Short_note_04

the conditional probability of a given b denoted by

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Unformatted text preview: se?) The conditional probability of A given B, denoted by P(A|B) is calculated by P(A|B) = . Hence P(A|B) is NOT same as P(B|A) P( |B) = 1 – P(A|B) | | P( For independent events A and B, P( P(A|B) = P(A), P(B|A) = P(B) If A and B are independent events then are independent events, are independent events For n independent events , , ,…, , the probability of their intersection can be written as P( ) = P( We do NOT have such a rule for union of n independent events But we can do a little trick to get a rule: P( …… ) = P(at least one of them occurs) = 1 – P(none of them occurs) = 1 - P( = 1 – P( … ) = 1 – P( = 1 – ( 1 – P( … ) So, when , , [Because they are independent events] … (1 P( ,…, P( ) = P( P( …… ) = 1 – ( 1 – P( … (1 P( Warning: Do NOT use above two results when , ,…, Bayes Theorem: For two events A and B, | P(B|A) = | (| ) If we have n disjoint non-empty events P( , , P( Then by Bayes theorem, for any other event B: P( |B) = | | | | Warning: Do NOT use this result when , ,…, P( P( ,…, such that...
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This document was uploaded on 03/21/2014.

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