Chapter 3: Probability Topics
3.4 Contingency Tables
Learning Objectives
By the end of this section, the student should be able to:
- Use a given contingency table to compute probabilities
- Complete a contingency table given starting information about a sample
Sometimes, when the probability problems are complex, it can be helpful to visualize the situation. Contingency tables, tree diagrams and Venn diagrams are some tools that can help us visualize and solve conditional probabilities. Tree and Venn diagrams will be covered in the next section.
Contingency Tables
A contingency table provides a way of portraying data that can facilitate calculating probabilities, particularly conditional probabilities. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner.
Example
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.
a. Find P(Person is a car phone user).
Solution
a. [latex]\frac{\text{number of car phone users}}{\text{total number in study}}\text{ }=\text{ }\frac{305}{755}[/latex]
b. Find P(person had no violation in the last year).
Solution
b. [latex]\frac{\text{number that had no violation}}{\text{total number in study}}\text{ }=\text{ }\frac{685}{755}[/latex]
c. Find P(Person had no violation in the last year AND was a car phone user).
Solution
c. [latex]\frac{280}{755}[/latex]
d. Find P(Person is a car phone user OR person had no violation in the last year).
Solution
d. [latex]\left(\frac{305}{755}\text{ }+\text{ }\frac{685}{755}\right)\text{ }-\text{ }\frac{280}{755}\text{ }=\text{ }\frac{710}{755}[/latex]
e. Find P(Person is a car phone user GIVEN person had a violation in the last year).
Solution
e. [latex]\frac{25}{70}[/latex] (The sample space is reduced to the number of persons who had a violation.)
f. Find P(Person had no violation last year GIVEN person was not a car phone user)
Solution
f. [latex]\frac{405}{450}[/latex] (The sample space is reduced to the number of persons who were not car phone users.)
Your Turn!
[link] 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(athlete stretches before exercising)?
- What is P(athlete stretches before exercising | no injury in the last year)?
Solution
- P(athlete stretches before exercising) = [latex]\frac{350}{800}[/latex] = 0.4375
- P(athlete stretches before exercising | no injury in the last year) = [latex]\frac{295}{514}[/latex] = 0.5739
Key Takeaways
[link] shows a random sample of 200 cyclists broken down by type of bicycle and type of route they prefer. Let M = mountain bike and H = hilly path.
| Lake Path | Hilly Path | Wooded Path | Total | |
|---|---|---|---|---|
| Road | 45 | 38 | 27 | 110 |
| Mountain | 26 | 52 | 12 | 90 |
| Total | 71 | 90 | 39 | 200 |
- Out of the mountain bike cyclists, what is the probability that the cyclist prefers a hilly path?
- Are the events “prefers a mountain bike” and “preferring the hilly path” independent events?
Your Turn!
[link] relates the weights and heights of a group of individuals participating in an observational study.
| Weight/Height | Tall | Medium | Short | Totals |
|---|---|---|---|---|
| Obese | 18 | 28 | 14 | |
| Normal | 20 | 51 | 28 | |
| Underweight | 12 | 25 | 9 | |
| Totals |
Complete the contingency table by finding the total for each row and column. Then use the table to find the probability that a randomly chosen individual from this group is:
- Tall.
- Obese and Tall.
- Tall given that the individual is Obese.
- Obese given that the individual is Tall.
- Tall and Underweight.
- Are the events Obese and Tall independent?
Your Turn!
Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the Cat is [latex]\frac{1}{5}\text{}[/latex]. If he goes out the second door, the probability he gets caught by Alissa is [latex]\frac{1}{4}[/latex]. The probability that Alissa catches Muddy coming out of the third door is [latex]\frac{1}{2}[/latex]. It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is [latex]\frac{1}{3}[/latex].
| Caught or Not | Door One | Door Two | Door Three | Total |
|---|---|---|---|---|
| Caught | [latex]\frac{1}{15}[/latex] | ________ | ________ | ________ |
| Not Caught | [latex]\frac{4}{15}[/latex] | ________ | ________ | ________ |
| Total | ____ | ____ | ____ | 1 |
The first column of the contingency table has been filled in for you. The first entry is computed as [latex]P(\text{caught }|\text{ Door 1}) P(\text{Door 1}) = \frac{1}{5} \cdot \frac{1}{3} = \frac{1}{15}[/latex]. Similarly, the first entry in the second row is computed as [latex]P(\text{not caught }|\text{ Door 1}) P(\text{Door 1}) = (1-P(\text{caught }|\text{ Door 1})) P(\text{Door 1}) = \frac{4}{5} \cdot \frac{1}{3} = \frac{4}{15}.[/latex]
Complete the rest of the table, verifying that the totals sum to 1 as probabilities should. Then use the table to answer the following questions.
- What is the probability that Alissa does not catch Muddy?
- What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa?
References
“Blood Types.” American Red Cross, 2013. Available online at http://www.redcrossblood.org/learn-about-blood/blood-types (accessed May 3, 2013).
Data from the National Center for Health Statistics, part of the United States Department of Health and Human Services.
Data from the United States Senate. Available online at www.senate.gov (accessed May 2, 2013).
Haiman, Christopher A., Daniel O. Stram, Lynn R. Wilkens, Malcolm C. Pike, Laurence N. Kolonel, Brien E. Henderson, and Loīc Le Marchand. “Ethnic and Racial Differences in the Smoking-Related Risk of Lung Cancer.” New England Journal of Medicine, 2013. Available online at http://www.nejm.org/doi/full/10.1056/NEJMoa033250 (accessed May 2, 2013).
“Human Blood Types.” United Blood Services, 2011. Available online at http://www.unitedbloodservices.org/learnMore.aspx (accessed May 2, 2013).
Samuel, T. M. “Strange Facts about RH Negative Blood.” eHow Health, 2013. Available online at http://www.ehow.com/facts_5552003_strange-rh-negative-blood.html (accessed May 2, 2013).
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.
