I am trying to solve the following problem:
A box contains N balls: $N_1\ white, N_2\ black,\ and\ N_3\ red\ (N = N_1 + N_2 + N_3).$ A random sample of n balls is selected from the box (without replacement). Let $Y_1, Y_2, and\ Y_3$ denote the number of white, black, and red balls, respectively, observed in the sample. Define $c = (N − n)/(N − 1)$, and $p_i = N_i/N,\ for\ i = 1, 2, 3.$ Assume $3 ≤ n ≤ min{\{N_1, N_2, N_3}\}$.
How to deduce the marginal distributions of $Y_1, Y_2, and\ Y_3,$ and compute their respective means and variances.and find their $Cov(Y_1, Y_2)$. and $Y_i \sim Hypergeometry(N,N_i,n)$.
Also $Cov(Y_1,Y_2)= E(Y_1Y_2)-E(Y_1)E(Y_2)$. But How could I find $E(Y_1Y_2)$?
Any suggestion or help would be highly appreciated.
The probability of selecting $y_1$ of $N_1$ white balls in a sample of $n$ from $N$ (where $3\leq n< N_1$ is: $$\mathsf P(Y_1=y_1) = \dfrac{\binom{N_1}{y_1}\binom{N-N_1}{n-y_1}}{\binom{N}{n}}$$
Which is a Hypergeometric Distribution, whose mean and variance you should know; or you can obtain by using indicators.
Let $Y_{1,i}$ be the indicator that the $i$-th ball in the sample of $n$ is white ( $Y_{1,i}$ is $1$ if the i-th ball is white, otherwise it is $0$ ). Then: $$\begin{align}\mathsf E(Y_1) =&~ \sum_{i=1}^{n}\mathsf E(Y_{1,i}) \\[2ex] \mathsf E(Y_1^2) =&~ \sum_{i=1}^n\sum_{j=1}^n \mathsf E(Y_{1,i}Y_{1,j}) \\[1ex] =&~ \sum_{i=1}^n \mathsf E(Y_{1,i}^2)+ 2\mathop{\sum\sum}\limits_{1\leq i<j\leq n} \mathsf E(Y_{1,i}Y_{1,j}) \end{align}$$
Can you complete to find the mean and variance of each colour's count?
You can find the covariance between two colours by the same technique; noting that $\mathsf E(Y_{1,i}Y_{2,i})=0$ because any ball can only be one colour.