Question

3. A device has three components A, B and C such as A and B are “backupping” each other: the device still works if A is failed or if B is failed, but it doesn’t work if both A and B are failed. Component C is crucial: if it is broken, then the device doesn’t work. During a certain period, the components A, B and C have probabilities 0.35, 0.40 and 0.20 to fail, correspondingly. Failures of the components are independent of each other.

   1. a) What is the probability that the device works the whole period?

   2. b) If the device doesn’t work, what is the probability that (at least) component C is failed?

Answer 1

Pr (component C fails) = P(C) = 0.2

Pr (component C works) = 1 - P(C) = 0.8

P(A) = 0.35; P(B) = 0.4; P(C) = 0.2 where P(A), P(B) & P(C) are probabilities that respective component will fail.

Pr (Either of A or B works) = 1 - Pr (both A & B fails) = 1-P(A)*P(B) = 1-0.35*0.4 = 0.86

a) Pr that device works whole time = Pr(C works)* Pr(Either A or B works) = 0.8*0.86 = 0.688

b) P(device doesn't work/device fails) = 1 - P(device works) = 1-0.688 = 0.312

So, P(component C failed/device failed) = P(device fails & C fails)/P(device fails) = 0.2/0.312 = 0.64

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