Question

Question 4 [20 marks] A system consists of five components in two branches as shown in the following diagram: C-D-E- In other words, the system works if components A and B work or components C, D, and E work. Assume that the components fail independently with the following probabilities: P(A fails) = P(B fails) = 0.1 and P(C fails) = P(D fails) = P(E fails) = 0.2. (a) What is the probability that the system works? (b) Given that the system works, what is the probability that component A does not work? (c) Given the system does not work, what is the probability that component A also does not work?

Answer 1

a) Probability that the subsystem AB works = Probability that both A and B works = (1 - 0.1)² = 0.81

Probability that the subsystem CDE works = Probability that all 3 of them works = (1 - 0.2)³ = 0.512

Probability that the system works
= 1 - Probability that the system does not work
= 1 - Probability that both the subsystems AB and CDE dont work
= 1 - (1 - 0.81)*(1 - 0.512)
= 0.90728

Therefore 0.90728 is the probability that the system works.

b) Given that system works, probability that A does not work is computed here as: (Using bayes theorem)
= Probability that A does not work and subsystem CDE works / Probability that the system works
= 0.1*0.512 / 0.90728
= 0.0564

Therefore 0.0564 is the required probability here.

c) Given that the system does not work, probability that A does not work is computed here as: (Using bayes theorem )
= Probability that A does not work * Probability that subsystem CDE dont work / Probability that the system does not work
= 0.1*(1 - 0.512) / (1 - 0.90728)
= 0.5263

Therefore 0.5263 is the required probability here.