for number 1. A. B. C.. i think i did A i think ok. i NEED box and line diagram

Question

for number 1. A. B. C.. i think i did A i think ok. i NEED box and line diagram in same style of format in second pic as well.

1. Consider the following system: 90 90 or Determine the probability that the system will operate under cach of these conditions a. The system as shown. b. Each system component has a backup with a probability of 90 and a switch that is 100 percent reliable. c. Backups with.90 probability and a switch that is 99 percent reliable. 2. A product is composed of four parts. In order for the product to function properly in a given situa- tion, each of the parts must function. Two of the parts have a 96 probability of functioning, and tweo have a probability of .99. What is the overall probability that the product will function properly? 3. A system consists of three identical components. In order for the system to perform as intended, all of the components must perform. Each has the same probability of performance. If the system is to have a .92 probability of performing, what is the minimum probability of performing needed by each of the individual components? 4. A product engineer has developed the following equation for the cost of a system component: C = (10P), where C is the cost in dollars and P is the probability that the component will operate

Explanation / Answer

Determine the following probabilities :

a) The system as shown :

System Operational Probability = 0.90 * 0.90 = 0.81

b) Each system component has a backup with a probability of 0.90 and switch that is 100% reliable.

Each component of the System has an Operational Probability computed based on a back-up.

Each component is a parallel configuration of the System component and back-up so that theparallel

configuration has

Operational Probability = {1 – ((1 – 0.9)*(1 – 0.9))}.

And since there are two of the components the total Operational Probability is a multiplier of both which is:

System Operational Probability = {1 – ((1 – 0.9)*(1 – 0.9))}* {1 – ((1 – 0.9)*(1 – 0.9))} = 0.9801

c) Backups with 0.90 probability and switch that is 99% reliable.

Each component has

Operational Probability ={1 – ((1 – 0.9)*(1 – 0.9)) = 0.99.

And since the back-up switch  is only 0.99 reliable, the back-up switch is actually a “series” element.

This means that for each component there is a back-up in parallel and that there is also a “series”switch.

Component Operation Probability = [Operational Probability of “series” Switch] * [OperationalProbability of Component/Back-up]

= [((1- 0.9)*(1 -0.99))] * [(1-((1-0.9)*(1-0.9)))]= 0.98901

System Operational Probability = 0.98901 * 0.98901 = (approx) 0. 9783

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