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Probability Topics

Contingency Tables

OpenStaxCollege

A contingency table provides a way of portraying data that can facilitate calculating probabilities. The table helps in determining conditional probabilities quite easily. 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.

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).

a. \frac{\text{number of car phone users}}{\text{total number in study}}\text{ }=\text{ }\frac{305}{755}

b. Find P(person had no violation in the last year).

b. \frac{\text{number that had no violation}}{\text{total number in study}}\text{ }=\text{ }\frac{685}{755}

c. Find P(Person had no violation in the last year AND was a car phone user).

c. \frac{280}{755}

d. Find P(Person is a car phone user OR person had no violation in the last year).

d. \left(\frac{305}{755}\text{ }+\text{ }\frac{685}{755}\right)\text{ }-\text{ }\frac{280}{755}\text{ }=\text{ }\frac{710}{755}

e. Find P(Person is a car phone user GIVEN person had a violation in the last year).

e. \frac{25}{70} (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)

f. \frac{405}{450} (The sample space is reduced to the number of persons who were not car phone users.)

Try it

[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
  1. What is P(athlete stretches before exercising)?
  2. What is P(athlete stretches before exercising|no injury in the last year)?
  1. P(athlete stretches before exercising) = \frac{350}{800} = 0.4375
  2. P(athlete stretches before exercising|no injury in the last year) = \frac{295}{514} = 0.5739

[link] shows a random sample of 100 hikers and the areas of hiking they prefer.

Hiking Area Preference
Sex The Coastline Near Lakes and Streams On Mountain Peaks Total
Female 18 16 ___ 45
Male ___ ___ 14 55
Total ___ 41 ___ ___

a. Complete the table.

a.

Hiking Area Preference
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. Are the events “being female” and “preferring the coastline” independent events?

Let F = being female and let C = preferring the coastline.

  1. Find P(F AND C).
  2. Find P(F)P(C)

Are these two numbers the same? If they are, then F and C are independent. If they are not, then F and C are not independent.

b.

P(F AND C) = \frac{18}{100} = 0.18
P(F)P(C) = \left(\frac{45}{100}\right)\left(\frac{34}{100}\right) = (0.45)(0.34) = 0.153

P(F AND C) ≠ P(F)P(C), so the events F and C are not independent.

c. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let M = being male, and let L = prefers hiking near lakes and streams.

  1. What word tells you this is a conditional?
  2. Fill in the blanks and calculate the probability: P(___|___) = ___.
  3. Is the sample space for this problem all 100 hikers? If not, what is it?

c.

The word ‘given’ tells you that this is a conditional.
P(M|L) = \frac{25}{41}
No, the sample space for this problem is the 41 hikers who prefer lakes and streams.

d. Find the probability that a person is female or prefers hiking on mountain peaks. Let F = being female, and let P = prefers mountain peaks.

  1. Find P(F).
  2. Find P(P).
  3. Find P(F AND P).
  4. Find P(F OR P).

d.

P(F) = \frac{45}{100}
P(P) = \frac{25}{100}
P(F AND P) = \frac{11}{100}
P(F OR P) = \frac{45}{100} + \frac{25}{100}\frac{11}{100} = \frac{59}{100}
Try It

[link] shows a random sample of 200 cyclists and the routes they prefer. Let M = males and H = hilly path.

Gender Lake Path Hilly Path Wooded Path Total
Female 45 38 27 110
Male 26 52 12 90
Total 71 90 39 200
  1. Out of the males, what is the probability that the cyclist prefers a hilly path?
  2. Are the events “being male” and “preferring the hilly path” independent events?
  1. P(H|M) = \frac{52}{90} = 0.5778
  2. For M and H to be independent, show P(H|M) = P(H)

    P(H|M) = 0.5778, P(H) = \frac{90}{200} = 0.45

    P(H|M) does not equal P(H) so M and H are NOT independent.

