Probability exercise - system reliability

Question

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) = ?\% $

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

EDIT - adding context

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

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

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?

Answer 1

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

Answer 2

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:

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

$x = 0.104828$