Chapter 15, Problem 41E

(a)

To determine

To find: the probability that the researcher is detained for further testing.

Answer to Problem 41E

20%

Explanation of Solution

Given:

  $\begin{array}{c}P\left(\text{driver is drunk}\right)=0.12\\ P\left(\text{police make right choice}\right)=0.80\end{array}$

The police are taking the right decision 80% of the time, it implies that 20% of the time they are taking the wrong decision. Therefore, the probability that detained even though you have not been drinking is 20%

(b)

To determine

To find: the probability that any mention driver would be detained.

Answer to Problem 41E

27.2%

Explanation of Solution

Given:

  $\begin{array}{c}P\left(\text{driver is drunk}\right)=0.12\\ P\left(\text{police make right choice}\right)=0.80\end{array}$

Calculation:

There are two events in which a driver may be detained. Either the driver is sober and the police take the wrong choice, or the driver is drunk and the police take the right choice. The two variables are independent, although the police take the right choice 80% of the time whether the driver is sober or not, and the drivers are drunk 12% of the time regardless of the police. Therefore,

P {driver is sober and police take wrong choice}

  $\begin{array}{c}=P\left(\text{driver is sober}\right)\times P\left(\text{police make wrong choice}\right)\\ =\left(100\%-12\%\right)\times \left(20\%\right)\\ =17.6\%\end{array}$

P (driver is drunk and police make right choice)

  $\begin{array}{c}=P\left(\text{driver is drunk}\right)\times P\left(\text{police make right choice}\right)\\ =\left(12\%\right)\times \left(80\%\right)\\ =9.6\%\end{array}$

Therefore, the probability that a driver is detained is 17.6% + 9.6% = 27.2%

(c)

To determine

To find: the probability that a driver who is detained has actually been drinking.

Answer to Problem 41E

35.29%

Explanation of Solution

Given:

  $\begin{array}{c}P\left(\text{driver is drunk}\right)=0.12\\ P\left(\text{police make right choice}\right)=0.80\end{array}$

Calculation:

Knowing that P (driver is drunk | driver is detained), that is, the probability that the driver is drunk mention that he is detained.

  $\begin{array}{c}=\frac{P\left(\text{Driver is drunk and driver is detained}\right)}{P\left(\text{driver is detained}\right)}\\ =\frac{9.6\%}{27.2\%}\\ =35.29\%\end{array}$

(d)

To determine

To find: the probability that a driver who was released had actually been drinking.

Answer to Problem 41E

3.30%

Explanation of Solution

Given:

  $\begin{array}{c}P\left(\text{driver is drunk}\right)=0.12\\ P\left(\text{police make right choice}\right)=0.80\end{array}$

Calculation:

P (driver is drunk and driver was released)

  $\begin{array}{c}=P\left(\text{Driver is drunk and police made wrong choice}\right)\\ =P\left(\text{driver is drunk}\right)\times P(\text{police made wrong choice})\\ =(12\%)\times (20\%)\\ =2.4\%\\ P\left(\text{driver was released}\right)\\ =\left(100\%-P\left(\text{driver was detained}\right)\right)\\ =100\%-27.2\%\\ =72.8\%\end{array}$

So P(driver is drunk | driver was released) =2.4% / 72.8% = 3.30%