- #1
r0bHadz
- 194
- 17
- Homework Statement
- A certain system is based on two independent modules, A and B. A failure of any module causes a failure of the whole system. The lifetime of each module has a Gamma distribution, with parameters α λ given in the table: [COMPONENT A: α:3 λ:1] [COMPONENT B: α:2 λ:2]
a)What is the probability that the system works at least 2 years without a failure? (I already have the answer for this, which is .06195, estimated to .062)
b)Given that the system failed during the first 2 years, what is the probability that it failed due to the failure of component B (but not component A)?
- Relevant Equations
- gamma density [itex]f(x) = \dfrac{(λ^α)(x^{(α-1)})(e^{(-λx)}) }{Γ(α)}[/itex]
I'm lost. First one was easy to calculate, second one is harder.
I have:
P{a fails before 2 yrs} = .323325
P{b fails before 2 yrds} = .90844
P{system doesn't fail for 2 years or longer} = .062
P{system does fail before 2 years} = .938
P{A and B fail before 2 yrs} = .29372
P{before 2 years A fails but B doesn't} = .0296
P{before 2 years A doesn't fail but B fails} = .6147
so in answering B, I assume it is asking me:
P{System failed due to B and not A | system failed during first 2 yrs} =
(.6147*.938)/.938
which = .6147 but this is not the right answer. I don't know where my logic is wrong.
I have:
P{a fails before 2 yrs} = .323325
P{b fails before 2 yrds} = .90844
P{system doesn't fail for 2 years or longer} = .062
P{system does fail before 2 years} = .938
P{A and B fail before 2 yrs} = .29372
P{before 2 years A fails but B doesn't} = .0296
P{before 2 years A doesn't fail but B fails} = .6147
so in answering B, I assume it is asking me:
P{System failed due to B and not A | system failed during first 2 yrs} =
(.6147*.938)/.938
which = .6147 but this is not the right answer. I don't know where my logic is wrong.
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