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I'm curious what the breakdown of how the transition happens per the formula below. I get how $P(A \cap B) = P(A\mid B)P(B)$ which is the famous conditional probability. But am totally lost when there are three sets involved. Thanks!!

$$P(A\cap B\cap C)=P(A)P(B\mid A)P(C\mid A\cap B)$$

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    Reverse the order of first term and last term: it will probably then be clear.2012-12-02
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    @AndréNicolas, nice use of words2016-01-19
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    @AndréNicolas, woow! That comment was enough! Solved it in seconds! :)2017-07-01
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    [This link may be further helpful for the users to come.](https://math.stackexchange.com/questions/1037327/extended-bayes-theorem-pa-b-c-d-constructing-a-bayesian-network)2018-03-27

2 Answers 2

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It’s just a double application of the two-event formula, first thinking of $A\cap B$ as a single event:

$$\begin{align*} P(A\cap B\cap C)&=P\big((A\cap B)\cap C\big)\\ &=P\big(C\mid(A\cap B)\big)P(A\cap B)\\ &=P\big(C\mid(A\cap B)\big)\Big(P(B\mid A)P(A)\Big)\\ &=P(A)P(B\mid A)P(C\mid A\cap B)\;. \end{align*}$$

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It is the same thing but you use the same rule two times.

$$P(A,B,C)=P(C\mid B,A)P(B,A)$$

Here assume $(A,B)=K$ then $P(A,B,C)=P(C\mid K)P(K)$ same with your rule. Then for the second case

$$P(A,B,C)=P(C\mid B,A)P(B,A)=P(C\mid B,A)P(B\mid A)P(A)$$

using again the same rule $P(B,A)=P(B\mid A)P(A)$.