The assumptions leading to the hypergeometric distribution are as follows:

1. The population or set to be sampled consists of $N$ individuals, objects, or elements (a finite population).
2. Each individual can be characterized as a success ($S$) or a failure ($F$), and there are $M$ successes in the population.
3. A sample of $n$ individuals is selected without replacement in such a way that each subset of size $n$ is equally likely to be chosen.

Prop. If $X$ is the number of $S$'s in a completely random sample of size $n$ drawn from a population consisting of $M$ $S$'s and $(N-M)$ $F$'s, then the probability distribution of $X$, called the hypergeometric distribution, is given by

\[
P(X=x)=h(x;n,M,N)=\frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}}
\]

for $x$ an integer satisfying $\max(0,n-N+M)\leqslant x\leqslant\min(n,M)$.

Notation:

$N$: number of individuals

Prop. The mean and variance of the hypergeometric rv $X$ having pmf $h(x;n,M,N)$ are 

\[
E(X)=n\cdot\frac{M}{N},\quad V(X)=\frac{N-n}{N-1}\cdot n\cdot\frac{M}{N}\cdot\biggl(1-\frac{M}{N}\biggr)
\]

The ratio $\frac{M}{N}$ is the proportion of $S$'s in the population. If we replace $M/N$ by $p$ in $E(X)$ and $V(X)$, we get

\[
\begin{aligned}
E(X)&=np,\\
V(X)&=\frac{N-n}{N-1}\cdot np(1-p)=\frac{N-n}{N-1}\cdot npq
\end{aligned}
\]

It shows that the means of the binomial and hypergeometric rv's are equal, whereas the variances of the two rv's differ by the factor $\frac{N-n}{N-1}$, often called the finite population correction factor. 

This factor is less than 1, so the hypergeometric variables has smaller variance than does the binomial rv. The correction factor can be written

\[
\dfrac{1-\frac{n}{N}}{1-\frac{1}{N}},
\]

which is approximately 1 when $n$ is small relative to $N$.

Remark:

An easy way to remember the scope of $x$ is use the formula

\[
\binom{N-M}{n-x}=\binom{N-M}{N-M-(n-x)}=\binom{N-M}{x-(n-N+M)}.
\]

Hence, to make the equation meaningful, $x$ should satisfies $n-N+M\leqslant x$.

References:

Proposition in section 5 of Chapter 3, BOOK:

《Probability and Statistics For Engineering and The Sciences》(Fifth Edtion) P.129
Author: Jay L. Devore

Show that $E(X)=np$ when $X$ is a binomial random variable.

[Hint: First express $E(X)$ as a sum with lower limit $x=1$. Then factor out $np$, let $y=x-1$ so that the sum is from $y=0$ to $n-1$, and show that the sum equals $1$.]
 

The variance is $V(X)=np(1-p)=npq$, and the Standard Deviation (SD) of $X$ is $\sigma_X=\sqrt{npq}$ (where $q=1-p$).

The proof can be seen in Question 32. (Where the variance is denoted by $D(X)$.)

(a) Show that $b(x;n,1-p)=b(n-x;n,p)$.

(b) Show that $B(x;n,1-p)=1-B(n-x-1;n,p)$. [Hint: At most $x$ $S$'s is equivalent to at least $(n-x)$ $F$'s.]

 

Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, $60\%$ can be repaired whereas the other $40\%$ must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?

[中文]

$20\%$的某種類型的電話在保修期間需要維修。其中$60\%$可以修好，其餘$40\%$必須更換新裝置。如果一家公司購買了其中十部手機，那麼在保修期內，正好有兩部電話被替換的可能性是多少？

[法语]

Vingt pour cent de tous les téléphones d'un certain type sont soumis au service pendant la garantie. De ce nombre, $60\%$ peuvent être réparés alors que les $40\%$ restants doivent être remplacés par de nouvelles unités. Si une entreprise achète dix de ces téléphones, quelle est la probabilité que exactement deux finissent par être remplacés sous garantie?
 

