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The question reads:

Suppose the lifetime of a component $T_i$ in hour is uniformly distributed on $[100, 200]$. Components are replaced as soon as one fails and assume that this process has been going on long enough to reach equilibrium. Suppose it is known that the current component has been in operation for exactly 90 hours. What is the probability that it will last at least 50 more hours?

Essentially, we are being asked to compute:

$$P(B_t \geq 50\ |\ A_t = 90),$$

where $B_t$ is the time remaining in the component's life and $A_t$ is the current age of the component. The computation can be rewritten as

$$P(B_t \geq 50\ |\ A_t = 90)=\frac{P(B_t \geq 50, A_t = 90)}{P(A_t = 90)}$$

by Bayes' Theorem. I know that $A_t$ and $B_t$ aren't independent, but I'm not seeing how to factor that into the computation. I know that both follow uniform distributions with mean $\mu=150$, but honestly I can't get much farther than that. I know how to calculate both $P(B_t \geq 50)$ and $P(A_t = 90)$, but I can't put the two together. Any help is greatly appreciated.

Thanks!

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    Your mistake is to believe that $A$ and $B$ are uniform. They are not. Do you know why $A+B$ is not distributed like $T_i$ and what is the distribution of $(A,B)$? (Incidentally, since $100\leqslant A+B\leqslant200$ with full probability, $\mathbb E(A)=\mathbb E(B)=150$ is impossible.)2012-11-26
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    Ah, you do make a good point. In that case, I'm thinking that the distribution of $A+B$ would be uniform with $E(A+B)=150$. Am I right in thinking that?2012-11-26
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    You are reasoning as if $A+B$ was distributed like $T_i$. This is not so. The correct distribution of $A+B$ is one of the first results in the theory of renewal processes. Have you been exposed to the [inspection time paradox](http://en.wikipedia.org/wiki/Renewal_theory#The_inspection_paradox)?2012-11-26
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    Yes, I have seen that in class but only briefly. How can I apply that to this problem?2012-11-26
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    Then did they explain to you the joint distribution of $(A,B)$?2012-11-26
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    Only a bit. As I gathered, the distribution of the total lifetime of the component $C_t$ can be thought of as the sum of the distributions of the current age $A_t$ and the residual life $B_t$, but I don't know much of anything else regarding this matter. I should add that I'm going to bed now (I've got class in 4 hours) so if you have a formal answer to my question, I'd appreciate it if you posted it and I can ask questions about it tomorrow if I have any. Thanks for your time!2012-11-26
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    See the answer below. Have a good night.2012-11-26

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