Asystem consists of five identical components connected in series as shown: as soon as one components fails, the entire system will fail. suppose each component has a lifetime that is exponentially distributed with λ = 0.01 and that components fail independently of one another. define events ai = {ith component lasts at least t hours}, i = 1, . . , 5, so that the ais are independent events. let x = the time at which the system fails—that is, the shortest (minim

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((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.) 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((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.)

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