Solutions-Tutorial-2

Solutions-Tutorial-2 - = 0.2873 3.116 Optional Exercises...

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AB103 – Statistical and Quantitative Methods Solutions - Tutorial 2 Chapter 3 (MBS) 3.6
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3.16 3.20
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3.36 + P (D) -
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3.62 c. Let A2 be the event that CVSA yields incorrect results for 2 out of 4 suspects CVSA can yield incorrect results for 2 suspects out of 4 in 6 {= 2 4 }different ways. P (A2) = 6* (0.98) 2 * (0.02) 2 = 0.0023 Since this probability is very low, we would doubt the claimed accuracy rate of the new lie detector.
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3.115 Define the following events: O1: Component 1 operates properly O2: Component 2 operates properly O3: Component 3 operates properly P(O1) = 1 – P(Component #1 fails) = 1 – 0.12 = 0.88 P(O2) = 1 – P(Component #2 fails) = 1 – 0.09 = 0.91 P(O3) = 1 – P(Component #3 fails) = 1 – 0.11 = 0.89 P(System operates properly) = P(O1 O2 O3) = P(O1) * P(O2) * P(O3) (since the 3 components operate independently) = 0.88 * 0.91 * 0.89 = 0.7127 P(System Fails) = 1 - P(System operates properly) = 1 – 0.7127 (from part a)
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Unformatted text preview: = 0.2873 3.116 Optional Exercises 3.18 3.26 a. P ( A c ) = P ( E 3 ) + P ( E 6 ) + P ( E 8 ) = 0.2 + 0.3 + 0.03 = 0.53 b. P ( B c ) = P ( E 1 ) + P ( E 7 ) + P ( E 8 ) = 0.10 + 0.06 + 0.03 = .19 c. P ( A c ∩ B ) = P ( E 3 ) + P ( E 6 ) = 0.2 + 0.3 = 0.5 d. P ( A ∪ B ) = P ( E 1 ) + P ( E 2 ) + P ( E 3 ) + P ( E 4 ) + P ( E 5 ) + P ( E 6 ) + P ( E 7 ) = .10 + .05 + .20 + .20 + .06 + .30 + .06 = .97 e. P ( A ∩ B ) = P ( E 2 ) + P ( E 4 ) + P ( E 5 ) = .05 + .20 + .06 = .31 f. P ( A c ∪ B c ) = P ( E 1 ) + P ( E 7 ) + P ( E 3 ) + P ( E 6 ) + P ( E 8 ) = .10 + .06 + .20 + .30 + 0.03 = .69 g. No. A and B are mutually exclusive if P ( A ∩ B ) = 0. Here, P ( A ∩ B ) = .31. 3.32 3.38 3.52 3.54 Thus P(B c | A) = 0.10 We require to determine P(A ∩ B c ) = P(A) * P(B c | A) = 0.9 * 0.10 = 0.09...
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Solutions-Tutorial-2 - = 0.2873 3.116 Optional Exercises...

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