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sec_2_4_part2_sol

# sec_2_4_part2_sol - (0 40(0 70(0 40(0 70(0 90(0 30 ≈...

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APPM 4/5570 Solutions to Problems from Section 2.5 55. Let L be the event that a randomly selected tick carries Lyme disease and let H be the event that a randomly selected tick carries HGE. We are given that P ( L ) = 0 . 16 , P ( H ) = 0 . 10 , and P ( L H | L H ) = 0 . 10 . We want to find P ( L | H ) = P ( L H ) P ( H ) . Note that 0 . 10 = P ( L H | L H ) = P (( L H ) ( L H )) P ( L H ) = P ( L H ) P ( L H ) which implies that P ( L H ) = 0 . 10 · P ( L H ) = 0 . 10 · [ P ( L ) + P ( H ) - P ( L H )] . So, P ( L H ) = 0 . 10 P ( L ) + 0 . 10 P ( H ) - 0 . 10 P ( L H ) which implies that 1 . 10 P ( L H ) = 0 . 10 P ( L ) + 0 . 10 P ( H ) = (0 . 10)(0 . 16) + (0 . 10)(0 . 10) . So, P ( L H ) = (0 . 10)(0 . 16) + (0 . 10)(0 . 10) 1 . 10 0 . 02363636 . The final answer is then P ( L | H ) = P ( L H ) P ( H ) 0 . 02363636 0 . 10 0 . 2364 . 60. Let L be the event that an aircraft has a locator and let D be the event that an aircraft is discovered. We are given that P ( D ) = 0 . 70 , P ( L | D ) = 0 . 60 , P ( L 0 | D 0 ) = 0 . 90 Note then that we also now know that P ( D 0 ) = 0 . 30 , P ( L 0 | D ) = 0 . 40 , P ( L | D 0 ) = 0 . 10 Using Bayes Rule, we get:

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a. P ( D 0 | L ) = P ( L | D 0 ) · P ( D 0 ) P ( L | D ) · P ( D ) + P ( L | D 0 ) · P ( D 0 ) = (0 . 10)(0 . 30) (0 . 60)(0 . 70) + (0 . 10)(0 . 30) 0 . 06667 b. P ( D | L 0 ) = P ( L 0 | D ) · P ( D ) P ( L 0 | D ) · P ( D ) + P ( L 0 | D 0 ) · P ( D
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Unformatted text preview: (0 . 40)(0 . 70) (0 . 40)(0 . 70) + (0 . 90)(0 . 30) ≈ . 50901 62. Let B be the event that the purchasher has a basic model, let D be the event that the purchaser has a deluxe model. Let E be the event that the purchaser has an extended warranty. We are given that P ( B ) = 0 . 40 (and hence we know that P ( D ) = 0 . 60) and we are given that P ( E | B ) = 0 . 30 , P ( E | D ) = 0 . 50 . So, we know that P ( E | B ) = 0 . 70 , P ( E | D ) = 0 . 50 . The answer, by Bayes Rule, is then P ( B | E ) = P ( E | B ) · P ( B ) P ( E | B ) · P ( B ) + P ( E | D ) · P ( D ) = (0 . 30)(0 . 40) (0 . 30)(0 . 40) + (0 . 50)(0 . 60) ≈ . 2857...
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