2.1: Applications

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At this point in time, we have gone though a lengthy discussion of definitions, axioms, theorems and proofs. With these tools we have developed, we are now in a position to answer some basic questions concerning probability. However, although these questions may be basic, do not underestimate their importance! Our first example is a sample question from The Society of Actuaries' Exam P. (For those unfamiliar with actuarial science, an excellent resource is https://www.beanactuary.org/why-actuarial-science/. Be sure to click on the salary tab!)

Caution

Please do not be overwhelmed by the details in our answer to the next question! There are a lot of moving parts and with practice, we will be able to answer these questions quite concisely.

Example $\PageIndex{1}$

A survey of a group's viewing habits over the last year revealed the following information:

28% watched gymnastics
29% watched baseball
19% watched soccer
14% watched gymnastics and baseball
12% watched baseball and soccer
10% watched gymnastics and soccer
8% watched all three sports.

Calculate the percentage of the group that watched none of the three sports during the last year.

Source: https://www.soa.org/globalassets/assets/files/edu/edu-exam-p-sample-quest.pdf

Solution 1

Since this is our first example, we will discuss everything in great detail. Allow us to make a few comments. When reading the above question, we notice that there is an immediate disconnect between the theory we have been learning and the way that the question is phrased. You see, at this point in time, our framework of probability is phrased in terms of sets and events. In the above question, there are no sets mentioned to us! And so the first part of this problem is to translate the word problem into a question about the probability of an event. How do we do this?

Whenever the question mentions a percentage or a probability, that is the probability of some event! And so we should introduce notation. In the above problem, let $ E_G = \{ \text{a person watched gymnastics} \} $, let $ E_B = \{ \text{a person watched baseball} \} $ and let $ E_S = \{ \text{a person watched soccer} \} $.

In order to translate the above percentages in terms of events, there is one last remark to make. This remark is that the English word of "and" corresponds to the set theoretic operation of intersection while the word "or" corresponds to the set theoretic operation of union. Thus, when the problem states "14% watched gymnastics and baseball" this means that the event $ \{ \text{watched gymnastics and baseball} \} $ can be written as $ E_G \cap E_B $.

Thus we have the following:

Given	Translation
28% watched gymnastics	$ P(E_G) = 0.28 $
29% watched baseball	$ P(E_B) = 0.29 $
19% watched soccer	$ P(E_S) = 0.19 $
14% watched gymnastics and baseball	$ P(E_G \cap E_B) = 0.14 $
12% watched baseball and soccer	$ P(E_B \cap E_S) = 0.12 $
10% watched gymnastics and soccer	$ P(E_G \cap E_S) = 0.10 $
8% watched all three sports.	$ P(E_G \cap E_B \cap E_S) = 0.08 $

We are asked to find the percentage of the group that watched none of the three sports during the last year. Again, we must translate what the question is asking us to be in terms of set. When the question says none of the three sports this means the person did NOT watch gymnastics AND the person did NOT watch baseball AND the person did NOT watch soccer.

This means we are asked to find $ P(E_G^c \cap E_B^c \cap E_S^c) $. This is now a question that we can answer with the framework we have developed. Let us study the expression $ P(E_G^c \cap E_B^c \cap E_S^c) $ and see if we can relate this to any relevant theorems.

A close inspection of the expression leads us to think of the DeMorgan's Laws. By the General DeMorgan's Laws, we know that

\[ P(E_G^c \cap E_B^c \cap E_S^c) = P \big( (E_G \cup E_B \cup E_S)^c \big) \nonumber\ \]

We see that we have the probability of a complement and so perhaps The Probability of a Complement Rule comes to mind. Piecing everything together yields the following solution:

\begin{align*}
P(E_G^c \cap E_B^c \cap E_S^c) &= P \big( (E_G \cup E_B \cup E_S)^c \big) \\
&= 1 - P(\underbrace{E_G \cup E_B \cup E_S}_{\text{prob. of union of events}}) \\
&= 1 - \big( P(E_G) + P(E_B) + P(E_S) - P(E_G \cap E_B) - P(E_B \cap E_S) - P(E_G \cap E_S) + P(E_G \cap E_B \cap E_S) \big) \\
&= 1 - (0.28 + 0.29 + 0.19 - 0.14 - 0.12 - 0.10 +0.08 ) \\
&= 1 - 0.48 \\[6pt]
&= 0.52
\end{align*}

Allow us to summarize four important remarks we made in our above solution.

Remark 1

To tackle a word problem, we often have to introduce suitable notation. Whenever the question states a percentage or probability, there is some set/event being described. Thus, we should introduce some capital letter to denote that event. Example: We let $ E_G = \{ \text{a person watched gymnastics} \} $.

