Hypergeometric Distribution : probability that more than half is good To simplify the context, let's say that 34 % of people are ugly. haha...
lets take a sample of 15 people. (n = 15)
a) What is the probability that 3 or less out of the 15 are ugly ?
I went on and computed $P(0) + P(1) + P(2) + P(3)$ and it gives the same result I found with this calculator
b) - Find the average of ugly people in the sample ?
I used $n {K\over N}$ and found : $9.9$%


*

*Find the variation


I used $n{K\over N}{(N-K)\over N}{N-n\over N-1}$ and found $0.306*0.859$.
what should be my answer to the variation question ? is it 0.2623 or 26.23 % ?
c) What is the probability that more than half of the sample are ugly ?
Here is my question : Is there a faster way to find C, than doing : $P(0) + ... + P(8)$ 
 A: Here's the problem:  if you say 34% of people are ugly, then this represents a population proportion.  In order to use a hypergeometric distribution model, your sample would have to have an exact, known number of people who are ugly; e.g., 15 people are selected, 5 of which are ugly, and 10 of which are not.  Of those 15 people sampled, you select $m$ without replacement.  What is the probability that you obtain $3$ or fewer ugly people out of the $m$ selected?  This would be hypergeometrically distributed.  But in your case, 34% of 15 is 5.1, not 5; furthermore, you are asking about the distribution of the total number of ugly people in the sample of 15.  Therefore the hypergeometric model does not apply.
Instead, what you have is a binomial model.  We suppose that the population of individuals from which you have taken the sample is large enough that the sampling can be considered to have taken place with replacement: i.e., the number of ugly people in the sample follows a binomial distribution with parameters $n = 15$, $p = 0.34$.  Then the probability that no more than $3$ people are found to be ugly is simply $$\Pr[X \le 3] = \sum_{k=0}^3 \binom{15}{k} (0.34)^k (1 - 0.34)^{15-k}.$$
The mean and variance of the binomial distribution is $${\rm E}[X] = np = 15(0.34) = 5.1, \quad {\rm Var}[X] = np(1-p) = 3.366.$$
As for finding the probability that more than half of the sample is ugly--i.e., $X \ge 8$, we would have to compute $$\Pr[X \ge 8] = \sum_{k=8}^{15} \binom{15}{k} (0.34)^k (1-0.34)^{15-k}.$$  There isn't really a shorter way for such a small sample size.  If the sample size were larger, such that $np \ge 10$ and $n(1-p) \ge 10$, then we could use a Normal approximation to the binomial.  But in this case, we shouldn't.
