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Lecturenotes4 - STAT 430/510 Lecture 4 STAT 430/510...

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STAT 430/510 Lecture 4 STAT 430/510 Probability Hui Nie Lecture 4 June 1st, 2009
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STAT 430/510 Lecture 4 Review Properties of Probability P ( E c ) = 1 - P ( E ) If E F , then P ( E ) P ( F ) P ( E S F ) = P ( E ) + P ( F ) - P ( EF )
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STAT 430/510 Lecture 4 Inclusion-exclusion Identity Inclusion-exclusion Identity P ( E 1 [ E 2 [ · · · [ E n ) = n X i = 1 P ( E i ) - X i 1 < i 2 P ( E i 1 E i 2 ) + · · · + ( - 1 ) r + 1 X i 1 < i 2 < ··· < i r P ( E i 1 E i 2 · · · E i r ) + · · · + ( - 1 ) n + 1 P ( E i 1 E i 2 · · · E i n )
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STAT 430/510 Lecture 4 Example A diagnostic test for the AIDS virus has probability of 0.005 of producing a false positive. If the 140 employees of a medical clinic are tested and all are free of AIDS, what is the probability that at least one false positive will occur? It is reasonable to assume that the test results for different individuals are independent. The probability of a negative=1-0.005=0.995 P ( at least one positive ) = 1 - P ( no positives ) = 1 - P ( 140 negatives ) = 1 - 0 . 995 140 = 0 . 504
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STAT 430/510 Lecture 4 Sample Spaces with Equally Likely Outcomes For many experiments, it is natural to assume that all outcomes in the sample space are equally likely to occur.
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