Let two natural integers $A$ and $n$ such that $n \leqslant A$ and $p$ a real number between 0 and 1. We assume $pA \in \mathbb{N}$ and we denote $q = 1-p$. Let $X$ be a discrete real random variable. We say that $X$ follows the hypergeometric distribution with parameters $n, p$ and $A$ when $$\left\{\begin{array}{l} X(\Omega) \subset \llbracket 0, n \rrbracket, \\ \mathbb{P}(X = k) = \frac{\binom{pA}{k}\binom{qA}{n-k}}{\binom{A}{n}} \quad \text{for all } k \in \llbracket 0, n \rrbracket. \end{array}\right.$$ Verify that we have indeed defined a probability distribution.

Let two natural integers $A$ and $n$ such that $n \leqslant A$ and $p$ a real number between 0 and 1. We assume $pA \in \mathbb{N}$ and we denote $q = 1-p$. Let $X$ be a random variable such that $X \hookrightarrow \mathcal{H}(n, p, A)$, i.e. $$\mathbb{P}(X = k) = \frac{\binom{pA}{k}\binom{qA}{n-k}}{\binom{A}{n}} \quad \text{for all } k \in \llbracket 0, n \rrbracket.$$ We recall that, for all non-zero natural integers $k$ and $N$, $k\binom{N}{k} = N\binom{N-1}{k-1}$. Calculate the expectation of $X$.

Let two natural integers $A$ and $n$ such that $n \leqslant A$ and $p$ a real number between 0 and 1. We assume $pA \in \mathbb{N}$ and we denote $q = 1-p$. Let $X$ be a random variable such that $X \hookrightarrow \mathcal{H}(n, p, A)$, i.e. $$\mathbb{P}(X = k) = \frac{\binom{pA}{k}\binom{qA}{n-k}}{\binom{A}{n}} \quad \text{for all } k \in \llbracket 0, n \rrbracket.$$ Show that the sequence $(\mathbb{P}(X = k))_{k \in \mathbb{N}}$ is hypergeometric. Deduce an expression of the generating function of $X$ using a hypergeometric function.

We consider two urns each containing $A$ balls of which $pA$ are white and $qA$ are black. We draw simultaneously, in an equiprobable manner, $n$ balls from the first urn. We denote $Y$ the number of white balls obtained. Prove that $Y \hookrightarrow \mathcal{H}(n, p, A)$.

Let $X$ be a random variable following the distribution $\mathcal{H}(n, p, A)$, i.e. $$\mathbb{P}(X = k) = \frac{\binom{pA}{k}\binom{qA}{n-k}}{\binom{A}{n}} \quad \text{for all } k \in \llbracket 0, n \rrbracket.$$ We fix $n$ and $p$. Let $k \in \llbracket 0, n \rrbracket$. Show that $$\lim_{A \to +\infty} \mathbb{P}(X = k) = \binom{n}{k} p^k (1-p)^{n-k}.$$

Let $X$ be a random variable following the distribution $\mathcal{H}(n, p, A)$. We fix $n$ and $p$. We have shown that $$\lim_{A \to +\infty} \mathbb{P}(X = k) = \binom{n}{k} p^k (1-p)^{n-k}.$$ Interpret this result in connection with those obtained for the expectation and variance of $Y$.

A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let $X$ be the number of white balls, among the drawn balls. If $\sigma ^ { 2 }$ is the variance of $X$, then $100 \sigma ^ { 2 }$ is equal to $\_\_\_\_$.

There rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable X denote the number of rotten apples. If $\mu$ and $\sigma ^ { 2 }$ represent mean and variance of X , respectively, then $10 \left( \mu ^ { 2 } + \sigma ^ { 2 } \right)$ is equal to
(1) 20
(2) 250
(3) 25
(4) 30

From a lot of 12 items containing 3 defectives, a sample of 5 items is drawn at random. Let the random variable X denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $X$ is $\frac { m } { n }$, where $\operatorname { gcd } ( m , n ) = 1$, then $n - m$ is equal to $\_\_\_\_$