Difficult probability problem I am stuck in this question $33$ miners are trapped in a mine. There´s an elevator that takes $10$ minutes to go down and another $10$ minutes to pick up a miner and return to the surface. One of the miners will have respiratory problems that will lead him to death in case he is not dealt with before $13$ minutes, and the time that he will get sick is a random variable that is uniformly distributed $(0,20)$. If the miner gets sick before the elevator goes down then he is chosen; if he is not sick when the elevator goes down then any miner will be chosen with the same probability. What is the probability that the miner is out in time with this method? I am afraid I haven´t been able make any progress. I don´t know how to start it. Any suggestions/ideas will be really appreciated :)

We classify according on the time at which the sickly prisoner would have gone up before he got sick.

The complete time frame is $[0,660]$ let $20k$ denote the time in which the miner would have been selected. then the miner dies if and only if he gets sick in a point in the time frame $(20l,20l+17)$ with $l
But we need the sum from $k=0$ to $33$ and divide by $34$ so the probability is $\frac{17\cdot33\cdot34}{20\cdot33\cdot2\cdot34}=\frac{17}{40}\approx 42.5\%$

So the probability he survives is $57.5\%$