How to show that this stat statement is true Someone made this statement:
1 out of 1000 corresponds to a proportion of 0,1%. In this case you 
have a chance of 37% of having 0 outcome, significance level 63%. 
3 out of 1000 translating to a proportion of 0,3% has a 5% chance to hit 0. 

Are the above statement correct and how one arrives on proving this is indeed correct.
The above statement goes something like this:
So lets assume that I observe 0 cases out of 1000. Can I say that assuming hypothetically 1 dead out of 1000 would give me a 37% chance of observing 0?
Or otherwise what is the probability of observing 0 cases if one assumes (historically) 1/1000, under 63% confidence and 95% confidence level. 
Solution, is this correct?
dbinom(0, 1000, prob = 1/1000) # gives ~36%

and
dbinom(0, 1000, prob = 3/1000) # gives ~5%

 A: Ah. Well, this seems to be a very badly formulated way of saying:

We perform 1000 independent trials, each with a success probability of 0.001. Then we have a chance of 37% to observe no successes at all.

This is a true statement, pbinom(0,1000,1/1000). Note how the statement in the question does not mention the number of trials, only the proportion of successes - but without the number of trials, we can't say anything about the probability of observing $k$ successes.
If we increase the success probability to 0.003, then yes, we only have a chance of about 5% to observe no success at all, pbinom(0,1000,3/1000).
"significance level 63%" - our situation has nothing to do with statistical significance.
A: Another potential way of solving this problem is with a Poisson Distribution.
In this case $\lambda$ is 1/1000.  So the probability of of observing 0 events with a single trial is:
ppois(0, 1/1000)
#[1] 0.9990005

Thus it is 99.9% chance that no events will occur.  Now if we are interested in testing a 1000 independent trials then the overall probability is:
ppois(0, 1/1000)^1000
#[1] 0.3678794

So a 36.7% chance that there will be no events in 1000 random, independent trials.  
Now repeating for the $\lambda=$ 3/1000 probability case:
ppois(0, 3/1000)
#[1] 0.9970045

#drawing 1000 possible options
ppois(0, 3/1000)^1000
#[1] 0.04978707

Now the chance of having 0 events per 1000 trials fall to ~5%.
