So If B=B1UB2U.....UBn would Pr(A|B)=Pr(A|B1)Pr(B1)+Pr(A|B2)Pr(B2)+.....+Pr(A|Bn)Pr(Bn)?(adsbygoogle = window.adsbygoogle || []).push({});

I haven't found a formula for this but it makes intuitive sense.

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# Decompostion of conditional probabilities

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