Definition 4.1.1.
A probability is a numerical value between 0 and 1 that measures the chance, or likelihood, that a specific event will occur.
Section 4.1 Introduction to Probability
Decision makers, including those in the business world, are greatly influenced by uncertainty. Probability provides a valuable tool in quantifying uncertainty, leading to greater success in decision-making.
Before jumping into some examples, let’s discuss some key terms for our study of probability concepts.
Definition 4.1.2.
- An experiment results in a specific outcome e.g., a coin flip (the experiment) results in a “heads” or a “tails” (the outcome).
- An event is a collection of outcome(s) of an experiment, e.g., a die roll results in “an odd number”. More specifically, a simple event is an event with a single outcome.
-
The sample space for an experiment consists of all possible outcomes e.g., \(S=\{\text{heads},\text{tails}\}\text{.}\)
-
The outcomes must be collectively exhaustive and mutually exclusive.
- Collectively exhaustive means every simple event is included
- Mutually exclusive means events that cannot occur at the same time.
-
Exercise 4.1.3.
A single die is rolled.
(a)
Define the sample space.
Answer.
\begin{equation*}
S=\{1,2,3,4,5,6\}
\end{equation*}
(b)
Define the event “an odd number is rolled”. What is the probability of this event?
Answer.
\begin{equation*}
A=\{1,3,5\}\Rightarrow P(A)=\frac{3}{6}=0.5
\end{equation*}
Exercise 4.1.4.
A pair of dice are rolled.
(a)
Define the sample space. (How many total possible outcomes are there?)
Answer.
\begin{equation*}
S=\{(1,1),(1,2),(2,1),(2,2),(3,1),(3,2),(3,3),...\}
\end{equation*}
There are 36 total possible outcomes.
(b)
Define the event “doubles are rolled”. What is the probability of this event?
Answer.
\begin{equation*}
B=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}
\end{equation*}
\begin{equation*}
P(B)=\frac{6}{36}=\frac{1}{6}
\end{equation*}
(c)
Define the event “the sum of the dice is 10”. What is the probability of this event?
Answer.
\begin{equation*}
C=\{(4,6),(6,4),(5,5)\}
\end{equation*}
\begin{equation*}
P(C)=\frac{3}{36}=\frac{1}{12}
\end{equation*}
Exercise 4.1.5.
(Donnelly 4.8)
The table below shows the frequencies of executives’ salary ranges at a particular organization. Table 4.1.6.
| Salary Range | Frequency |
| Under $60,000 | 2 |
| $60,000 to under $70,000 | 5 |
| $70,000 to under $80,000 | 7 |
| $80,000 to under $90,000 | 4 |
| $90,000 to under $100,000 | 6 |
| $100,000 or more | ? |
| Total: | 26 |
(a)
How many executives at this organization earned $100,000 or more per year?
Answer.
\(26-24=2\)
(b)
What is the probability of randomly selecting an executive who earned $100,000 or more per year?
Answer.
\begin{equation*}
\frac{2}{26}=\frac{1}{13}
\end{equation*}
(c)
What is the probability of randomly selecting an executive who earned $60,000 or more per year?
Answer.
\begin{equation*}
\frac{24}{26}=\frac{12}{13}
\end{equation*}
(d)
What is the probability of randomly selecting an executive who earned between $70,000 and $90,000 per year?
Answer.
\begin{equation*}
\frac{11}{26}
\end{equation*}
Exercise 4.1.7.
Basic Properties of Probability:
- If \(P(A)=1\text{,}\) then event \(A\) is certain to occur.
- If \(P(A)=0\text{,}\) then event \(A\) is certain to NOT occur.
- The probability of any event must range between 0 and 1 (inclusive).
- The sum of all probabilities of simple events in a sample space is equal to 1.
Definition 4.1.8.
The complement, \(A'\) (or \(A^c\)), is defined as all outcomes in the sample space that are not in \(A\text{.}\)
The complement rule:
\begin{equation*}
P(A)+P(A')=1
\end{equation*}
Exercise 4.1.9.
Draw one card from a standard deck of 52 playing cards. (Here is a link to a description of a deck of cards:
https://intranet.missouriwestern.edu/cas/wp-content/uploads/sites/17/2020/05/Standard-Deck-of-Cards.pdf)(a)
How many total outcomes are in the sample space for this experiment?
Answer.
\(13\cdot 4=52\)
(b)
What is the probability of drawing a “face card”?
Answer.
\begin{equation*}
\frac{12}{52}=\frac{3}{13}
\end{equation*}
(c)
What is the probability of drawing “a card divisible by 3”?
Answer.
\begin{equation*}
\frac{12}{52}=\frac{3}{13}
\end{equation*}
(d)
What is the probability of drawing “a card NOT divisible by 3”?
