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Replacing <MATH> A \bigcap B </MATH> with D
 
Replacing <MATH> A \bigcap B </MATH> with D
 +
 +
<MATH> P(D)=P(D \bigcap C)+P(D \bigcap C^c) </MATH>

Revision as of 20:16, 15 September 2008

The theorem of total probalility states that

$ P(A)=P(A|C)P(C)+P(A|C^c)P(C^c) $


$ P(A|B)=P(A|B \bigcap C)P(C|B)+P(A|B \bigcap C^c)P(C^c|B) $


$ P(A|B)=P(A|B \bigcap C)P(B \bigcap C)/P(B) +P(A|B \bigcap C^c)P(B \bigcap C^c)/P(B) $

Since $ P(A|B)=P(A \bigcap B)/P(B), P(A \bigcap B)=P(A|B)P(B) $


$ P(A|B)P(B)=P(A|B \bigcap C)P(B \bigcap C)+P(A|B \bigcap C^c)P(B \bigcap C^c) $


$ P(A \bigcap B)=P(A \bigcap B \bigcap C)+P(A \bigcap B \bigcap C^C) $

Replacing $ A \bigcap B $ with D

$ P(D)=P(D \bigcap C)+P(D \bigcap C^c) $

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