An electrical system consists of 4 components with each probability of functioning as above. What is the probability that the system works? Assume the components fail independently.
An electrical system consists of 4 components with each probability of functioning as above. What is the probability that the system works? Assume the components fail independently.
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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![**Title: Probability of an Electrical System Functioning**
An electrical system consists of four components, labeled A, B, C, and D. The system is represented in a diagram as follows:
- **Component A** has a probability of functioning of 0.95.
- **Component B** has a probability of functioning of 0.7.
- **Component C** has a probability of functioning of 0.8.
- **Component D** has a probability of functioning of 0.9.
The diagram shows that components B and C are arranged in parallel, whereas components A and D are arranged in series with the parallel combination of B and C.
**Explanation of the Diagram:**
- The system starts with Component A.
- Then, the system splits into two paths that run parallel: one path contains Component B, and the other contains Component C.
- Both paths converge before reaching Component D.
**Question:**
What is the probability that the entire system functions? Assume the components fail independently.
**Solution Steps:**
1. **Calculate the Probability of the Parallel Section (B and C):**
The probability that at least one of the parallel components (B or C) works:
\[
P(\text{B or C works}) = 1 - (1 - 0.7) \times (1 - 0.8)
\]
\[
= 1 - (0.3 \times 0.2) = 1 - 0.06 = 0.94
\]
2. **Calculate the Probability of the Entire Series System (A, B/C, D):**
Since A, the parallel configuration of B/C, and D are in series, the probability that the system functions is the product of each of these probabilities:
\[
P(\text{System works}) = P(A) \times P(\text{B or C works}) \times P(D)
\]
\[
= 0.95 \times 0.94 \times 0.9
\]
\[
= 0.8037
\]
Thus, the probability that the system works is 0.8037.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdac418c7-7b26-43d7-bc46-18689988fbfb%2F6b79923c-be40-4585-ac41-b50c28fbd62a%2Fg18obl8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Probability of an Electrical System Functioning**
An electrical system consists of four components, labeled A, B, C, and D. The system is represented in a diagram as follows:
- **Component A** has a probability of functioning of 0.95.
- **Component B** has a probability of functioning of 0.7.
- **Component C** has a probability of functioning of 0.8.
- **Component D** has a probability of functioning of 0.9.
The diagram shows that components B and C are arranged in parallel, whereas components A and D are arranged in series with the parallel combination of B and C.
**Explanation of the Diagram:**
- The system starts with Component A.
- Then, the system splits into two paths that run parallel: one path contains Component B, and the other contains Component C.
- Both paths converge before reaching Component D.
**Question:**
What is the probability that the entire system functions? Assume the components fail independently.
**Solution Steps:**
1. **Calculate the Probability of the Parallel Section (B and C):**
The probability that at least one of the parallel components (B or C) works:
\[
P(\text{B or C works}) = 1 - (1 - 0.7) \times (1 - 0.8)
\]
\[
= 1 - (0.3 \times 0.2) = 1 - 0.06 = 0.94
\]
2. **Calculate the Probability of the Entire Series System (A, B/C, D):**
Since A, the parallel configuration of B/C, and D are in series, the probability that the system functions is the product of each of these probabilities:
\[
P(\text{System works}) = P(A) \times P(\text{B or C works}) \times P(D)
\]
\[
= 0.95 \times 0.94 \times 0.9
\]
\[
= 0.8037
\]
Thus, the probability that the system works is 0.8037.
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