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!