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 19 probability of failure complete parts (a) through (c) below (a) would it be unusual to observe one component fail? two components? it be unusual to observe one component fail, since the probability that one component fails, is than 0 05. it be unusual to observe two components fail, since the probability that two components fail, is than 0.05 (type integers or decimals. do not round.) (b) what is the probability that a parallel structure with 2 identical components will succeed? (note: for a parallel structure with 2 identical components, the system can succeed if at least one of the components succeeds) (round to four decimal places as needed.)

Explanation:

Step1: Probability of one component failing

The probability that one component fails is given as \(P(\text{fail}) = 0.19\). Since \(0.19>0.05\), it would not be unusual to observe one component fail.

Step2: Probability of two components failing

Since components fail independently, the probability that two components fail is \(P(\text{both fail})=0.19\times0.19 = 0.0361\). Since \(0.0361<0.05\), it would be unusual to observe two components fail.

Step3: Probability of system success (for part b)

The probability that a component succeeds is \(P(\text{succeed})=1 - 0.19=0.81\). The probability that the system fails (both components fail) is \(0.19\times0.19 = 0.0361\). Then the probability that the system succeeds is \(P(\text{system succeed})=1 - P(\text{both fail})\). So \(P(\text{system succeed})=1-0.0361 = 0.9639\)

Answer:

For part (a):
It would not be unusual to observe one component fail, since the probability that one component fails, \(0.19\), is greater than \(0.05\). It would be unusual to observe two components fail, since the probability that two components fail, \(0.0361\), is less than \(0.05\).
For part (b): \(0.9639\)