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Suppose the diagram of an electrical system is as shown. What is the probability that the system works? Assume the components fail independently. 0.94 A 0.6 B 0.6 C 0.9 D

Suppose the diagram of an electrical system is as shown. What is the probability that the system works? Assume the components fail independently. 0.94 A 0.6 B 0.6 C 0.9 D

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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### Understanding Probabilities in an Electrical System #### Problem Statement: Suppose the diagram of an electrical system is as shown. What is the probability that the system works? Assume the components fail independently. #### System Diagram: The provided diagram represents an electrical system with four components (A, B, C, and D), each having a specific probability of functioning correctly: - Component A is operational with a probability of 0.94. - Component B is operational with a probability of 0.6. - Component C is operational with a probability of 0.6. - Component D is operational with a probability of 0.9. The system layout is as follows: - A horizontal line from the left enters component A. - After component A, the line splits into two parallel paths: - The upper path goes through component B. - The lower path goes through component C. - These two paths then rejoin before passing through component D. - Finally, the line exits from the right side of component D. This setup suggests parallel paths with components B and C having to operate independently, and series paths involving components A and D. #### Calculation: The probability that the system works is determined by calculating the probabilities for components arranged in series and parallel configurations, using their individual operational probabilities and combining them accordingly. **Series and Parallel Configurations:** 1. **Parallel Subsystem (B and C):** - For the parallel subsystem (either B or C needs to work for the path to work), we calculate the combined probability: - Failure of parallel system: \( (1 - P_B) \times (1 - P_C) \) - \( (1 - 0.6) \times (1 - 0.6) = 0.4 \times 0.4 = 0.16 \) - Success of parallel system: \( 1 - 0.16 = 0.84 \) 2. **Series Configuration (A, Parallel Subsystem, and D):** - For the entire series path to work: - \( P_{System} = P_A \times P_{Parallel Subsystem} \times P_D \) - \( 0.94 \times 0.84 \times 0.9 \) **Result:** \[ P_{System} = 0.94 \times 0.84 \times 0.9 =
Transcribed Image Text:### Understanding Probabilities in an Electrical System #### Problem Statement: Suppose the diagram of an electrical system is as shown. What is the probability that the system works? Assume the components fail independently. #### System Diagram: The provided diagram represents an electrical system with four components (A, B, C, and D), each having a specific probability of functioning correctly: - Component A is operational with a probability of 0.94. - Component B is operational with a probability of 0.6. - Component C is operational with a probability of 0.6. - Component D is operational with a probability of 0.9. The system layout is as follows: - A horizontal line from the left enters component A. - After component A, the line splits into two parallel paths: - The upper path goes through component B. - The lower path goes through component C. - These two paths then rejoin before passing through component D. - Finally, the line exits from the right side of component D. This setup suggests parallel paths with components B and C having to operate independently, and series paths involving components A and D. #### Calculation: The probability that the system works is determined by calculating the probabilities for components arranged in series and parallel configurations, using their individual operational probabilities and combining them accordingly. **Series and Parallel Configurations:** 1. **Parallel Subsystem (B and C):** - For the parallel subsystem (either B or C needs to work for the path to work), we calculate the combined probability: - Failure of parallel system: \( (1 - P_B) \times (1 - P_C) \) - \( (1 - 0.6) \times (1 - 0.6) = 0.4 \times 0.4 = 0.16 \) - Success of parallel system: \( 1 - 0.16 = 0.84 \) 2. **Series Configuration (A, Parallel Subsystem, and D):** - For the entire series path to work: - \( P_{System} = P_A \times P_{Parallel Subsystem} \times P_D \) - \( 0.94 \times 0.84 \times 0.9 \) **Result:** \[ P_{System} = 0.94 \times 0.84 \times 0.9 =
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