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 $Y \hookrightarrow \mathcal{H}(n, p, A)$.

We consider an urn containing $A$ balls of which $pA$ are white and $qA$ are black. We draw simultaneously $n$ balls from the urn. We number from 1 to $pA$ each of the white balls and, for any natural integer $i \in \llbracket 1, pA \rrbracket$, we define $$Y_i = \begin{cases} 1 & \text{if the ball numbered } i \text{ was drawn,} \\ 0 & \text{otherwise.} \end{cases}$$ For $1 \leqslant i < j \leqslant pA$, prove that the random variable $Y_i Y_j$ follows a Bernoulli distribution whose parameter we will specify.

We consider an urn containing $A$ balls of which $pA$ are white and $qA$ are black. We draw simultaneously $n$ balls from the urn. We number from 1 to $pA$ each of the white balls and, for any natural integer $i \in \llbracket 1, pA \rrbracket$, we define $$Y_i = \begin{cases} 1 & \text{if the ball numbered } i \text{ was drawn,} \\ 0 & \text{otherwise.} \end{cases}$$ Let $Y = \sum_{i=1}^{pA} Y_i$ be the number of white balls drawn. Deduce the value of the variance of $Y$. Compare it to that of $Z$ (where $Z \sim \mathcal{B}(n,p)$).

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 $\_\_\_\_$

Q86. From a lot of 10 items, which include 3 defective items, a sample of 5 items is drawn at random. Let the random variable X denote the number of defective items in the sample. If the variance of X is $\sigma ^ { 2 }$, then $96 \sigma ^ { 2 }$ is equal to $\_\_\_\_$

Q90. 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 $\_\_\_\_$ t

ANSWER KEYS

\begin{tabular}{|l|l|l|l|} \hline 1. (2) & 2. (1) & 3. (3) & 4. (2) \hline 9. (2) & 10. (2) & 11. (4) & 12. (4) \hline 17. (3) & 18. (2) & 19. (3) & 20. (3) \hline 25. (93) & 26. (160) & 27. (100) & 28. (10) \hline 33. (3) & 34. (2) & 35. (2) & 36. (3) \hline 41. (4) & 42. (3) & 43. (1) & 44. (3) \hline