(2 intermediate revisions by the same user not shown)
Line 17: Line 17:
  
 
<MATH>P(A \bigcap B)=P(A \bigcap B \bigcap C)+P(A \bigcap B \bigcap C^C) </MATH>
 
<MATH>P(A \bigcap B)=P(A \bigcap B \bigcap C)+P(A \bigcap B \bigcap C^C) </MATH>
 +
 +
Replacing <MATH> A \bigcap B </MATH> with D
 +
 +
<MATH> P(D)=P(D \bigcap C)+P(D \bigcap C^c) </MATH>
 +
 +
<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.

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin