Chapter 16, Problem 22E

(a)

To determine

To Explain: that two contracts are independent.

Explanation of Solution

Given:

  $\begin{array}{c}P\left(O\right)=0.8\\ P\left(T\right)=1-0.8\\ =0.2\\ P\left(T|O\right)=0.2\\ P\left(T|O\text{'}\right)=0.2\\ P\left(T|O\text{'}\right)=0.3\end{array}$

Calculation:

O = getting the contract 1

T = getting the contract 2

  

Bothe the contracts are not independent

  $\begin{array}{c}P\left(T\right)=P\left(O\cap T\right)+P\left(O\text{'}\cap T\text{'}\right)\\ =0.16+0.06\\ =0.22\\ P\left(T|O\right)=0.2\end{array}$

Although $P\left(T\right)\ne P\left(T|O\right)$ both the contracts are not independent

(b)

To determine

To find: the probability for getting both contracts.

Answer to Problem 22E

16%

Explanation of Solution

Given:

  $\begin{array}{c}P\left(O\right)=0.8\\ P\left(T\right)=1-0.8\\ =0.2\\ P\left(T|O\right)=0.2\\ P\left(T|O\text{'}\right)=0.2\\ P\left(T|O\text{'}\right)=0.3\end{array}$

Calculation:

O = getting the contract 1

T = getting the contract 2

  

probability for getting both the contracts is

  $\begin{array}{c}P\left(T\cap O\right)=\left(0.2\right)\left(0.8\right)\\ =0.16\end{array}$

Therefore, there is 16% possibility that get both contracts

(c)

To determine

To find: the probability for getting no contracts.

Answer to Problem 22E

14%

Explanation of Solution

Given:

  $\begin{array}{c}P\left(O\right)=0.8\\ P\left(T\right)=1-0.8\\ =0.2\\ P\left(T|O\right)=0.2\\ P\left(T|O\text{'}\right)=0.2\\ P\left(T|O\text{'}\right)=0.3\end{array}$

Calculation:

O = getting the contract 1

T = getting the contract 2

  

probability that both get no contract is

  $\begin{array}{c}P\left(O\text{'}\cap T\text{'}\right)=\left(0.2\right)\left(0.7\right)\\ =0.14\end{array}$

Therefore, there is 14% possibility that get no contract.

(d)

To determine

To find: the probability model for X.

Explanation of Solution

Given:

  $\begin{array}{c}P\left(O\right)=0.8\\ P\left(T\right)=1-0.8\\ =0.2\\ P\left(T|O\right)=0.2\\ P\left(T|O\text{'}\right)=0.2\\ P\left(T|O\text{'}\right)=0.3\end{array}$

Calculation:

O = getting the contract 1

T = getting the contract 2

  

X is the number of contracts

the probability models

X	0   $\left(O\text{'}\cap T\text{'}\right)$	1   $\left(O\text{'}\cap T\right)\text{or}\left(O\cap T\text{'}\right)$	2   $\left(O\cap T\right)$
P(X)	0.14	0.06 + 0.64 = 0.7	0.16

(e)

To determine

To find: the expected value and standard deviation.

Answer to Problem 22E

The expected and standard deviation is 0.39 and 0.83 respectively.

Explanation of Solution

Given:

  $\begin{array}{c}P\left(O\right)=0.8\\ P\left(T\right)=1-0.8\\ =0.2\\ P\left(T|O\right)=0.2\\ P\left(T|O\text{'}\right)=0.2\\ P\left(T|O\text{'}\right)=0.3\end{array}$

Formula used:

  $\begin{array}{l}E\left(X\right)={\displaystyle \sum xP\left(X=x\right)}\\ \text{Var}\left(X\right)={\displaystyle \sum {(x-\mu )}^{2}P\left(x\right)}\\ SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\end{array}$

Calculation:

O = getting the contract 1

T = getting the contract 2

  

The expected value is

  $\begin{array}{c}\mu =E\left(X\right)\\ ={\displaystyle \sum xP\left(x\right)}\\ =0\left(0.14\right)+1\left(0.7\right)+2\left(0.16\right)\\ =0.7+0.32\\ =0.39\end{array}$

Finding the variance of X.

  $\begin{array}{c}\mathrm{var}\left(X\right)={\displaystyle \sum {\left(x-\mu \right)}^{2}P\left(x\right)}\\ ={\left(0-0.39\right)}^2\left(0.14\right)+{\left(1-0.39\right)}^2\left(0.7\right)+{\left(2-0.39\right)}^2\left(0.16\right)\\ =0.1521\left(0.14\right)+0.3721\left(0.7\right)+2.5921\left(0.16\right)\\ =0.021+0.26+0.41\\ =0.69\end{array}$

  $\begin{array}{c}SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\\ =\sqrt{0.69}\\ =0.83\end{array}$