Math Problem Statement

Consider the system of components connected as in the accompanying picture. Components 1 and 2 are connected in parallel, so that subsystem works if and only if either 1 or 2 works; since 3 and 4 are connected in series, that subsystem works if and only if both 3 and 4 work. If components work independently of one another and P(component i works) = 0.82 for i = 1, 2 and = 0.68 for i = 3, 4, calculate P(system works). (Round your answer to four decimal places.)

Solution

In this problem, you have a system of four components connected in a combination of parallel and series configurations. Here's a breakdown of the system's structure:

  1. Components 1 and 2 are connected in parallel. The parallel subsystem works if either 1 or 2 works.
  2. Components 3 and 4 are connected in series. The series subsystem works if both 3 and 4 work.

The goal is to calculate the probability that the entire system works, which depends on both the parallel and series subsystems working independently.

Step 1: Probability that the parallel subsystem works

For components 1 and 2 connected in parallel, the subsystem works if either component works. The formula for the probability that at least one of two components works in parallel is:

P(parallel works)=1P(both fail)P(\text{parallel works}) = 1 - P(\text{both fail})

The probability that both components fail is:

P(both fail)=P(component 1 fails)×P(component 2 fails)P(\text{both fail}) = P(\text{component 1 fails}) \times P(\text{component 2 fails})

Since the probability that a component works is 0.82 for both 1 and 2, the probability that each component fails is:

P(component 1 fails)=10.82=0.18P(\text{component 1 fails}) = 1 - 0.82 = 0.18 P(component 2 fails)=10.82=0.18P(\text{component 2 fails}) = 1 - 0.82 = 0.18

Thus, the probability that both components fail is:

P(both fail)=0.18×0.18=0.0324P(\text{both fail}) = 0.18 \times 0.18 = 0.0324

So, the probability that the parallel subsystem works is:

P(parallel works)=10.0324=0.9676P(\text{parallel works}) = 1 - 0.0324 = 0.9676

Step 2: Probability that the series subsystem works

For components 3 and 4 connected in series, the subsystem works if both components work. The probability that both components work is:

P(series works)=P(component 3 works)×P(component 4 works)P(\text{series works}) = P(\text{component 3 works}) \times P(\text{component 4 works})

The probability that each component works is 0.68 for both 3 and 4. So, the probability that the series subsystem works is:

P(series works)=0.68×0.68=0.4624P(\text{series works}) = 0.68 \times 0.68 = 0.4624

Step 3: Probability that the entire system works

The entire system works if both the parallel subsystem (components 1 and 2) and the series subsystem (components 3 and 4) work. Since these two subsystems are independent, the probability that the entire system works is:

P(system works)=P(parallel works)×P(series works)P(\text{system works}) = P(\text{parallel works}) \times P(\text{series works})

Substituting the values:

P(system works)=0.9676×0.4624=0.4475P(\text{system works}) = 0.9676 \times 0.4624 = 0.4475

Thus, the probability that the system works is approximately:

0.4475\boxed{0.4475}

Final Answer:

The probability that the system works is 0.4475\boxed{0.4475}, rounded to four decimal places.

Would you like more details on any step or have further questions?

Here are 5 related questions to think about:

  1. What is the probability that the parallel subsystem fails?
  2. How does the system reliability change if component 1's probability is reduced?
  3. How would the system work probability change if components 3 and 4 were in parallel instead of series?
  4. What would happen if all components had different probabilities of working?
  5. How would adding a third component in parallel with components 1 and 2 affect the system's overall reliability?

Tip: Understanding the difference between series and parallel systems is key when calculating the overall reliability in complex systems!

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Math Problem Analysis

Mathematical Concepts

Probability
Series and Parallel Systems
Independent Events

Formulas

P(system works) = P(parallel works) × P(series works)
P(parallel works) = 1 - P(both fail)
P(series works) = P(component 3 works) × P(component 4 works)

Theorems

Multiplication Rule for Independent Events
Complement Rule

Suitable Grade Level

Grades 10-12