Average Employment Time (Poisson Distribution)

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I'm working on the following problem:

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I'm having a real hard time finding any information regarding of how I should approach the problem. From what I see, am I correct that this might require the Poisson Distribution to be solved? My attempt to solve the problem included finding the total number of employees at the beginning of the period (2,418 at the beginning of the year + 744 hired, which totals 3162), at the end of the year (2418), and averaging the result to yield on average 2790 employees within one year. But I'm stuck on how I should assume the time here. For the second part (b), would I be right stating that it's just $\dfrac{50}{6}\cdot 57 = 475$? I would really appreciate your help, I'm quite stuck.

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(a) Hint - If you work for the company, the odds of leaving within a year are $\frac{744}{2418}$. So your odds of still being there are $\frac{1674}{2418}$.

If you are still there after a year, the odds of leaving and staying the next year are the same. That is, the odds of you being there after $2$ years is $\frac{1674}{2418}^2$.

You can work with fractional years. If 744 people leave per year, that works out to around $2$ per day.

You can see that pattern, and figure out the odds after n years. How big does n have to be to make the odds $50$%?

(b) looks right to me.