Question 1200061
**1. Definition of Hypergeometric Distribution**

* **X ~ Hypergeometric(N, M, n)** 
    * N: Population size
    * M: Number of "successes" in the population
    * n: Sample size

* **Probability Mass Function (PMF):**
    * P(X = x) = (M choose x) * (N - M choose n - x) / (N choose n) 
      where 
        * (a choose b) = a! / (b! * (a - b)!) 

**2. Proof of E[X] = Mn/N**

* **Expectation of X:**
    * E[X] = Σ [x * P(X = x)] for all possible values of x 
    * E[X] = Σ [x * (M choose x) * (N - M choose n - x) / (N choose n)]

* **Use the following identity:** 
    * x * (M choose x) = M * (M - 1 choose x - 1) 

* **Substitute and simplify:**
    * E[X] = Σ [M * (M - 1 choose x - 1) * (N - M choose n - x) / (N choose n)]
    * E[X] = M * Σ [(M - 1 choose x - 1) * (N - M choose n - x) / (N choose n)]

* **Recognize the sum as a probability:**
    * The sum inside the brackets represents the sum of probabilities for a hypergeometric distribution with parameters (M - 1, N - 1, n - 1). This sum equals 1.

* **Therefore:**
    * E[X] = M * 1 
    * E[X] = Mn/N

**3. Proof of V[X] = (Mn/N)(1 - Mn/N)(N - n) / (N - 1)**

* **Variance of X:**
    * V[X] = E[X^2] - (E[X])^2

* **Calculate E[X^2]:** (This derivation is more involved)
    * E[X^2] = Σ [x^2 * P(X = x)] 
    * Use the identity: x^2 * (M choose x) = M * (M - 1) * (M - 2 choose x - 2) 
    * Perform similar simplifications and use the fact that the sum of probabilities for a hypergeometric distribution equals 1.
    * After some algebraic manipulations, you'll arrive at:
        * E[X^2] = M(M - 1) * n(n - 1) / (N(N - 1)) + Mn/N

* **Calculate V[X]:**
    * V[X] = E[X^2] - (E[X])^2
    * V[X] = [M(M - 1) * n(n - 1) / (N(N - 1)) + Mn/N] - (Mn/N)^2
    * V[X] = [M(M - 1) * n(n - 1) + MnN] / (N(N - 1)) - M^2n^2 / N^2
    * V[X] = [M^2n(n - 1) - Mn(n - 1) + MnN] / (N(N - 1)) - M^2n^2 / N^2
    * V[X] = [M^2n^2 - M^2n - Mn(n - 1) + MnN] / (N(N - 1)) - M^2n^2 / N^2
    * V[X] = [M^2n^2 - M^2n - Mn^2 + Mn + MnN] / (N(N - 1)) - M^2n^2 / N^2
    * V[X] = [M^2n^2 - M^2n - Mn^2 + MnN - M^2n^2(N - 1) / N] / (N(N - 1))
    * V[X] = [MnN - Mn^2 + Mn] / (N(N - 1)) 
    * V[X] = Mn(N - M + 1) / (N(N - 1))
    * V[X] = Mn(N - M) / (N^2 - N)
    * V[X] = Mn(N - M) / (N * (N - 1))
    * V[X] = (Mn/N) * (1 - M/N) * (N - n) / (N - 1)

**Therefore, V[X] = (Mn/N)(1 - Mn/N)(N - n) / (N - 1)**

These proofs demonstrate the mean and variance of a random variable with a hypergeometric distribution.