An urn contains 30 balls, of which 10 are red and
8 are blue. From this urn, 12 balls are randomly
withdrawn. Let X denote the number of red and Y
the number of blue balls that are withdrawn. Find
Cov(X,Y) by defining appropriate indicator (that is,
Bernoulli) random variables
$X_i,Y_j$ such that
$X=\sum_{i=1}^{10} X_i$,$Y=\sum_{j=1}^{8} Y_j$
My attempt is:
Let $X_{i} = \begin{cases}1 & \text{ if i-th red ball is withdrawn } \\
0 & \text{ otherwise }\end{cases}$
$X=\sum_{i=1}^{10} X_i$=number of red balls withdrawn.
Let $Y_{i} = \begin{cases}1 & \text{ if i-th blue ball is withdrawn } \\
0 & \text{ otherwise }\end{cases}$
$Y=\sum_{i=1}^{8} Y_i$=number of blue balls withdrawn.
$$cov(X,Y)= cov (\sum_{i=1}^{10} X_i,\sum_{j=1}^{8} Y_j)=\sum_{i=1}^{10} \sum_{j=1}^{8}cov(X_i, Y_j)$$
with $$\begin{align}cov(X_i,Y_j) &= E[X_i*Y_j]-E[X_i]*E[Y_j]\\ &=p(X_i=1,Y_j=1)-p(X_i=1)*p(Y_j=1)\end{align}$$
I don't understand why
$$p(X_i=1,Y_j=1)=\binom{28}{10} /\binom{30}{12} $$ and $$p(X_i=1)=p(Y_j=1)=12/30$$
that are the solution according to my book ($cov(X,Y)=80*[\binom{28}{10} /\binom{30}{12}-( \frac{12}{30})^2]=- \frac{96}{145}$.
Could someone help me?
Best Answer
$\binom{30}{12}$ is total number of different sets of 12 balls from the set of 30 balls. $\binom{28}{10}$ is a number of sets of 12 balls, which contains $i$-th red ball and $j$-th blue ball, since we must add 10 new balls from 28 other balls. So, the probability that $i$-th red ball and $j$-th blue ball are taken is equal to $\binom{28}{10}/\binom{30}{12}$.
In the same manner the probability that one fixed ball is taken is equal to $$ \binom{29}{11}/\binom{30}{12}=12/30. $$