3.2: The Addition Rules of Probability
- Page ID
- 25652
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Your favorite professional basketball team either won or lost their last game. Winning and losing are mutually exclusive events.
\(\text{A}\) and \(\text{B}\) are mutually exclusive events (or disjoint events) if they cannot occur at the same time. This means that \(\text{A}\) and \(\text{B}\) do not share any outcomes and \(P(\text{A AND B}) = 0\).
For example, suppose the sample space
\[S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}. \nonumber\]
Let \(\text{X} = \{1, 2, 3, 4, 5\}, \text{Y} = \{4, 5, 6, 7, 8\}\), and \(\text{Z} = \{7, 9\}\). Events X and Y both have 4 and 5. Thus, \(\text{X AND Y} = \{4, 5\}\).
Because there are two shared outcomes from the sample space \(S\), the probability of X AND Y is \[P(\text{X AND Y}) = \dfrac{2}{10} \nonumber\] .
Since \(P(\text{X AND Y})\) is not equal to zero, \(\text{X}\) and \(\text{Y}\) are not mutually exclusive.
However, events \(\text{X}\) and \(\text{Z}\) have no outcomes (numbers) in common. So, \(P(\text{X AND Z}) = \frac{0}{10}=0\). Therefore, \(\text{X}\) and \(\text{Y}\) are mutually exclusive.
The probability of two mutually exclusive events \(A\) OR \(B\) (two events that share no outcomes) is
\(P(A \text{ OR } B ) = P(A) + P(B) \)
The probability of two non-mutually exclusive events \(A\) OR \(B\) (two events that share outcomes) is
\(P(A \text{ OR } B ) = P(A) + P(B) - P(A \text{ AND } B) \)
Using the example from above, where the sample space \(S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\) and events \(\text{X} = \{1, 2, 3, 4, 5\}, \text{Y} = \{4, 5, 6, 7, 8\}\), and \(\text{Z} = \{7, 9\}\).
Since events \(\text{X}\) and \(\text{Z}\) are mutually exclusive then the probability of \(X\) OR \(Z\)
\[ P(X \text{ OR } Z)=P(X) + P(Z) = \frac{5}{10} + \frac{2}{10} = \frac{7}{10}.\]
Since events \(\text{X}\) and \(\text{Y}\) are not mutually exclusive then the probability of \(X\) OR \(Y\)
\[ P(X \text{ OR } Y)=P(X) + P(Y) - P(X \text{ AND } Y) = \frac{5}{10} + \frac{5}{10} - \frac{2}{10} = \frac{8}{10}.\]
Below we will see our first contingency table, which is a table with categories in both the horizontal and vertical direction. As you see below, the contingency table describes cell phone users versus speeding violations.
Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:
Speeding violation in the last year | No speeding violation in the last year | Total | |
---|---|---|---|
Cell phone user | 25 | 280 | 305 |
Not a cell phone user | 45 | 405 | 450 |
Total | 70 | 685 | 755 |
The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755.
Calculate the following probabilities using the table.
- Find \(P(\text{Person is a cell phone user})\).
- Find \(P(\text{person had no violation in the last year})\).
- Find \(P(\text{Person had no violation in the last year AND was a cell phone user})\).
- Find \(P(\text{Person is a cell phone user OR person had no violation in the last year})\).
Answer
- \(\dfrac{\text{number of cell phone users}}{\text{total number in study}}\) = \(\dfrac{305}{755}\)
- \(\dfrac{\text{number that had no violation}}{\text{total number in study}} = \dfrac{685}{755}\)
- \(\dfrac{280}{755}\)
- \(\left(\dfrac{305}{755} + \dfrac{685}{755}\right) - \dfrac{280}{755} = \dfrac{710}{755}\)
Table shows the number of athletes who stretch before exercising and how many had injuries within the past year.
Injury in last year | No injury in last year | Total | |
---|---|---|---|
Stretches | 55 | 295 | 350 |
Does not stretch | 231 | 219 | 450 |
Total | 286 | 514 | 800 |
- What is \(P(\text{athlete stretches before exercising})\)?
- What is \(P(\text{athlete stretches before exercising and no injury in the last year})\)?
- What is \(P(\text{athlete stretches before exercising or no injury in the last year})\)?
