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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)
For independent events A and B,
P(A|B) = P(A), P(B|A) = P(B) If A and B are independent events then
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(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.
- Spring '14