Consider a system of five identical components connected in series, as illustrated in Figure 2.3. 2 3 4 1 5 A system of five components connected in a series Figure 2.3 Denote a component that fails by F and one that doesn't fail by S (for success) Let A be the event that the system fails. For A to occur, at least one of the individual components must fail. Copyright O Cengage Learning. All rights reserved. Outcomes in A include SSFSS (1, 2, 4, and 5 all work, but 3 does not), FFSSS, and so on. There are in fact 31 different outcomes in A. However, A', the event that the system works, consists of the single outcome SSSSS We will see in Section 2.5 that if 90% of all such components do not fail and different components fail independently of one another, then P(A') P(SSSSS) .95 = 59. Thus P(A) 1 - .59 = .41; so among a large number of such systems, roughly 41% will fail Copyright O Cengage Learning. All rights reserved
Consider a system of five identical components connected in series, as illustrated in Figure 2.3. 2 3 4 1 5 A system of five components connected in a series Figure 2.3 Denote a component that fails by F and one that doesn't fail by S (for success) Let A be the event that the system fails. For A to occur, at least one of the individual components must fail. Copyright O Cengage Learning. All rights reserved. Outcomes in A include SSFSS (1, 2, 4, and 5 all work, but 3 does not), FFSSS, and so on. There are in fact 31 different outcomes in A. However, A', the event that the system works, consists of the single outcome SSSSS We will see in Section 2.5 that if 90% of all such components do not fail and different components fail independently of one another, then P(A') P(SSSSS) .95 = 59. Thus P(A) 1 - .59 = .41; so among a large number of such systems, roughly 41% will fail Copyright O Cengage Learning. All rights reserved
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.2: Arithmetic Sequences
Problem 68E
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I would like a little more explanation on that please. That is an example given in a lecture. How did they come up with 31 different outcomes in A?
P(A) = 1 – .59 = .41; so among a large number of such systems, roughly 41% will fail.
What do they mean by that?
Thank You.
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