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 \frac{1}{5}\text{} and the probability he is not caught is \frac{4}{5}\text{}. If he goes out the second door, the probability he gets caught by Alissa is \frac{1}{4} and the probability he is not caught is \frac{3}{4}. The probability that Alissa catches Muddy coming out of the third door is \frac{1}{2} and the probability she does not catch Muddy is \frac{1}{2}. It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is \frac{1}{3}.

Door Choice
Caught or Not Door One Door Two Door Three Total
Caught \frac{1}{15}\text{} \frac{1}{12}\text{} \frac{1}{6}\text{} ____
Not Caught \frac{4}{15} \frac{3}{12} \frac{1}{6} ____
Total ____ ____ ____ 1
  • The first entry \frac{1}{15}=\left(\frac{1}{5}\right)\left(\frac{1}{3}\right) is P(Door One AND Caught)
  • The entry
    \frac{4}{15}=\left(\frac{4}{5}\right)\left(\frac{1}{3}\right) is P(Door One AND Not Caught)

Verify the remaining entries.

a. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.

a.

Door Choice
Caught or Not Door One Door Two Door Three Total
Caught \frac{1}{15}\text{} \frac{1}{12}\text{} \frac{1}{6}\text{} \frac{19}{60}
Not Caught \frac{4}{15} \frac{3}{12} \frac{1}{6} \frac{41}{60}
Total \frac{5}{15} \frac{4}{12} \frac{2}{6} 1

b. What is the probability that Alissa does not catch Muddy?

b. \frac{41}{60}

c. What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa?

c. \frac{9}{19}

[link] contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the U.S.

United States Crime Index Rates Per 100,000 Inhabitants 2008–2011
Year Robbery Burglary Rape Vehicle Total
2008 145.7 732.1 29.7 314.7
2009 133.1 717.7 29.1 259.2
2010 119.3 701 27.7 239.1
2011 113.7 702.2 26.8 229.6
Total

TOTAL each column and each row. Total data = 4,520.7

  1. Find P(2009 AND Robbery).
  2. Find P(2010 AND Burglary).
  3. Find P(2010 OR Burglary).
  4. Find P(2011|Rape).
  5. Find P(Vehicle|2008).

a. 0.0294, b. 0.1551, c. 0.7165, d. 0.2365, e. 0.2575

Try It

[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
  1. Find the total for each row and column
  2. Find the probability that a randomly chosen individual from this group is Tall.
  3. Find the probability that a randomly chosen individual from this group is Obese and Tall.
  4. Find the probability that a randomly chosen individual from this group is Tall given that the idividual is Obese.
  5. Find the probability that a randomly chosen individual from this group is Obese given that the individual is Tall.
  6. Find the probability a randomly chosen individual from this group is Tall and Underweight.
  7. Are the events Obese and Tall independent?
Weight/Height Tall Medium Short Totals
Obese 18 28 14 60
Normal 20 51 28 99
Underweight 12 25 9 46
Totals 50 104 51 205
  1. Row Totals: 60, 99, 46. Column totals: 50, 104, 51.
  2. P(Tall) = \frac{50}{205} = 0.244
  3. P(Obese AND Tall) = \frac{18}{205} = 0.088
  4. P(Tall|Obese) = \frac{18}{60} = 0.3
  5. P(Obese|Tall) = \frac{18}{50} = 0.36
  6. P(Tall AND Underweight = \frac{12}{205} = 0.0585
  7. No. P(Tall) does not equal P(Tall|Obese).

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 United States Senate. Available online at www.senate.gov (accessed May 2, 2013).

Haiman, Christopher A., Daniel O. Stram, Lynn R. Wilkens, Malcom C. Pike, Laurence N. Kolonel, Brien E. Henderson, and Loīc Le Marchand. “Ethnic and Racial Differences in the Smoking-Related Risk of Lung Cancer.” The 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.” Unite 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).

“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).

Chapter 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 probabilites that have multiple dependent variables.

