# Probability exercise - system reliability

Probability of failure of each component in subsystem A & C is 0.1. The failure of each component is independent of each other component and both components of subsystem B have the same unknown failure rate X. The pobability a system failure was due to a failure of the parallel subsystem B is 10 %. What is the probability of failure of the components of subsystem B?

SOLUTION

$$P (S|B) = 10\%$$

$$P (B) = ?\%$$

$$P (S|B) = \frac{P (B|S)\times P(S_2)}{P(B)} = \frac{\frac{0.001x^2}{0.001-0.009x^2} \times 0.001}{x^2} = \frac{0.001-0.009x^2}{0.001x^2}= 0.1$$

$$P (B|S) = \frac{P (S|B)\times P(B)}{P(S_1)} = \frac{0.001x^2}{0.001-0.009x^2}$$

$$P(S_1)=[P(A)+P(B)-P(A)P(B)]\times P(C) = (0.1 + x^2 - 0.1x^2)\times 0.01 = 0.001 - 0.009x^2$$

Assuming B has failed: $$P(S_2)=P(C)\times P(A)=0.0001$$

then

$$0.001-0.009x^2 = 0.0001x^2$$ $$x^2 = 0.1109$$ $$x = 0.33$$

Is this correct?

Reliability of B subsystem $$= 1-x^2$$

Reliability of A-B Subsystem $$= 0.9*(1-x^2) = .9(1-x^2)$$

Reliability of C subsystem $$= 1-.1^2 = 0.99$$

Reliability of ABC system $$= 1- (1-.9(1-x^2)).01$$

Probability of ABC failure $$= 0.01(.1+.9x^2)$$

Probability of ABC failure due to B failure $$= .01x^2$$

Applying Bayes' theorem:

$$\dfrac{.01x^2}{0.01(.1+.9x^2)} = 0.1$$

$$x = 0.104828$$

It appears to me that 10% is not the system failure rate, it's the failure probability of subsystem B. So you're being given a problem with extraneous information.

"The [probability] a system failure was due to a failure of the parallel subsystem B is 10 %."

If this is a fault diagram, then the components of B being parallel imply redundancy.

Since x represents the probability of failure of each component of B. For the subsystem to fail, both must fail. The probability of that is just x^2 if we assume that the events are independent (no common cause failures). Solving x^2=.1 yields x=.316

• "The [probability] a system failure was due to a failure of the parallel subsystem B is 10 %." - does this not mean that given that the subsystem B failed, the whole system failed? The probability of that is 0.10? Nov 8 '18 at 16:15
• Oh I see what you mean. I read it a little differently, without the "was". So I think you interpreted it as meaning B is responsible for 10% of all system failures. The question is poorly worded. When a question has poor use of English articles, I assume they also have poor conjugation of verbs. I interpreted the phrase to mean that subsystem B has a 10% chance of causing a system failure in a given time period. Ordinarily in reliability analysis terminology would focus on failure rates rather than probabilities. Nov 8 '18 at 16:33
• Ah ok. Thats the wording of an old exam, as it was given to us (=the students). Would my solution to the problem be wrong? Or can it be argued the way i solved it? Nov 8 '18 at 16:41
• You have P(S|B) = 10% but under the interpretation that 10% of system failures also involve B failing, you should have P(B|S) = 10%. Note that it is literally impossible for the system to fail solely "due to B failing", since C could keep the system working, so a more refined interpretation is that 10% of system failures include B in a failed state. P(S|B) = P(C)=.1^2 because the only way that the system can fail when B is failed is if C also fails. This really is a poorly worded question! Anyway under the refined interpretation, I get x=.209 Nov 8 '18 at 18:02
• I should also add that P(S) is just .1^2(1-(1-.1)(1-x^2)), and you have to use Bayes' theorem. Nov 8 '18 at 18:04