3.6 Conditional Probability

Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:

1. What is the probability that a randomly selected person is a cell phone user given that they had no speeding violations in the last year?
2. If a randomly selected person does not have a cell phone, what is the probability they had a speeding violation last year?
3. What is the probability that someone with a cell phone did not have a speeding violation last year?

Solution:

1. The given event is “no speeding violations,” so we restrict the table to just the column involving “no speeding violations.”  With this restriction, the table would look like this:
   

   	No speeding violation in the last year
   Cell phone user	280
   Not a cell phone user	405
   Total	685

   Now, we want to find the probability a person is a cell phone user in this restricted sample space:

   [latex]\begin{eqnarray*}  P(\mbox{cell phone}|\mbox{no violations}) & = & \frac{\mbox{number of cell phone users in restricted sample space}}{\mbox{total number in restricted sample space}} \\ & = & \frac{280}{685} \\ \\ \end{eqnarray*}[/latex]

2. The given event is “no cell phone,” so we restrict the table to just the row involving “no cell phone.”  With this restriction, the table would look like this:
   

   	Speeding violation in the last year	No speeding violation in the last year	Total
   Not a cell phone user	45	405	450

   Now, we want to find the probability a person has a speeding violation in the last year in this restricted sample space:

   [latex]\begin{eqnarray*}  P(\mbox{violation}|\mbox{no cell phone}) & = & \frac{\mbox{number of violations in restricted sample space}}{\mbox{total number in restricted sample space}} \\ & = & \frac{45}{450} \\ \\ \end{eqnarray*}[/latex]

3. The given event is “cell phone,” so we restrict the table to just the row involving “cell phone.”  With this restriction, the table would look like this:
   

   	Speeding violation in the last year	No speeding violation in the last year	Total
   Cell phone user	25	280	305

   Now, we want to find the probability a person does not have a speeding violation in the last year in this restricted sample space:

   [latex]\\ \begin{eqnarray*} P(\mbox{no violations}|\mbox{cell phone}) & = & \frac{\mbox{number with no violations in restricted sample space}}{\mbox{total number in restricted sample space}} \\ & = & \frac{280}{305} \end{eqnarray*}[/latex]

At a local language school, 40% of the students are learning Spanish, 20% of the students are learning German and 8% of the students are learning both Spanish and German.

1. What is the probability that a randomly selected student is learning Spanish given that they are learning German?
2. What is the probability that a randomly selected Spanish student is learning German?

Solution:

1. [latex]\displaystyle{P(\mbox{Spanish}|\mbox{German})  =   \frac{P(\mbox{Spanish and German})}{P(\mbox{German})}  =  \frac{0.08}{0.2} =  0.4}[/latex]
2. [latex]\displaystyle{P(\mbox{German}|\mbox{Spanish}) = \frac{P(\mbox{Spanish and German})}{P(\mbox{Spanish})}= \frac{0.08}{0.4} = 0.2}[/latex]

There are 50 students enrolled in the second year of a business degree program.  During this semester, the students have to take some elective courses.  18 students decide to take an elective in psychology, 27 students decide to take an elective in philosophy, and 10 students decide to take an elective in both psychology and philosophy.

1. What is the probability that a student takes an elective in psychology given that they take an elective in philosophy?
2. If a student takes an elective in psychology, what is the probability that they take an elective in philosophy?

Solution:

1. [latex]\displaystyle{P(\mbox{psychology}|\mbox{philosohpy})  =  \frac{P(\mbox{psychology and philosophy})}{P(\mbox{philosophy})} = \frac{\frac{10}{50}}{\frac{27}{50}} = 0.3704}[/latex]
2. [latex]\displaystyle{P(\mbox{philosophy}|\mbox{psychology}) = \frac{P(\mbox{psychology and philosophy})}{P(\mbox{philosophy})} = \frac{\frac{10}{50}}{\frac{18}{50}} = 0.5556}[/latex]

At a local language school, 40% of the students are learning Spanish, 20% of the students are learning German and 8% of the students are learning both Spanish and German.  Are the events “Spanish” and “German” independent?  Explain.

Solution:

To check for independence, we need to check two probabilities:  [latex]P(\mbox{Spanish})[/latex] and [latex]P(\mbox{Spanish}|\mbox{German})[/latex].  If these probabilities are equal, the events are independent.  If the probabilities are not equal, the events are dependent.

From the information provided in the question, [latex]P(\mbox{Spanish})=0.4[/latex].  Previously, we calculated [latex]P(\mbox{Spanish}|\mbox{German})[/latex]:

[latex]\begin{eqnarray*} \\ P(\mbox{Spanish}|\mbox{German}) & = &  \frac{P(\mbox{Spanish and German})}{P(\mbox{German})}  \\ & = &  \frac{0.08}{0.2} \\ & = &0.4 \\ \\ \end{eqnarray*}[/latex]

We can see that [latex]P(\mbox{Spanish})=P(\mbox{Spanish}|\mbox{German})[/latex].  Because these two probabilities are equal, the events “Spanish” and “German” are independent.  This means that the probability a student is taking Spanish does not affect the probability a student is taking German.

Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:

Are the events “cell phone user” and “speeding violation in the last year” independent?  Explain.

Solution:

To check for independence, we need to check two probabilities:  [latex]P(\mbox{cell phone})[/latex] and [latex]P(\mbox{cell phone}|\mbox{speeding violation})[/latex].  If these probabilities are equal, the events are independent.  If the probabilities are not equal, the events are dependent.

[latex]\begin{eqnarray*} \\ P(\mbox{cell phone}) & = & \frac{305}{755} \\ & = & 0.4040 \\  \\ P(\mbox{cell phone}|\mbox{speeding violation})  & = & \frac{25}{70} \\ & = & 0.03571 \\  \\ \end{eqnarray*}[/latex]

We can see that [latex]P(\mbox{cell phone}) \neq P(\mbox{cell phone}|\mbox{speeding violation})[/latex].  Because these probabilities are not equal, the events “cell phone user” and “speeding violation” are dependent.  This means that the probability a person is a cell phone user does affect the probability the person had a speeding violation in the last year.