Chapter 7.3, Problem 7.65E

(a)

To determine

To construct: a tree diagram to represent this situation.

Explanation of Solution

Given:

  $\begin{array}{l}{\text{A}}_{\text{1}}\text{=Regular gas}\\ {\text{A}}_2\text{=Midgrade gas}\\ {\text{A}}_3\text{=Premium gas}\\ \text{B = Paid with credit card}\\ {\text{B}}^{\text{c}}\text{= not paid with credit card}\end{array}$

  $\begin{array}{l}\text{P}\left({\text{A}}_{\text{1}}\right)=0.88\\ \text{P}\left({\text{A}}_2\right)=0.02\\ \text{P}\left({\text{A}}_3\right)=0.1\\ \text{P}\left(B|{\text{A}}_{\text{1}}\right)=0.28\\ \text{P}\left(B|{\text{A}}_2\right)=0.34\\ \text{P}\left(B|{\text{A}}_3\right)=0.42\end{array}$

Graph:

Tree diagram:-

(b)

To determine

To Calculate: the probability that the customer paid with a credit card.

Answer to Problem 7.65E

0.2952

Explanation of Solution

Calculation:

General multiplication rule

  $\begin{array}{c}P\left(B\cap {\text{A}}_{\text{1}}\right)=\text{P}\left({\text{A}}_{\text{1}}\right)\times \text{P}\left(B|{\text{A}}_{\text{1}}\right)\\ =0.88\times 0.28\\ =0.2464\\ P\left(B\cap {\text{A}}_2\right)=\text{P}\left({\text{A}}_2\right)\times \text{P}\left(B|{\text{A}}_2\right)\\ =0.02\times 0.34\\ =0.0068\end{array}$

  $\begin{array}{c}P\left(B\cap {\text{A}}_3\right)=\text{P}\left({\text{A}}_3\right)\times \text{P}\left(B|{\text{A}}_3\right)\\ =0.10\times 0.42\\ =0.042\end{array}$

Addition rule for mutually exclusive event

  $\begin{array}{c}P\left(B\right)=P\left(B\cap {\text{A}}_{\text{1}}\right)+P\left(B\cap {\text{A}}_2\right)+P\left(B\cap {\text{A}}_3\right)\\ =0.2464+0.0068+0.042\\ =0.2952\end{array}$

(c)

To determine

To Calculate: the probability that she or he bought premium gas.

Answer to Problem 7.65E

0.1423

Explanation of Solution

Formula used:

Conditional Probability

  $P\left({\text{A}}_3|B\right)=\frac{P\left(B\cap {\text{A}}_3\right)}{P\left(B\right)}$

Calculation:

  $\begin{array}{c}P\left({\text{A}}_3|B\right)=\frac{P\left(B\cap {\text{A}}_3\right)}{P\left(B\right)}\\ =\frac{0.042}{0.2952}\\ =0.1423\end{array}$