Q.7.13 - A First Course in Probability

Question

The nine players on a basketball team consist of $2$ centers, $3$ forwards, and $4$ backcourt players. If the players are paired up at random into three groups of size $3$ each, find (a) the expected value and (b) the variance of the number of triplets consisting of one of each type of player. 

Step-by-Step Solution

Answer

From the given information the expected value and the variance of the number of triplets consisting of one each type of player will be 

$$\begin{gathered} \text{a)Expected value is}E\left[\sum _{i=1}^3{X}_i\right]=\frac{6}{7}. \\ \text{b) Variance,}\mathrm{Var}\left(\sum _{i=1}^3{X}_i\right)=0.6367 \end{gathered}$$

Step 1: Given Information (part a)

 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

Step 2: Explanation (part a)

Expected value is 

$E\left[\sum _{i=1}^3{X}_i\right]=\sum _{i=1}^{3}E\left[X_i\right]$

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

=$\frac{6}{7}$

$\text{Therefore,}E\left[\sum _{i=1}^3{X}_i\right]=\frac{6}{7}$

Step 3: Final Answer (part a)

The expected value is $\frac{6}{7}$

Step 4: Given Information (part b)

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

Step 5: Explanation (part b)

Now variance is, 

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

=$\frac{10}{49}$

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

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

$=P[X_i=1]P[X_j=1\mid X_i=1]$

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

$=\frac{6}{70}$

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

Therefore, the variance of $\sum _x$, is

$\mathrm{Var}\left(\sum _{i=1}^3{X}_i\right)=\sum _{i=1}^3\left(X_i\right)+2\sum \sum _{j=1}\mathrm{Cov}(X_i,X_j)$

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

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

$=0.6367$

$\text{Therefore,}\mathrm{Var}\left(\sum _{i=1}^3{X}_i\right)=0.6367$

Step 6 : Final Answer (part b)

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