Unformatted text preview: . 12) = exp(1 . 32) ≈ . 267 6 Additional Problem Since P [ { ω i } ] = 0 . 25 for i = 1 , 2 , 3 , 4, we have that P ( A ) = P ( B ) = P ( C ) = 0 . 25 + 0 . 25 = 0 . 5 . To show that A is independent of B, a is independent of C, and B is independent of C: P ( A ∩ B ) = P ( { ω 2 } ) = 0 . 25 which is in fact = (0 . 5)(0 . 5) = P ( A ) P ( B ) P ( A ∩ C ) = P ( { ω 1 } ) = 0 . 25 which is in fact = (0 . 5)(0 . 5) = P ( A ) P ( C ) P ( B ∩ C ) = P ( { ω 3 } ) = 0 . 25 which is in fact = (0 . 5)(0 . 5) = P ( B ) P ( C ) But the ±nal condition for independence does not hold: P ( A ∩ B ∩ C ) = P ( ∅ ) = 0 which is NOT (0 . 5)(0 . 5)(0 . 5) = P ( A ) P ( B ) P ( C ) . Thus { A,B,C } is not a family of independent events. 1...
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 Fall '11
 SerapSavari
 Probability, Probability theory, Cinlar Exercise

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