Question

four component system Assume A, B, C, and D function independently. If the probabilities that A, B, C, and D fail are 0.1, 0.2, 0.05, and 0.3 respectively, what is the probability that the system functions?

Answer 1

Answer:

then the probability of failure goes between 0.00003 (0.003%) and 0.5212 (52.12%) depending on the system configuration

Step-by-step explanation:

the solution depends on the system configuration, that is , if some component ( lets say A) is run in parallel from other , or is in series

if a component is run in parallel then the system fails only if all the components in parallel fails

but if the system is connected in series , the system will fail only if one of the components the serie fails.

Therefore denoting the events A= fails A , B= fails B , C= fails C , D= fails D , we have:

- lower bound of probability of failure = all components are in parallel

probability of failure P(A∩B∩C∩D)=P(A)*P(B)*P(C)*P(D)= 0.1 * 0.2 * 0.05 * 0.3 = 0.00003 (0.003%)

- upper bound of probability of failure = all components are in parallel

probability of failure P(A∪B∪C∪D)= P(A) + P(B) + P(C) +P(D) - P(A ∩ B) - P(A ∩ C) - P(A ∩ D)- P(B ∩ C) - P(B ∩ D) - P(C ∩ D) + P(A ∩ B ∩ C) + P(A ∩ B ∩ D) + P(A ∩ C ∩ D) + P(B ∩ C ∩ D) - P(A ∩ B ∩ C ∩ D) = (P(A) + P(B) + P(C) +P(D)) - ( P(A)*P(B) + P(A)*P(C) + P(A)*P(D) + P(B)*P(C) + P(B)*P(D) + P(C)*P(D) ) + P(A)*P(B)*P(C)  + P(A)*P(B)*P(D)+  P(A)*P(C)*P(D)+  P(B)*P(C)*P(D) -  P(A)*P(B)*P(C)*P(D)

replacing values

P(A∪B∪C∪D)= 0.5212 (52.12%)

then the probability of failure goes between 0.00003 (0.003%) and 0.5212 (52.12%) depending on the system configuration