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<MATH> P(D)=P(D \bigcap C)+P(D \bigcap C^c) </MATH> | <MATH> P(D)=P(D \bigcap C)+P(D \bigcap C^c) </MATH> | ||
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| + | <MATH> P(D)=P(D|C)P(C)+P(D|C^c)P(C^c) </MATH> | ||
| + | |||
| + | Thus, it is proved. | ||
Latest revision as of 19:17, 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) $
$ P(D)=P(D|C)P(C)+P(D|C^c)P(C^c) $
Thus, it is proved.