Use the following information to answer the next four exercises.[link] shows a random sample of musicians and how they learned to play their instruments.GenderSelf-taughtStudied in SchoolPrivate InstructionTotalFemale12382272Male19241558Total316237130Find P(musician is a female).

Find P(musician is a male AND had private instruction).

P(musician is a male AND had private instruction) = \frac{15}{130} = \frac{3}{26} = 0.12

Find P(musician is a female OR is self taught).

Are the events “being a female musician” and “learning music in school” mutually exclusive events?

P(being a female musician AND learning music in school) = \frac{38}{130} = \frac{19}{65} = 0.29

P(being a female musician)P(learning music in school) = \left(\frac{72}{130}\right)\left(\frac{62}{130}\right) = \frac{4,464}{16,900} = \frac{1,116}{4,225} = 0.26

No, they are not independent because P(being a female musician AND learning music in school) is not equal to P(being a female musician)P(learning music in school).

Bringing It Together

Use the following information to answer the next seven exercises. An article in the New England Journal of Medicine, reported about a study of smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans, and 7,650 Whites. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 Whites. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 Whites. Of the people smoking at least 31 cigarettes per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 Whites.

Complete the table using the data provided. Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day.

Smoking Levels by Ethnicity
Smoking Level African American Native Hawaiian Latino Japanese Americans White TOTALS
1–10
11–20
21–30
31+
TOTALS

Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day.

\frac{35,065}{100,450}

Find the probability that the person was Latino.

In words, explain what it means to pick one person from the study who is “Japanese American AND smokes 21 to 30 cigarettes per day.” Also, find the probability.

To pick one person from the study who is Japanese American AND smokes 21 to 30 cigarettes per day means that the person has to meet both criteria: both Japanese American and smokes 21 to 30 cigarettes. The sample space should include everyone in the study. The probability is \frac{4,715}{100,450}.

In words, explain what it means to pick one person from the study who is “Japanese American OR smokes 21 to 30 cigarettes per day.” Also, find the probability.

In words, explain what it means to pick one person from the study who is “Japanese American GIVEN that person smokes 21 to 30 cigarettes per day.” Also, find the probability.

To pick one person from the study who is Japanese American given that person smokes 21-30 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 21-30 cigarettes per day. The probability is \frac{4715}{15,273}.

Prove that smoking level/day and ethnicity are dependent events.

Homework

Use the information in the [link] to answer the next eight exercises. The table shows the political party affiliation of each of 67 members of the US Senate in June 2012, and when they are up for reelection.

Up for reelection: Democratic Party Republican Party Other Total
November 2014 20 13 0
November 2016 10 24 0
Total

What is the probability that a randomly selected senator has an “Other” affiliation?

0

What is the probability that a randomly selected senator is up for reelection in November 2016?

What is the probability that a randomly selected senator is a Democrat and up for reelection in November 2016?

\frac{10}{67}

What is the probability that a randomly selected senator is a Republican or is up for reelection in November 2014?

Suppose that a member of the US Senate is randomly selected. Given that the randomly selected senator is up for reelection in November 2016, what is the probability that this senator is a Democrat?

\frac{10}{34}

Suppose that a member of the US Senate is randomly selected. What is the probability that the senator is up for reelection in November 2014, knowing that this senator is a Republican?

The events “Republican” and “Up for reelection in 2016” are ________

  1. mutually exclusive.
  2. independent.
  3. both mutually exclusive and independent.
  4. neither mutually exclusive nor independent.

d

The events “Other” and “Up for reelection in November 2016” are ________

  1. mutually exclusive.
  2. independent.
  3. both mutually exclusive and independent.
  4. neither mutually exclusive nor independent.

[link] gives the number of suicides estimated in the U.S. for a recent year by age, race (black or white), and sex. We are interested in possible relationships between age, race, and sex. We will let suicide victims be our population.

