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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. – André Nicolas Dec 2 '12 at 0:37
@AndréNicolas, nice use of words – Riley Jan 19 at 9:28
up vote 9 down vote accepted

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

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