How do i prove: $$ P(A ∪ B ∪ C) = P(A)+ P(A^c∩B)+P(A^c∩B^c∩C) $$ knowing that: $$ P(A∪B∪C)=P(A)+P(B)+P(C)-P(B∩C)-P(A∩B)-P(A∩C)+P(A∩B∩C) $$ thank you in advance.
probability set theory expansion
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probability
elementary-set-theory
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0You could also prove the statement $A\cup B\cup C=A\sqcup (A^c\cap B)\sqcup (A^c\cap B^c\cap C)$ directly by element chasing and mentioning why $A,A^c\cap B,$ and $A^c\cap B^c\cap C$ are in fact disjoint (*here $E\sqcup F$ denotes the disjoint union of $E$ and $F$, i.e. it is $E\cup F$ while calling to attention that $E\cap F=\emptyset$*). The right side should be a big hint as to how to set up the cases. – 2017-02-20
1 Answers
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HINT
Note that $$P(A^c∩B)=P(B)-P(A∩B)$$ And $$P(A^c∩B^c)=P((A \cup B)^c)$$ I think you can continue from here.
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0thank you very much, i didn't knew those properties so it was a little bit difficult. But know i already solve it. Thank you again. – 2017-02-20