Question

For a parallel structure of identical components, the system can succeed if at least one of the components succeeds. Assume that components fail independently of each other and that each component has a 0.17 probability of failure.
Would it be unusual to observe one component fail?
Would it be unusual to observe two components fail?
What is the probability that a parallel structure with 2 identical components will succeed?
How many components would be needed in the structure so that the probability the system will succeed is greater than 0.9999?

Answer 1

Answer:

a.) To observe one component fail it would not be unusual

b.) it would be unusual to have two components fail ( for a significance level of 0.05)

c.) A parallel structure will succeed with 2 identical components with a probability of 0.9711.

d.) The structures needed will be 6.

Step-by-step explanation:

It is given that failure of the components is independent of each other.

a.) The failure of each component is with a probability of 0.17.

   In general we assume a significance level of 0.05 and since the probability of failure of one event is greater than the significance level.

Therefore we can say that it is not unusual to observe one component fail.

b.) The probability that two components fail is = (0.17)×(0.17) = 0.0289.

   Here we can see that the probability of failure of two events is less than the significance level of 0.05.

So for a significance level of 0.05 it would be unusual to have two components fail.

For a significance level of 0.01 it would not be unusual for two components to fail.

c.) A parallel structures having 2 identical components  to succeed the probability = 1  - P(2 components failing)

                  = 1 - 0.0289

                   = 0.9711

d.) It is required to find the number of components that would be need in the structure so that system succeeds with a probability greater than 0.9999.

So we can write

                            1 - $(0.17)^n$ > 0.9999

                     ⇒  $(0.17)^n$ > 0.0001

                     ⇒   n  > $\frac{\log{0.0001}}{\log{0.17}} = 5.198$

       Therefore the number of structures needed will be 6.