Baye's Theorem

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Main Question or Discussion Point

P(B/A) = P(A/B).P(B) / P(A)

Later we expand P(A) as P(A/B).P(B) + P(A/B).P(B) ... B is complement of B

I don't understand how we can expand P(A) like that. Doesn't that assume that A ℂ B?
 

Answers and Replies

  • #2
marcusl
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Think about what it is saying. The probability that A happens is the probability that A happens given that B happens plus the probability that A happens given that B doesn't happen. Both cases are needed to cover all possibilities.
 
  • #3
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Well basically what my book says is that : -
P(A) = P(A/B1).P(B1) + P(A/B2).P(B2) + ... + P(A/Bn).P(Bn)

Doesn't this assume that B1 U B2 ... U Bn is a super-set of A?
 
  • #4
D H
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Doesn't this assume that B1 U B2 ... U Bn is a super-set of A?
Of course.

Your text should have specified that B1, B2, ···, Bn are a set of mutually disjoint subsets of the universe U of possible outcomes and that B1B2 ∪ ··· ∪ Bn=U. The set A must be a subset of this universe of outcomes U; otherwise it doesn't even make sense to talk about P(B1|A).
 
  • #5
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Of course.

Your text should have specified that B1, B2, ···, Bn are a set of mutually disjoint subsets of the universe U of possible outcomes and that B1B2 ∪ ··· ∪ Bn=U. The set A must be a subset of this universe of outcomes U; otherwise it doesn't even make sense to talk about P(B1|A).
It didn't. Anyways, thank you. This clears my doubt.
 

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