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) = 0.86 for i = 1, 2 and = 0.75 for i = 3, 4, calculate P(system works). (Round your answer to four decimal places.)

Solution

In this system, we have two subsystems of components connected differently: one in parallel and one in series. Let’s break the problem into smaller parts to find the probability that the entire system works.

Subsystem 1 (Parallel Connection of Components 1 and 2)

For components connected in parallel, the subsystem works if either component 1 or component 2 works. The probability that at least one of them works can be found using the formula:

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

The probability that both components fail is the product of their individual failure probabilities:

P(both fail)=(1P(Component 1 works))×(1P(Component 2 works))P(\text{both fail}) = (1 - P(\text{Component 1 works})) \times (1 - P(\text{Component 2 works}))

Given that P(Component 1 works)=0.86P(\text{Component 1 works}) = 0.86 and P(Component 2 works)=0.86P(\text{Component 2 works}) = 0.86, the failure probability for each component is 10.86=0.141 - 0.86 = 0.14. So,

P(both fail)=0.14×0.14=0.0196P(\text{both fail}) = 0.14 \times 0.14 = 0.0196

Thus, the probability that the parallel subsystem works is:

P(Parallel subsystem works)=10.0196=0.9804P(\text{Parallel subsystem works}) = 1 - 0.0196 = 0.9804

Subsystem 2 (Series Connection of Components 3 and 4)

For components connected in series, the subsystem works only if both components 3 and 4 work. The probability that both components work is the product of their individual working probabilities:

P(Series subsystem works)=P(Component 3 works)×P(Component 4 works)P(\text{Series subsystem works}) = P(\text{Component 3 works}) \times P(\text{Component 4 works})

Given that P(Component 3 works)=0.75P(\text{Component 3 works}) = 0.75 and P(Component 4 works)=0.75P(\text{Component 4 works}) = 0.75, we have:

P(Series subsystem works)=0.75×0.75=0.5625P(\text{Series subsystem works}) = 0.75 \times 0.75 = 0.5625

System Probability

The entire system works if both subsystems work, i.e., the parallel subsystem (components 1 and 2) works and the series subsystem (components 3 and 4) works. Therefore, the probability that the entire system works is the product of the two subsystem probabilities:

P(System works)=P(Parallel subsystem works)×P(Series subsystem works)P(\text{System works}) = P(\text{Parallel subsystem works}) \times P(\text{Series subsystem works})

Substitute the calculated values:

P(System works)=0.9804×0.5625=0.5511P(\text{System works}) = 0.9804 \times 0.5625 = 0.5511

Thus, the probability that the system works is approximately:

0.5511\boxed{0.5511}

Let me know if you would like more details or further clarification.

Here are some related questions to explore:

  1. How would the system probability change if components 1 and 2 had different probabilities?
  2. What if components 3 and 4 were connected in parallel instead of series?
  3. How does the system reliability change if the probability of each component working increases?
  4. What is the system failure probability, given the same setup?
  5. How would the calculation change if the components were not independent?

Tip: For series connections, the reliability is always lower because all components must work, whereas in parallel connections, having more components increases the chance of success.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Series and Parallel Systems
Independent Events

Formulas

P(Parallel subsystem works) = 1 - P(both fail)
P(Series subsystem works) = P(Component 3 works) × P(Component 4 works)
P(System works) = P(Parallel subsystem works) × P(Series subsystem works)

Theorems

Multiplication rule for independent events
Probability of failure for parallel components
Product rule for series components

Suitable Grade Level

Undergraduate level