In reply to Jack B:
Jack, it sounds like you know more stats than me; I have to admit I can't remember what a Poisson distribution is. But in the spirit of UKC, let me have a crack at it anyway!
It seems like a reasonable assumption that a given miner in a given period of hours underground has a certain probability of dying, independently of anything else. Let's set
p = probability of a given miner dying during a given 400 hour period underground.
This will be the answer to the first question for miner A. It's also useful to define
q = probability that a given miner does not die during a given 400 hour period.
Obviously, q = 1 - p. For miner B, we want to know the probability, p', that a given miner dies during an 800 hour period underground. Since not dying during an 800 hour period is the same as not dying during the first 400 hours AND not dying during the second 400 hours, the probability, q', that he
doesn't die during these 800 hours is equal to the probability that he doesn't die during the first 400 hours multiplied by the probability that he doesn't die during the second 400 hours. So
q' = q^2
or
1 - p' = (1 - p)^2
or
p' = 1 - (1 - p)^2.
Similarly, 1200 = 3*400 hours spent down t'pit will be your last with probability
p'' = 1 - (1 - p)^3.
You might be able to spot a pattern now and guess that your chances of death after 1 = (1/400)*400 hour is
p''' = 1 - (1 - p)^(1/400).
In fact this is right because it is equivalent to
(1 - p) = (1 - p''')^400,
which just says that to survive 400 hours you have to not die during the first hour and not die during the second hour and so on right through to the 400th hour.
In general, your chances of surviving t hours is given by
p_t = 1 - (1 - p)^(t/400).
Re-writing this as
q_t = q^(t/400),
it begins to look like your formula for the Poisson distribution, i.e. exponential decay, but with a different base.
Anyway, to answer the original question, all you have to do is figure out what p is and you're done.
I'm not certain how to interpret the data we're given, but taking the expected number of deaths for a cohort of 900 miners each working 400 hours to be equal to 1, we get
expected number of deaths for a given miner working 400 hours = 1/900
p = probability of death for a given miner working 400 hours = 1/900.
Then the answers are
miner A : 0.0011111
miner B : 0.0022210
miner C : 0.0033296
miner D : 0.0000028
The guy who works 1 million hours should really get a will drawn up
miner E : 0.9397194.
These answers are really close to yours, Jack, but different. I don't know if that's because our models are different or because we're reading the input data differently, or one of us has made a mistake. Any ideas?