A company that produces fine crystal knows from experience that $10\%$ of its goblets have cosmetic flaws and must be classified as "seconds."

• (a) Among six randomly selected goblets, how likely is it that only one is a second?
• (b) Among six randomly selected goblets, what is the probability that at least two are seconds?
• (c) If goblets are examined one by one, what is the probability that at most five must be selected to find four that are not seconds?
   

Definition. Given a binomial experiment consisting of $n$ trials, the binomial random variable(二项随机变量) $X$ associated with this experiment is defined as

\[
X=\text{the number of S's among the}\ n\ \text{trials}
\]

Here we use S to denote the outcome H(heads) and F to denote the outcome T(tails).

We write $X\sim\mathrm{Bin}(n,p)$ to indicate that $X$ is a binomial rv based on $n$ trials with success probability $p$.

We denote the pmf of a binomial rv $X$ by $b(x;n,p)$. The cdf is denoted by
\[
P(X\leqslant x)=B(x;n,p)=\sum_{y=0}^{x}b(y;n,p)\quad x=0,1,2,\ldots,n.
\]

Here

\[
b(x;n,p)=\begin{cases}
\binom{n}{x}p^x(1-p)^{n-x},&x=0,1,2,\ldots,n\\
0,&\text{otherwise}.
\end{cases}
\]

Calculate the following probabilities:

• (a) $B(4; 10, .3)$
• (b) $b(4; 10, .3)$
• (c) $b(6; 10, .7)$
• (d) $P(2\leqslant X\leqslant 4)$ when $X\sim\mathrm{Bin}(10,.3)$
• (e) $P(2\leqslant X)$ when $X\sim\mathrm{Bin}(10,.3)$
• (f) $P(X\leqslant 1)$ when $X\sim\mathrm{Bin}(10,.7)$
• (g) $P(2 < X < 6)$ when $X\sim\mathrm{Bin}(10,.3)$
   

Suppose $E(X)=5$ and $E[X(X-1)]=27.5$. What is

• $E(X^2)$ ? [Hint: $E[X(X-1)]=E[X^2-X]=E(X^2)-E(X)$.]
• $V(X)$ ?
• The general relationship among the quantities $E(X)$, $E[X(X-1)]$, and $V(X)$.

Using the definition of variance, prove that
\[
V(aX+b)=a^2\cdot\sigma_X^2
\]

[Hint] With $h(X)=aX+b$, $E[h(X)]=a\mu+b$ where $\mu=E(X)$.

The expected value of $X$ measures where the probability distribution is centered. We will use the variance of $X$ to measure the amount of variability in (the distribution of) $X$.

Let $X$ have pmf $p(x)$ and expected value $\mu$. Then the variance of $X$ ($X$ 的方差), denoted by $V(X)$ or $\sigma_X^2$, or just $\sigma^2$, is defined by
\[
V(X):=\sum_{D}(x-\mu)^2\cdot p(x)=E\bigl[(X-\mu)^2\bigr]
\]
Prove that
\[
V(X)=E(X^2)-(E(X))^2.
\]

i.e.,

\[V(X)=\sigma^2=\biggl[\sum_{D}x^2\cdot p(x)\biggr]-\mu^2\]
 

Let $X$ have pmf
\[
p(x)=\begin{cases}
\frac{k}{x^2}, & x=1,2,3,\ldots\\
0, & \text{otherwise}
\end{cases}
\]
where $k$ is chosen so that $\sum_{x=1}^{\infty}\frac{k}{x^2}=1$. So the expected value of $X$ is
\[
\mu=E(X)=\sum_{x=1}^{\infty}x\cdot\frac{k}{x^2}=k\sum_{x=1}^{\infty}\frac{1}{x}
\]
What is the value of $k$?

Answer: The value of $k$ is equal to $\frac{6}{\pi^2}$.

Please see the Question 20, or search $\frac{\pi^2}{6}$ in this site.