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Find the expected value if the nine players on a basketball team consisting of $2$centers, $3$forwards, and $4$ backcourt players. If the players are paired up at random into three groups of size $3$ each

Expected value is

$E\left[\underset{i=1}{\overset{3}{鈭�}}{X}_i\right]=\underset{i=1}{\overset{3}{鈭�}}E\left[X_i\right]$

$=\frac{2}{7}+\frac{2}{7}+\frac{2}{7}$

=$\frac{6}{7}$

$\text{Therefore,}E\left[\underset{i=1}{\overset{3}{鈭�}}{X}_i\right]=\frac{6}{7}$

The expected value is$\frac{6}{7}$

The variance of the number of triplets consisting of one of each type of player.

Now variance is,

$\mathrm{Var}鈦�\left(X_i\right)=\left(\frac{2}{7}\right)(1鈭�\frac{2}{7})$

=$\frac{10}{49}$

$\text{Now, for}i鈮�j\text{, the value of}E(X_i,X_j)\text{is,}$

$E(X_i,X_j)=P[X_i=1,X_j=1]$

$=P[X_i=1]P[X_j=1鈭�X_i=1]$

$=\frac{\left(\begin{array}{l}2\\ 1\end{array}\right)\left(\begin{array}{l}3\\ 1\end{array}\right)\left(\begin{array}{l}4\\ 1\end{array}\right)}{\left(\begin{array}{l}9\\ 3\end{array}\right)}脳\frac{\left(\begin{array}{l}1\\ 1\end{array}\right)\left(\begin{array}{l}2\\ 1\end{array}\right)\left(\begin{array}{l}3\\ 1\end{array}\right)}{\left(\begin{array}{l}6\\ 3\end{array}\right)}$

$=\frac{6}{70}$

Find the variance of the number of triplets consisting of one of each type of player.

Therefore, the variance of $\underset{x}{鈭�}$, is

$\mathrm{Var}鈦�\left(\underset{i=1}{\overset{3}{鈭�}}{X}_i\right)=\underset{i=1}{\overset{3}{鈭�}}\left(X_i\right)+2鈭�\underset{j=1}{鈭�}\mathrm{Cov}鈦�(X_i,X_j)$

$=\left(\frac{10}{49}\right)脳3+2\left(\begin{array}{l}3\\ 2\end{array}\right)(\frac{6}{70}鈭�\frac{2^2}{7^2})$

$=\frac{30}{49}+6(\frac{6}{70}鈭�\frac{4}{49})$

$=0.6367$

$\text{Therefore,}\mathrm{Var}鈦�\left(\underset{i=1}{\overset{3}{鈭�}}{X}_i\right)=0.6367$

The variance of the number of triplets consisting of one of each type of player. is$=0.6367$