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Rule of elimination

Linty Fresh

Junior Member
Joined
Sep 6, 2005
Messages
58
So I'm trying to teach myself about Bayes' Theorem, and in deriving it, we total up P(A ? B[sub:26r8gu1y]i[/sub:26r8gu1y]) x P(B[sub:26r8gu1y]i[/sub:26r8gu1y]) for all of event B to get P (A). My question is how we get the rule of elimination that allows us to figure this out? Is it a complicated problem, or am I missing something really obvious? How do we get P(A)=sum(P(A ? B))?

Thanks!
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,688
Linty Fresh said:
So I'm trying to teach myself about Bayes' Theorem, and in deriving it, we total up P(A ? B[sub:3ldh7e8j]i[/sub:3ldh7e8j]) x P(B[sub:3ldh7e8j]i[/sub:3ldh7e8j]) for all of event B to get P (A). My question is how we get the rule of elimination that allows us to figure this out? Is it a complicated problem, or am I missing something really obvious? How do we get P(A)=sum(P(A ? B))?
What you are missing is the concept of a partition of the space.
The collection \(\displaystyle \{B_1,B_2,\cdots,B_n\}\) is a collection of pair-wise disjoint non-empty subsets such that \(\displaystyle \bigcup\limits_n {B_n } = S\).
That is, the partition ‘fills up’ the whole space.
Therefore, the event \(\displaystyle A\) is also partitioned by that collection: \(\displaystyle A = \bigcup\limits_n {\left( {A \cap B_n } \right)}\)
Because they are pair-wise disjoint, we can simply add the probability of the parts.
So \(\displaystyle P(A) = \sum\limits_n {P\left( {A \cap B_n } \right)} = \sum\limits_n {P\left( {A|B_n } \right)P\left( {B_n } \right)}\)
 
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