Answer.
\begin{equation*}
1-\frac{12}{52}=\frac{40}{52}=\frac{10}{13}
\end{equation*}
(e)
What is the probability of drawing a card that is “a face card and a card divisible by 3”?
Answer.
0
(f)
What is the probability of drawing a card that is “a red card or a black card”?
Answer.
1
Definition 4.1.10.
- A contingency table shows the number of occurrences of events that are classified by two categorical variables.
- Marginal probability is a simple probability found in the margins, or row and column totals, of a contingency table.
Exercise 4.1.11.
(Donnelly 4.19)
A local car dealership currently has 36 used GM, Ford, and Toyota vehicles on the lot that can be classified as either cars or trucks. The following data are available:
- Twenty-six vehicles are cars
- Eleven vehicles are GMs
- Fifteen vehicles are Fords
- Three vehicles are both Toyotas and trucks
- Fourteen vehicles are both Fords and cars
(a)
Create a contingency table that summarizes the data.
| GM | Ford | Toyota | Total | |
|---|---|---|---|---|
| Car | ||||
| Truck | ||||
| Total |
Answer.
| GM | Ford | Toyota | Total | |
|---|---|---|---|---|
| Car | 5 | 14 | 7 | 26 |
| Truck | 6 | 1 | 3 | 10 |
| Total | 11 | 15 | 10 | 36 |
(b)
What is the probability that a randomly selected vehicle is a Toyota?
Answer.
\begin{equation*}
\frac{10}{36}\approx 0.278
\end{equation*}
(c)
What is the probability that a randomly selected vehicle is a truck?
Answer.
\begin{equation*}
\frac{10}{36}\approx 0.278
\end{equation*}
Definition 4.1.12.
The intersection of Events A and B represents the number of times when events A and B occur at the same time
- Notation: \(A\cap B\)
- Notation: \(A\cup B\)
Union and Intersection (Made in GeoGebra by Vaibhav Zade)
Link to GeoGebra:
https://www.geogebra.org/m/grxxwm4c
Definition 4.1.14.
The joint probability of two events is the probability of the intersection of two events.
The addition rule for probabilities is used to calculate the probability of the union of events. It depends on knowing whether or not two events are mutually exclusive.
- For mutually exclusive events,\begin{equation*} P(A\cup B)=P(A)+P(B) \end{equation*}
- For events that are not mutually exclusive,\begin{equation*} P(A\cup B)=P(A)+P(B)-P(A\cap B) \end{equation*}
Exercise 4.1.15.
Use the contingency table from Exercise 4.1.11 to answer the following questions.
| GM | Ford | Toyota | Total | |
|---|---|---|---|---|
| Car | 5 | 14 | 7 | 26 |
| Truck | 6 | 1 | 3 | 10 |
| Total | 11 | 15 | 10 | 36 |
(a)
What is the probability that a randomly selected vehicle is either a Ford or a car?
Answer.
\begin{equation*}
\frac{15}{36}+\frac{26}{36}-\frac{14}{36}=\frac{27}{36}=0.75
\end{equation*}
(b)
What is the probability that a randomly selected vehicle is a GM truck?
Answer.
\begin{equation*}
\frac{6}{36}\approx 0.167
\end{equation*}
Exercise 4.1.16.
Definition 4.1.17.
Conditional probability is the probability that event A will occur given that event B has occurred.
This probability is denoted by \(P(A|B)\text{,}\) and is “the probability of A given B”.
Exercise 4.1.19.
- \(\frac{P(A\cap B)}{P(B)}\)
- \(\frac{P(A\cap B)}{P(A)}\)
- \(\frac{P(A\cup B)}{P(A)}\)
- \(\frac{P(A\cup B)}{P(B)}\)
Which of the following is the formula used to calculate \(P(A|B)\text{?}\)
Definition 4.1.20.
- Prior probability: the probability that an event will occur as determined without any additional information that could affect it
- Posterior probability: a revision of a prior probability using additional information
Exercise 4.1.21.
Use the contingency table from Exercise 4.1.11 to answer the following questions.
| GM | Ford | Toyota | Total | |
|---|---|---|---|---|
| Car | 5 | 14 | 7 | 26 |
| Truck | 6 | 1 | 3 | 10 |
| Total | 11 | 15 | 10 | 36 |
(a)
What is the probability that a randomly selected vehicle is a Toyota, given it is a car?
Answer.
\begin{equation*}
P(\text{Toyota}|\text{car})=\frac{P(\text{Toyota and car})}{P(\text{car})}=\frac{7/36}{26/36}=\frac{7}{26}\approx 0.269
\end{equation*}
(b)
What is the probability that a randomly selected vehicle is a truck, given it is a Ford?
Answer.
\begin{equation*}
P(\text{Truck}|\text{Ford})=\frac{P(\text{Truck and Ford})}{P(\text{Ford})}=\frac{1/36}{15/36}=\frac{1}{15}\approx 0.067
\end{equation*}
