Prove that? : $P(AuuBuuC)=P(A)+P(B)+P(C)-P(AnnB)-P(BnnC)-P(AnnC)+P(AnnBnnC)$

Question

$P(AuuBuuC)=P(A)+P(B)+P(C)-P(AnnB)-P(BnnC)-P(AnnC)+P(AnnBnnC)$

I can show that with diagram, but how to prove it with formulas?

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Answer 1 · 2018-08-07T17:35:48

Please refer to The Explanation.

$"Prerequisite : "P(AuuB)=P(A)+P(B)-P(AnnB)....(star)$ .

$P(AuuBuuC)=P(AuuD)," where, "D=BuuC$ ,

$=P(A)+P(D)-P(AnnD)..........[because, (star)]$ ,

$=P(A)+color(red)(P(BuuC))-color(blue)(P[Ann(BuuC)])$ ,

$=P(A)+color(red)(P(B)+P(C)-P(BnnC))-color(blue)(P(AnnB)uu(AnnC)),$

$=P(A)+P(B)+P(C)-P(BnnC)-color(blue){[P(AnnB)+P(AnnC)-P((AnnB)nn(AnnC)]$ ,

$=P(A)+P(B)+P(C)-P(AnnB)-P(BnnC)-P(AnnC)+P(AnnBnnC),$

as desired!