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~HW1_2_Sol

# ~HW1_2_Sol - 12 = exp-1 32 ≈ 267 6 Additional Problem...

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ECEN 646 Homework 1, Part 2 Solutions 4 C ¸inlar Exercise (4.7), Chapter 1 For a real root, we need to have that B 2 - 4 AC 0. Combinations of A, B, and C satisfying B 2 4 AC , and their probabilities are as follows: B = - 3 ,A = 1 ,C = 1, with probability (0 . 25)(0 . 4)(0 . 5) = 0 . 05. B = - 3 ,A = 1 ,C = 2, with probability (0 . 25)(0 . 4)(0 . 4) = 0 . 04. B = - 3 ,A = 2 ,C = 1, with probability (0 . 25)(0 . 6)(0 . 5) = 0 . 075. B = - 2 ,A = 1 ,C = 1, with probability (0 . 25)(0 . 4)(0 . 5) = 0 . 05. Note that probabilities multiply because A, B, and C are independent random variables. Because the four are disjoint, add their probabilities to get that P ( B 2 - 4 AC 0) = 0 . 215. 5 C ¸inlar Exercise (4.8), Chapter 1 The reliability of the equipment for 4000 hours is the probability that all three components last 4000 hours, ie., P ( { X 1 4000 } ∩ { X 2 4000 } ∩ { X 3 4000 } ) = P { X 1 4000 } P { X 2 4000 } P { X 3 4000 } = (1 - P { X 1 4000 } )(1 - P { X 2 4000 } )(1 - P { X 3 4000 } ) = exp( - 0 . 4) exp(
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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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