Remark 2

The word "or" corresponds to the set theoretic operation of union and the word "and" corresponds to the set theoretic operation of intersection. Example: 14% watched gymnastics and baseball means $ P(E_G \cap E_B) = 0.14 $.

Remark 3

In rephrasing an event, we should always ask ourselves "can we describe this event by using the words or/and? Example: when the question says none of the three sports this means the person did NOT watch gymnastics AND the person did NOT watch baseball AND the person did NOT watch soccer.

Remark 4

Similar to Remark 1, translate the requested probability in terms of your notation. Example: When the question says calculate the percentage of the group that watched none of the three sports this means to find $ P(E_G^c \cap E_B^c \cap E_S^c) $.

The solution discussed above required us to have a strong command of our theory. However, allow us to discuss the same problem from another perspective.

Example 1 (Revisited) $\PageIndex{1}$

A survey of a group's viewing habits over the last year revealed the following information:

28% watched gymnastics
29% watched baseball
19% watched soccer
14% watched gymnastics and baseball
12% watched baseball and soccer
10% watched gymnastics and soccer
8% watched all three sports.

Calculate the percentage of the group that watched none of the three sports during the last year.

Solution 2

This time, we will study the problem via a Venn diagram:

$myImage.png$

The key to working with a Venn diagram is to work inside-out. What does this mean?

Inside-out means we first start inside the Venn diagram, specifically the "inner most" level. That is, we first start in the intersection of all three events. According to the problem, 8% watched all three sports and so $ P(E_G \cap E_B \cap E_S) = 0.08 $. Thus, we have:

$SOA2.png$

We will now work "one level" outside. According to the problem, 14% watched gymnastics and baseball meaning $ P(E_G \cap E_B) = 0.14 $. Since the intersection between $E_G $ and $E_B $ must total 14%, and we have already accounted for 8%, then the remaining portion of the intersection must account for 6%.

$SOA3.png$

By a similar logic applied to the statements "12% watched baseball and soccer" and "10% watched gymnastics and soccer", we obtain the following:

$SOA4.png$

Informally speaking, we now move out another level in an attempt to find the remaining numbers in the regions that are marked off:

$SOA5.png$

Let us first deal with the region $E_G$. We note that $ P(E_G $ = 28\% \). So far, we have accounted for $ 6\% + 8\% + 2\% = 16\% $. Hence the remaining portion must account for $ 12\% $.

By a similar argument concerning the regions $E_B$ and $E_S$, we obtain the following:

$SOA_2ndLast.png$

Finally, we note that by Axiom 2, $ P(S) = 1 $ and so our entire diagram is to total to 100%. At this point in time, we see from the above diagram that we have accounted for 48%. Hence, 52% must be on the "outside" as shown below:

$SOALast.png$

Thus, none of the three sports corresponds to 52% which is the same solution as above.

Here are two exercises for us to try which will test our understanding of the above concepts:

Exercise $\PageIndex{2}$

Based off your viewing habits, Netflix determines that there is a 30% chance you will like the show Stranger Things, a 40% chance you will like Bridgerton and a 25% chance you will like both shows. What is the probability that you would like at least one show? (Try to answer this question formally via the theory we discussed and then informally via a Venn diagram).


Answer: 45%

Exercise $\PageIndex{3}$

According to Blockbuster,

10% of viewers like horror
30% of viewers like comedy
5% of viewers like romance
8% like horror and comedy
2% like horror and romance
4% like comedy and romance
1% like all three genres

What is the probability that a person likes at least one of the above genres? (Try to answer this question formally via the theory we discussed and then informally via a Venn diagram).


Answer: 32%

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Solution 1

Solution 2

Given	Translation
28% watched gymnastics	\( P(E_G) = 0.28 \)
29% watched baseball	\( P(E_B) = 0.29 \)
19% watched soccer	\( P(E_S) = 0.19 \)
14% watched gymnastics and baseball	\( P(E_G \cap E_B) = 0.14 \)
12% watched baseball and soccer	\( P(E_B \cap E_S) = 0.12 \)
10% watched gymnastics and soccer	\( P(E_G \cap E_S) = 0.10 \)
8% watched all three sports.	\( P(E_G \cap E_B \cap E_S) = 0.08 \)

Caution

Example \(\PageIndex{1}\)

Solution 1

Remark 1

Remark 2

Remark 3

Remark 4

Example 1 (Revisited) \(\PageIndex{1}\)

Solution 2

Exercise \(\PageIndex{2}\)

Exercise \(\PageIndex{3}\)