4.3: Addition Rules for Probabilities
- Page ID
- 48915
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Understand the two addition rules for probability to calculate the chance of either one event or another occurring.
- Apply the addition rule for mutually exclusive events by summing their probabilities.
- Apply the addition rule for events that can occur together by adding their probabilities and subtracting the overlap to avoid double-counting.
When computing the probability that at least one of several events occurs, addition rules can be applied to simplify the calculation. For example, if a card is selected from a standard deck, what is the chance that the card is a jack or a queen? This probability can be computed rapidly by computing the probability of getting a jack and a queen separately and then adding the two probabilities. This process can be done because the events are mutually exclusive. In other words, the events have no outcomes in common.
Mutually exclusive events (disjoint) are events with no outcomes in common.
- Roll a single die. The events are to get a 2 or a 5.
- Select a single card. The event is to get a jack or a queen.
- A room has 5 math majors, 12 chemistry majors, 6 biology majors, and 4 physics majors. A single person is selected, and no person is double-majoring. The person is a math major or a biology major.
Solution
None of the events has outcomes in common.
Addition Rule I
For two events A and B that are mutually exclusive, the probability that A or B will occur is P(A or B) = P(A) + P(B).
Use the addition rule I to compute the following probabilities. Write the final answers as fractions or decimals rounded to three places.
- Roll a single die. What is the chance of getting a 2 or a 5?
- Select a single card. What is the chance the card is a jack or a queen?
- A room has 5 math majors, 12 chemistry majors, 6 biology majors, and 4 physics majors. A single person is selected, and no person is double-majoring. What is the chance that the person is a math or biology major?
Solutions
- P( 2 or 5) = \(\dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6} = \dfrac{1}{3} = 0.333\)
- P(ace or king) = \(\dfrac{4}{52} + \dfrac{4}{52} = \dfrac{8}{52} = \dfrac{2}{13} = 0.154\)
- P(math major or biology major) = \(\dfrac{5}{27} + \dfrac{6}{27} = \dfrac{11}{27} = 0.407\)
Two events are not mutually exclusive if they can happen at the same time, meaning they have some outcomes in common.
When two or more events have outcomes in common, they are labeled non-mutually exclusive events. These types of events have one or more outcomes in common. Care must be taken not to over-count these outcomes. For example, suppose a card is selected from a deck. What is the chance that the card is a jack or a black card? If the addition rule 1 is applied, this would result in the wrong answer. The reason is that two blackjacks are being over-counted. To correctly compute the probability, these extra two blackjacks must be subtracted from the total number of outcomes. This process is outlined in Addition Rule II.
Addition Rule II
For two events A and B that are not mutually exclusive, the probability that A or B will occur is P(A or B) = P(A) + P(B) – P(A and B)
Use the addition rule II to compute the following probabilities. Write the final answers as fractions or decimals rounded to three place places.
- Select a single card. What is the chance the card is a jack or a black card?
- Select a single card. What is the chance the card is a king or a diamond?
- A room has 5 math majors, 12 chemistry majors, 6 biology majors, and 4 physics majors. In this group, 2 people are double-majoring in math and physics. A single person. What is the chance that the person is a math or physics major?
Solutions
- P( Jack or black) = \(\dfrac{4}{52} + \dfrac{26}{52} - \dfrac{2}{52} = \dfrac{28}{52} = \dfrac{7}{13} = 0.538\)
- P(king or diamond) = \(\dfrac{4}{52} + \dfrac{13}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13} = 0.308\)
- P(math major or physics major) = \(\dfrac{5}{27} + \dfrac{4}{27} - \dfrac{2}{27} = \dfrac{7}{27} = 0.259\)
Sometimes data is presented in a tabular form. This type of table is called a contingency table, and the table is used to show a relationship between two data types. The data are listed in rows and columns. We can treat the contingency table as two sets where the column data types represent one set, and the row data types represent another data type. To compute probabilities using a contingency table, the first step is to find the sum of each row and column. For example, students are categorized into those who don’t work, work part-time, and work full-time, along with the modality of the class they are taking for Math 165: regular, with support, hybrid, or online. The contingency table is displayed below for 120 randomly selected students.
