Probability, Area, and Equally Likely Outcomes

Solution

The sample space in this problem is \(S=\{\text{{blue}}, \text{{red}}, \text{{green}}, \text{{yellow}}\}\).  Since there are four colors, we know that \(n(S)=4\). 

Let \(E\) be the event "land on yellow."  Since there is only one yellow region, we know that \(n(\text{{yellow}})=1\).

To compute the probability, we use the theoretical formula: \[P(\text{{yellow}})=\dfrac{n(\text{{yellow}})}{n(S)}=\dfrac{{1}}{{4}}=0.25\]

Note

Notice that the probability is the same as the proportion of area that is yellow on the spinner.  This is not an accident.  There is a deep connection between area and probability.  We'll exploit that relationship in the next example, and explain it in more depth in Chapter 6.  But we have already seen this connection between areas and proportions already: the empirical rule!

$$\tag*{\(\blacksquare\)}$$

Solution

The sample space in this problem is still \(S=\{\text{{blue}}, \text{{red}}, \text{{green}}, \text{{yellow}}\}\).  But since all the regions aren't the same size, we cannot use the theoretical probability formula.  In this case, we appeal to area.  As I mentioned the last example, probability and area are deeply connected.  In this case, the probability is just the area of the spinner that is yellow: \[P(\text{{yellow}})=\dfrac{{1}}{{2}}.\]

How do we reconcile this with theoretical probability?  Notice that the smallest region is 1/8 of the spinner.  So let's subdivide the spinner so there are 8 equal subdivisions of the spinner:

Since every region is equally likely to occur now, the theoretical formula applies.  In this case, since there are 8 regions, the sample space is now 

S = { Region 1, Region 2, Region 3, Region 4,
          Region 5, Region 6, Region 7, Region 8 }
    

and we have that \(n(S)=8\).  Let \(E\) be the event "land on a yellow region." This means that \(n(\text{{yellow}})=4\) since there are 4 yellow regions.  Now, we can calculate the probability: \[P(\text{yellow})=\dfrac{n(\text{yellow})}{n(S)}=\dfrac{4}{8}=\dfrac{1}{2}=0.5\]

Notice we got the probability as before!

$$\tag*{\(\blacksquare\)}$$