Module 4 Exam Exam Page 1 A factory has eight safety systems. During an emergency, the probability of any one of the safety systems failing is .08. What is the probability that six or more safety systems will fail during an emergency? f(x) = ( (n!) / (x!(n-x)!) ) x ( (p^x) x ((1-p)^n-x)) ) n = 8 x = 6, 7, 8 (number of failures) p = 0.08 6 failures: n = 8 x = 6 p = 0.8 n-x = 8-6 = 2 ( (8!) / (6!(2)!) ) x ( (0.08^6) x ((1-0.08)^2)) ) = 6.2 x 10^-6 7 failures: n = 8 x = 7 p = 0.8 n-x = 8-7 = 1 ( (8!) / (7!(1)!) ) x ( (0.08^7) x ((1-0.08)^1)) ) = 1.54 x 10^-7 8 failures: n = 8 x = 8 p = 0.8 n-x = 8-8 = 0 ( (8!) / (8!(0)!) ) x ( (0.08^8) x ((1-0.08)^0)) ) = 1.68 x 10^-9 MATH 110 MODULE 4 EXAM QUESTIONS AND ANSWERS- PORTAGE LEARNING f(6) = 6.21 x 10^-6
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MATH 110 MODULE 4 EXAM QUESTIONS AND ANSWERS- PORTAGE LEARNING.
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MATH 110 MODULE 4 EXAM QUESTIONS AND ANSWERS- PORTAGE LEARNING. Module 4 Exam Exam Page 1 A factory has eight safety systems. During an emergency, the probability of any one of the safety systems failing is .08. What is the probability that six or more safety systems will fail during an emergency? f(x) = ( (n!) / (x!(n-x)!) ) x ( (p^x) x ((1-p)^n-x)) ) n = 8 x = 6, 7, 8 (number of failures) p = 0.08 6 failures: n = 8 x = 6 p = 0.8 n-x = 8-6 = 2 ( (8!) / (6!(2)!) ) x ( (0.08^6) x ((1-0.08)^2)) ) = 6.2 x 10^-6 7 failures: n = 8 x = 7 p = 0.8 n-x = 8-7 = 1 ( (8!) / (7!(1)!) ) x ( (0.08^7) x ((1-0.08)^1)) ) = 1.54 x 10^-7 8 failures: n = 8 x = 8 p = 0.8 n-x = 8-8 = 0 ( (8!) / (8!(0)!) ) x ( (0.08^8) x ((1-0.08)^0)) ) = 1.68 x 10^-9 MATH 110 MODULE 4 EXAM QUESTIONS AND ANSWERS- PORTAGE LEARNING f(6) = 6.21 x 10^-6 A factory has eight safety systems. During an emergency, the probability of any one of the safety systems failing is .08. What is the probability that six or more safety systems will fail during an emergency?
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