Race and Sex 1–14 15–24 25–64 over 64 TOTALS
white, male 210 3,360 13,610 22,050
white, female 80 580 3,380 4,930
black, male 10 460 1,060 1,670
black, female 0 40 270 330
all others
TOTALS 310 4,650 18,780 29,760

Do not include “all others” for parts f and g.

Fill in the column for the suicides for individuals over age 64.
Fill in the row for all other races.
Find the probability that a randomly selected individual was a white male.
Find the probability that a randomly selected individual was a black female.
Find the probability that a randomly selected individual was black
Find the probability that a randomly selected individual was male.
Out of the individuals over age 64, find the probability that a randomly selected individual was a black or white male.
  1. Race and Sex 1–14 15–24 25–64 over 64 TOTALS
    white, male 210 3,360 13,610 4,870 22,050
    white, female 80 580 3,380 890 4,930
    black, male 10 460 1,060 140 1,670
    black, female 0 40 270 20 330
    all others 100
    TOTALS 310 4,650 18,780 6,020 29,760
  2. Race and Sex 1–14 15–24 25–64 over 64 TOTALS
    white, male 210 3,360 13,610 4,870 22,050
    white, female 80 580 3,380 890 4,930
    black, male 10 460 1,060 140 1,670
    black, female 0 40 270 20 330
    all others 10 210 460 100 780
    TOTALS 310 4,650 18,780 6,020 29,760
  3. \frac{\text{22,050}}{\text{29,760}}
  4. \frac{\text{330}}{\text{29,760}}
  5. \frac{\text{2,000}}{\text{29,760}}
  6. \frac{\text{23,720}}{\text{29,760}}
  7. \frac{\text{5,010}}{\text{6,020}}

Use the following information to answer the next two exercises. The table of data obtained from www.baseball-almanac.com shows hit information for four well known baseball players. Suppose that one hit from the table is randomly selected.

NAME Single Double Triple Home Run TOTAL HITS
Babe Ruth 1,517 506 136 714 2,873
Jackie Robinson 1,054 273 54 137 1,518
Ty Cobb 3,603 174 295 114 4,189
Hank Aaron 2,294 624 98 755 3,771
TOTAL 8,471 1,577 583 1,720 12,351

Find P(hit was made by Babe Ruth).

  1. \frac{1518}{2873}
  2. \frac{2873}{12351}
  3. \frac{583}{12351}
  4. \frac{4189}{12351}

Find P(hit was made by Ty Cobb|The hit was a Home Run).

  1. \frac{4189}{12351}
  2. \frac{114}{1720}
  3. \frac{1720}{4189}
  4. \frac{114}{12351}

b

[link] identifies a group of children by one of four hair colors, and by type of hair.

Hair Type Brown Blond Black Red Totals
Wavy 20 15 3 43
Straight 80 15 12
Totals 20 215
  1. Complete the table.
  2. What is the probability that a randomly selected child will have wavy hair?
  3. What is the probability that a randomly selected child will have either brown or blond hair?
  4. What is the probability that a randomly selected child will have wavy brown hair?
  5. What is the probability that a randomly selected child will have red hair, given that he or she has straight hair?
  6. If B is the event of a child having brown hair, find the probability of the complement of B.
  7. In words, what does the complement of B represent?

In a previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data were compiled into the following table.

Shirt# ≤ 210 211–250 251–290 > 290
1–33 21 5 0 0
34–66 6 18 7 4
66–99 6 12 22 5

For the following, suppose that you randomly select one player from the 49ers or Cowboys.

Find the probability that his shirt number is from 1 to 33.
Find the probability that he weighs at most 210 pounds.
Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds.
Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds.
Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210 pounds.
  1. \frac{26}{106}
  2. \frac{33}{106}
  3. \frac{21}{106}
  4. \left(\frac{26}{106}\right) + \left(\frac{33}{106}\right)\left(\frac{21}{106}\right) = \left(\frac{38}{106}\right)
  5. \frac{21}{33}

Glossary

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.

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