Chapter 4, Problem 4.135SE

a.

To determine

To find: the probability distribution for $x$ ,the number of women chosen to be orchestra representatives.

Explanation of Solution

Given:

The board of directors of a major symphony orchestra has voted to create a committee for the purpose of handing employee complaints.

The committee consists of the president and the vice president of the symphony board and two orchestra representatives.

Calculation:

According to the question,

Let $x$ be the number of women chosen to be orchestra representatives.

Let $M_1,M_2,M_3,M_4$ represents the four men and $W_1,W_2$ represents the two women candidates for choice of orchestra representatives.

Therefore, the simple events are

  $\begin{array}{l}{M}_1{M}_2,M_1{M}_3,M_1{M}_4,M_1{W}_1,M_1{W}_2\\ M_2{M}_3,M_2{M}_4,M_2{W}_1,M_2{W}_2\\ M_3{M}_4,M_3{W}_1,M_3{W}_2\\ M_4{W}_1,M_4{W}_2\\ W_1{W}_2\end{array}$

Let $x$ represent $x$ are chosen as representatives in the table.

$x$	Associated simple events	$P\left(x\right)$
$0$	$M_1{M}_2,M_1{M}_3,M_1{M}_4,M_2{M}_3,M_2{M}_4,M_3{M}_4,$	$\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$15$}\right.=\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$
$1$	$M_1{W}_1,M_1{W}_2,M_2{W}_1,M_2{W}_2,M_3{W}_1,M_3{W}_2,M_4{W}_1,M_4{W}_2$	$\raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$15$}\right.$
$2$	$W_1{W}_2$	$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$15$}\right.$

b.

To determine

To find: the probability that both the orchestra representatives will be women.

Answer to Problem 4.135SE

  $P\left(\text{ women orchestra representatives }\right)=\frac{1}{15}$

Explanation of Solution

Given:

The board of directors of a major symphony orchestra has voted to create a committee for the purpose of handing employee complaints.

The committee consists of the president and the vice president of the symphony board and two orchestra representatives.

Calculation:

According to the solution of part a,

The total number of events is 15 and the sample space for women orchestra representatives is $W_1{W}_2$

Therefore, the probability that both the orchestra representatives will be women is:

  $P\left(\text{ women orchestra representatives }\right)=\frac{1}{15}$ .

c.

To determine

To find: the mean and variance for the random variable $x$ .

Answer to Problem 4.135SE

  $\mu =\frac{2}{3};{\sigma }^2=\frac{16}{45}$

Explanation of Solution

Given:

The board of directors of a major symphony orchestra has voted to create a committee for the purpose of handing employee complaints.

The committee consists of the president and the vice president of the symphony board and two orchestra representatives.

Calculation:

According to solution table in part a,

  $\begin{array}{l}P\left(\text{only for both the men}\right)=\frac{2}{5}\\ P\left(\text{both men and women}\right)=\frac{8}{15}\\ P\left(\text{only for both the women}\right)=\frac{1}{15}\end{array}$

Therefore, the mean and variance for the random variable $x$ is:

  $\begin{array}{l}\text{Mean, }\mu ={\displaystyle \sum _{x}P\left(x\right)}x\\ =\frac{2}{5}\left(0\right)+\frac{8}{15}\left(1\right)+\frac{1}{15}\left(2\right)\\ =\frac{8}{15}+\frac{2}{15}\\ =\frac{10}{15}=\frac{2}{3}\end{array}$

  $\begin{array}{l}\text{Variance, }{\sigma }^2={\displaystyle \sum _{x}P\left(x\right)}{x}^2-{\sigma }^2\\ =\frac{2}{5}{\left(0\right)}^2+\frac{8}{15}{\left(1\right)}^2+\frac{1}{15}{\left(2\right)}^2-{\left(\frac{2}{3}\right)}^2\\ =\frac{8}{15}+\frac{4}{15}-\frac{4}{9}\\ =\frac{12}{15}-\frac{4}{9}=\frac{16}{45}\end{array}$

If $A\text{ and }B$ are independent, then $P\left(A|B\right)=P\left(A\right)$ .

Hence, the mean and variance for the random variable $x$ is $\mu =\frac{2}{3};{\sigma }^2=\frac{16}{45}$