Answer
- \( P( \text{athlete stretches} ) = \dfrac{350}{800} = 0.4375 \)
- \( P( \text{athlete stretches AND no injury in the last year})=\dfrac{295}{800}=0.3688\)
- \( P( \text{athlete stretches OR no injury in the last year}) \\ = P(\text{athlete stretches }) + P(\text{no injurty in the last year}) - P(\text{athlete stretches AND no injury in the last year}) \\ = \dfrac{350}{800} + \dfrac{514}{800} - \dfrac{295}{800} \\= \dfrac{569}{800}=0.7113\)
Table shows a random sample of 100 hikers and the areas of hiking they prefer.
Sex | The Coastline | Near Lakes and Streams | On Mountain Peaks | Total |
---|---|---|---|---|
Female | 18 | 16 | ___ | 45 |
Male | ___ | ___ | 14 | 55 |
Total | ___ | 41 | ___ | ___ |
- Complete the table.
- Find the probability that a person is female or prefers hiking on mountain peaks. Let \(\text{F} =\) being female, and let \(\text{P} =\) prefers mountain peaks.
- Find \(P(\text{F})\).
- Find \(P(\text{M})\).
- Find \(P(\text{F AND M})\).
- Find \(P(\text{F OR M})\).
Answers
a.
Sex | The Coastline | Near Lakes and Streams | On Mountain Peaks | Total |
---|---|---|---|---|
Female | 18 | 16 | 11 | 45 |
Male | 16 | 25 | 14 | 55 |
Total | 34 |
41 |
25 | 100 |
b.
- \(P(\text{F}) = \dfrac{45}{100}\)
- \(P(\text{M}) = \dfrac{25}{100}\)
- \(P(\text{F AND M}) = \dfrac{11}{100}\)
- \(P(\text{F OR M}) = P(\text{F}) + P(\text{M}) - P(\text{F AND M}) =\dfrac{45}{100} + \dfrac{25}{100} - \dfrac{11}{100} = \dfrac{59}{100}\)
The complement of event \(\text{A}\) is denoted \(\text{A'}\) (read "A prime"). \(\text{A'}\) consists of all outcomes that are NOT in \(\text{A}\). Notice that
\[P(\text{A}) + P(\text{A′}) = 1. \nonumber\]
In other words,
\[P(\text{A′}) = 1- P(\text{A}) \nonumber\]
For example, let \(\text{S} = \{1, 2, 3, 4, 5, 6\}\) and let \(\text{A} = {1, 2, 3, 4}\). Then, \(\text{A′} = {5, 6}\) and \(P(A) = \frac{4}{6}\), \(P(\text{A′}) = \frac{2}{6}\), and
\[P(\text{A}) + P(\text{A′}) = \frac{4}{6} + \frac{2}{6} = 1. \nonumber\]
Exercise 3.2.38
Review
There are several tools you can use to help organize and sort data when calculating probabilities. Contingency tables help display data and are particularly useful when calculating probabilities that have multiple dependent variables.
Use the following information to answer the next four exercises. Table shows a random sample of musicians and how they learned to play their instruments.
Gender | Self-taught | Studied in School | Private Instruction | Total |
---|---|---|---|---|
Female | 12 | 38 | 22 | 72 |
Male | 19 | 24 | 15 | 58 |
Total | 31 | 62 | 37 | 130 |
Find P(musician is a female).
Find \(P(\text{musician is a male AND had private instruction})\).
Answer
\(P(\text{musician is a male AND had private instruction}) = \dfrac{15}{130} = \dfrac{3}{26} = 0.12\)
Find \(P(\text{musician is a female OR is self taught})\).
Answer
\(P(\text{musician is a female OR is self taught}) \\
= P(\text{musician is a female}) + P(\text{self taught}) - P(\text{musician is a female OR is self taught})\\
=\frac{72}{130} + \frac{31}{130} - \frac{12}{130}\\
=\frac{91}{130}\)
Are the events “being a female musician” and “learning music in school” mutually exclusive events?
Answer
The events are not mutually exclusive. It is possible to be a female musician who learned music in school.
References
- “United States: Uniform Crime Report – State Statistics from 1960–2011.” The Disaster Center. Available online at http://www.disastercenter.com/crime/ (accessed May 2, 2013).
Glossary
- mutually exclusive (or disjoint) events
- events that cannot happen at the same time
- contingency table
- the method of displaying a frequency distribution as a table with rows and columns to show how two variables may be dependent (contingent) upon each other; the table provides an easy way to calculate conditional probabilities.
- complement of an event
- The complement of event \(\text{A}\) consists of all outcomes that are NOT in \(\text{A}\).