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Consider a computer system which employs two copies $A$ and $B$ of some chip. A chip $C$ on reserve is used to replace either $A$ or $B$ whichever fails first. What is the probability that $A$ is still in service after the other two have failed when:

(a) the lifetime of each chip is exactly 10 minutes.

(b) the lifetime are $i (i= 1,2,3)$ with probability 1/3.

(c) the lifetimes are exponential with mean $1/\mu$.

I guess the answer is somehow related to the properties of Poisson process and cannot figure out how. Any help is useful. Thanks!

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For (a) the answer is obviouisly $0$, since after 10 minutes $A$ and $B$ have failed and $C$ has just started

For (b) the answer is $\dfrac{1}{27}$ as the only pattern leaving $A$ being in service after $C$ has failed is that $B$ has lifetime $1$, $C$ has lifetime $1$ and $A$ has lifetime $3$.

For (c), you could reagrd this as a Poisson process, but there is an easier approach. $A$ and $B$ are iid with exponential distributions so the probability that $B$ fails before $A$ is $\dfrac{1}{2}$; $C$ then starts, and since the exponential distribution is memoryless, the probably that $C$ then fails before $A$ is also $\dfrac{1}{2}$. So the answer is $\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}.$

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