discrete distribution sampling error Let's say all people on earth can be divided in five categories depending on their favorite fruit or vegetable (so discrete probability distribution) and we know the true distribution.
For example 2% like apples the most, 30% like bananas the most, 40% like cherries, 20% like dates and 8% like eggplants.
I have surveyed $N$ people and measured that there's $x_a$ of them like apples, $x_b$ like bananas and so on with $x_{c},x_d$ and $x_e$. 
How do I compute the probability of obtaining such set of measurements, given that I know the global distribution?
[This is not a textbook question, I just rephrased my real world problem as fruit and veggies, to save on domain-specific terminology.]
 A: If $N \ll M,$ where $M$ is the total number of people in the world, then you simply use the Multinomial Distribution as a perfectly valid approximation. (The assumption here is that the probability distribution doesn't change as you draw samples. It isn't exact because when you draw a sample, there is one less member of that group in the remaining population, so the probability distribution changes slightly.)
On the other hand, if $N$ and $M$ are of comparable size (which, I know, is extremely unrealistic if we're talking about surveys, but let's pretend it could be for the sake of completion) then you simply want to solve a combinatorics problem. Suppose there are $K$ groups, and $x_j$ is the number of members sampled from group $j$ where $j = 1, \dots, K.$ Also, let $M_j$ be the number of people in the world in group $j$, so that $\sum_{j=1}^K M_j = M,$ where $M$ is the number of people in the world. Then, if our observation is ${\bf x} = \{x_1, x_2, \dots, x_K\},$ with $\sum_j x_j = N,$ the number of possible ways to have gotten such an observation, assuming each individual in the world has an equal chance of being selected, is,
$$
n({\bf x}) = \prod_{j=1}^K {M_j \choose x_j}.
$$
The number of possible choices of any subgroup of $N$ people is simply
$$
{M \choose N}.
$$
Thus the exact probability of observing the outcomes given by $\bf x$ is
$$
P({\bf x}) = \frac{\prod_{j=1}^K {M_j \choose x_j} } {M \choose N}.
$$
This is the multivariate hypergeometric distribution. (Thanks @Glen_b for suggesting I include the name.)
Usually this won't be tractable to calculate for large values of $M$ and $N,$ so approximate factorial calculations (such as the Stirling approximation) should be used. However, suppose we take the original assumption that $N \ll M.$ Let's further assume that this is true for each subgroup of the population, i.e. $x_j \ll M_j \hspace{1mm} \forall j.$ Then,
$$
{M_j \choose x_j} = \frac{M_j!}{x_j! (M_j - x_j)!} \approx \frac{M_j^{x_j}}{x_j!},
$$
and similarly, 
$$
{M \choose N} \approx \frac{M^N}{N!}.
$$
So 
$$
P({\bf x}) \approx \frac{N!}{\prod_j x_j!} \prod_j \left(\frac{M_j}{M}\right)^{x_j} = \frac{N!}{\prod_j x_j!} \prod_j q_j^{x_j}
$$
where $q_j = M_j/M$ is the probability of selecting a member of group $j.$ Thus we have shown that the exact answer reduces to the approximate answer that I suggested at the beginning, i.e. the Multinomial distribution, when the sample size of each subgroup of the population is much smaller than the true size of each subgroup.
