Ask Question
27 November, 06:10

Suppose that $n$ components are connected in series. for each unit, there is a backup unit, and the system fails if and only if both a unit and its backup fail. assuming that all the units are independent and fail with probability $p$, what is the probability that the system works? for $n=10$ and $p=.05$ compare these results with those of example f in section 1.6

+5
Answers (1)
  1. 27 November, 07:38
    0
    For each i, P (Ai) = P (exponential random variable with λ = 0.01 is ≥ t)

    = 1 - P (exponential random variable with λ = 0.01 is < t)

    = 1 - P (exponential random variable with λ = 0.01 is ≤ t) since exponentials are continuous

    = 1 - [1 - e^ (-λt) ] at λ = 0.01

    = e^ (-0.01t).

    Since the entire system fails as soon as one component fails, X ≥ t if and only if all five of the Ai's occur (that is, the system lasts at least t hours if and only if all five of the components last at least t hours.

    P (X ≥ t) = P (all five Ai's occur)

    = P (A1) P (A2) ... P (A5) since the components are independent

    = [e^ (-0.01t) ]^5

    = e^ (-0.05t).

    F (t) = P (X ≤ t)

    = P (X < t) since X is a continuous random variable

    = 1 - P (X ≥ t)

    = 1 - e^ (-0.05t).

    The pdf of X is f (t) = F' (t) = 0.05e^ (-0.05t).

    Note that X itself is also exponentially distributed, but with rate parameter 0.05 instead of 0.01.

    (More generally, the lifetime of a system with components connected in series is exponentially distributed, with rate parameter equal to the sum of the rate parameters for the lifetimes of the individual components, assuming that the components' lifetimes are independent and exponentially distributed.)
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Suppose that $n$ components are connected in series. for each unit, there is a backup unit, and the system fails if and only if both a unit ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers