Probability distribution of rare condition count A population of t=310M experiences an environmental change that causes one in s=112M of them to acquire a particular condition c, each one independent of the others.
What is the probability p that exactly n members of the population will have the condition? 
(Here, M = million, n is any integer >=0, with special attention to very low integers.)  
I am attempting to demonstrate what efforts I've taken to try to find an answer on my own, with multiple approaches below.  Clearly, at least one is wrong.  A right answer, with good explanation about why it's right (especially if it explains why the others are wrong), would be helpful.  Thanks!
 A: 
What is the probability p that exactly n specific members of the population will have the condition?

Since each one is independent of the others, that's the probability that (each of those n specific members having the condition) AND (each of the others don't.)
$(\frac{1}{s})^{n}*(1-\frac{1}{s})^{t-n}$  
For example, for any three specific members, it's:  
$(\frac{1}{112M})^{3}*(1-\frac{1}{112M})^{310M-3} = 4.47*10^{-24}\%$.
which is a starting point toward answering the real question, 

What is the probability p that exactly any n members of the population will have the condition?  

To get this, we multiply the previous answer by the count of all the possible ways to pick three members out of the population, abbreviated COMBIN(310M,3):
$(\frac{1}{112M})^{3}*(1-\frac{1}{112M})^{310M-3}*COMBIN(310M,3) = 22.1\%$ = p for n=3.  
For n=0, p= 6.3%.
For n=1, p= 17.4%.
For n=2, p= 24.1%.
For n=3, p= 22.1%.
For n=4, p= 15.4%.
For n=5, p= 8.5%.
For n=6, p= 3.9%.
For n=7, p= 1.6%.
For n=8, p= 0.5%.
For n=9, p= 0.2%.
For n>=10, p= 0.0%.  
The general answer is:
$p(n) = (\frac{1}{s})^{n}*(1-\frac{1}{s})^{t-n}*COMBIN(t,n)$.
A: It is easiest to start thinking about this with n=0:

What is the probability that none of the population have the condition?  

Another way of saying that is,

What is the probability that all of the population do not have the condition?  

The probability that any 1 is free from the condition is $(1-\frac{1}{112M})$.
To find the probability of an AND combination of independent events, we can just multiply together the probabilities of each event.  Here, there are 310M events.
So: $((1-\frac{1}{112M})^{310M})$ = p = 6.3% for n=0.  

What is the probability that at least one member of the population has the condition?  

Subtract the prior result from 100%. To find the probability there are 1 or more,
$(1-((1-\frac{1}{112M})^{310M}))$= an 93.7% chance that there are 1 or more individuals with the condition.  
Now we have taken care of n=0, and entered a range where n>=1.
Given that at least one member of the population has c, what's the probability that another one does?
Since the events are independent, you simply reduce the population size by 1 and recompute.  Let's first consider  

What is the probability that at least two members of the population have the condition?  

$(1-((1-\frac{1}{112M})^{310M}))*(1-((1-\frac{1}{112M})^{310M-1}))$ = an 87.8% chance that there are 2 or more individuals with the condition.
In that equation, the first term represents the probability of having at least one with c, which is independent from the probability that one of the remaining population members also has c.  These two are multiplied together for the AND combination of independent events.  Continuing another step,  

What is the probability that at least three members of the population have the condition?  

$(1-((1-\frac{1}{112M})^{310M}))*(1-((1-\frac{1}{112M})^{310M-1}))*(1-((1-\frac{1}{112M})^{310M-2}))$ = an 82.3% chance that there are 3 or more individuals with the condition.  
For very small values of n compared to 310M (population size t), this is approximately
$(1-((1-\frac{1}{112M})^{310M}))^{n}$.  
Knowing this, it's easier to answer the original question.

What is the probability that exactly one member of the population has the condition?  

That would be the probability that at least one member of the population has the condition, minus the probability that two or more members of the population have the condition.
$ 93.7\% - 87.8\% = 5.9\% $ = p for n=1.

What is the probability that exactly two members of the population have the condition?  

That would be the probability that at least two members of the population have the condition, minus the probability that three or more members of the population have the condition.
$87.8\% - 82.3\% = 5.5\% $ = p for n=2.
Again, for n << t, 
p(n)=$(1-((1-\frac{1}{112M})^{310M}))^{n}-(1-((1-\frac{1}{112M})^{310M}))^{n+1}$.  
Or, more generally,  
p(n)=$(1-((1-\frac{1}{s})^{t}))^{n}-(1-((1-\frac{1}{s})^{t}))^{n+1}$.  
If someone can point out if/where this answer goes wrong, I'd appreciate it.  If you downvote the answer for being wrong, please upvote the question for showing effort.