| Regular | With Support | Hybrid | Online | Total | |
|---|---|---|---|---|---|
| Does not work | 15 | 14 | 5 | 3 | 37 |
| Works part-time | 9 | 16 | 10 | 5 | 40 |
| Works full-time | 7 | 8 | 13 | 15 | 43 |
| Total | 31 | 38 | 28 | 23 | 120 |
Table \(\PageIndex{1}\): Contingency Table Involving Modality of Classes and Work Status
Using the contingency table above, compute the probability for one randomly selected student. Write the final answers as fractions or decimals rounded to three places.
- The student works full-time or takes a class online.
- The student is part-time and takes a hybrid class.
- The student does not work or take a class with support.
Solutions
- P( works full-time or takes an online class) = \(\dfrac{43}{120} + \dfrac{23}{120} - \dfrac{15}{120} = \dfrac{51}{120} = 0.425\)
- P( part-time and takes a hybrid class) = \(\dfrac{10}{120} = 0.083\)
- P( Does not work or takes a class with support) =\(\dfrac{37}{120} + \dfrac{38}{120} - \dfrac{14}{120} = \dfrac{61}{120} = 0.508\)
Exercises
A single fair six-sided die is rolled.
- Event A: Rolling a 2
- Event B: Rolling a 5
Since a die can only land on one number at a time, these events are mutually exclusive.
What is the probability of rolling a 2 or a 5?
- Answer
-
These events are mutually exclusive; thus, we use the addition rule I.
P(2 or 5) = \(\dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6} = \dfrac{1}{3}\)
A psychology researcher is studying students’ preferred study methods. In a group of students:
- 35% prefer studying alone
- 25% prefer studying in a group
Assume that no student prefers both methods at the same time. What is the probability that a randomly selected student prefers studying alone OR in a group?
- Answer
-
These events are mutually exclusive; thus, we use the addition rule I.
P (studying alone or in a group) = P (Studying alone) + P (Studying in a group) = 0.35 + 0.25 = 0.6 = 60%
At a college, students in the humanities division were surveyed about their interests:
- 55% of students are interested in literature.
- 40% are interested in history.
- 20% are interested in both literature and history.
What is the probability that a randomly selected student is interested in literature or history?
- Answer
-
These events are mutually exclusive; thus, we use the addition rule II.
P (literature or history) = P (literature) + P (history) – P (literature and history) = 0.55 + 0.40 – 0.20 = 0.75 = 75%
A political science professor surveys 200 students about their interests in public policy and international relations.
The results are shown in the table below:
| International Relations | Not International Relations | Total | |
|---|---|---|---|
| Public Policy | 50 | 70 | 120 |
| Not Public Policy | 30 | 50 | 80 |
| Total | 80 | 120 | 200 |
What is the probability that a randomly selected student is interested in public policy or international relations?
- Answer
-
These events are mutually exclusive; thus, we use the addition rule II.
P (public policy or international relations) = P (public policy) + P (international relations) – P (public policy and international relations) = \(\dfrac{120}{200}+\dfrac{80}{200}-\dfrac{50}{200}\)
P (public policy or international relations) =\(\dfrac{180}{200}=\dfrac{9}{10}\)
A standard deck of 52 playing cards is used. The cards are categorized by whether they are a heart and whether they are a face card (jack, queen, king).
The table below summarizes the counts:
| Face Card | Not a Face Card | Total | |
|---|---|---|---|
| Heart | 3 | 10 | 13 |
| Not a Heart | 9 | 30 | 39 |
| Total | 12 | 40 | 52 |
What is the probability that a randomly selected card is a heart or a face card?
- Answer
-
These events are mutually exclusive; thus, we use the addition rule II.
P (heart or face card) = P (heart) + P (face card) – P (heart and face card) = \(\dfrac{13}{52}+\dfrac{12}{52}-\dfrac{3}{52}\)
P (public policy or international relations) =\(\dfrac{22}{52}=\dfrac{11}{26}\)