} \\ &=P(A|B,C)P(B|C)\end{align}$$
Intuitionally, you can think of it as a more complex form of $P(A,B)=P(A|B)P(B)$, since you add a global given random variable $C$ to each term here, i.e. it is true for any given set of events $\mathcal{E}$: (e.g. $\mathcal{E}=C,D,E...$)
$$P(A,B|\mathcal{E})=P(A|B,\mathcal{E})P(B|\mathcal{E})$$
