Unformatted text preview: se?) The conditional probability of A given B, denoted by P(AB) is
calculated by P(AB) = . Hence P(AB) is NOT same as P(BA)
P( B) = 1 – P(AB)


P(
For independent events A and B,
P(
P(AB) = P(A), P(BA) = 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(BA) =  ( ) If we have n disjoint nonempty 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...
View
Full Document
 Spring '14
 Conditional Probability, Probability, Probability theory

Click to edit the document details