Probability Topics
Contingency Tables
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[latexpage]
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.)
[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 exercisingno injury in the last year)?
 P(athlete stretches before exercising) = \(\frac{350}{800}\) = 0.4375
 P(athlete stretches before exercisingno 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.
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.
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.
 Find P(F AND C).
 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) ≠ 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.
 What word tells you this is a conditional?
 Fill in the blanks and calculate the probability: P(______) = ___.
 Is the sample space for this problem all 100 hikers? If not, what is it?
c.
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.
 Find P(F).
 Find P(P).
 Find P(F AND P).
 Find P(F OR P).
d.
[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 
 Out of the males, what is the probability that the cyclist prefers a hilly path?
 Are the events “being male” and “preferring the hilly path” independent events?
 P(HM) = \(\frac{52}{90}\) = 0.5778
 For M and H to be independent, show P(HM) = P(H)
P(HM) = 0.5778, P(H) = \(\frac{90}{200}\) = 0.45
P(HM) 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 \(\frac{1}{3}\).
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 lowerright corner entry is 1.
a.
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.
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
 Find P(2009 AND Robbery).
 Find P(2010 AND Burglary).
 Find P(2010 OR Burglary).
 Find P(2011Rape).
 Find P(Vehicle2008).
a. 0.0294, b. 0.1551, c. 0.7165, d. 0.2365, e. 0.2575
[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 
 Find the total for each row and column
 Find the probability that a randomly chosen individual from this group is Tall.
 Find the probability that a randomly chosen individual from this group is Obese and Tall.
 Find the probability that a randomly chosen individual from this group is Tall given that the idividual is Obese.
 Find the probability that a randomly chosen individual from this group is Obese given that the individual is Tall.
 Find the probability a randomly chosen individual from this group is Tall and Underweight.
 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 
 Row Totals: 60, 99, 46. Column totals: 50, 104, 51.
 P(Tall) = \(\frac{50}{205}\) = 0.244
 P(Obese AND Tall) = \(\frac{18}{205}\) = 0.088
 P(TallObese) = \(\frac{18}{60}\) = 0.3
 P(ObeseTall) = \(\frac{18}{50}\) = 0.36
 P(Tall AND Underweight = \(\frac{12}{205}\) = 0.0585
 No. P(Tall) does not equal P(TallObese).
References
“Blood Types.” American Red Cross, 2013. Available online at http://www.redcrossblood.org/learnaboutblood/bloodtypes (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 SmokingRelated 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_strangerhnegativeblood.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.GenderSelftaughtStudied 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 selfreported 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 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 2130 cigarettes per day, means that the person must fulfill both criteria and the sample space is reduced to those who smoke 2130 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 ________
 mutually exclusive.
 independent.
 both mutually exclusive and independent.
 neither mutually exclusive nor independent.
d
The events “Other” and “Up for reelection in November 2016” are ________
 mutually exclusive.
 independent.
 both mutually exclusive and independent.
 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.

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 
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  \(\frac{\text{22,050}}{\text{29,760}}\)
 \(\frac{\text{330}}{\text{29,760}}\)
 \(\frac{\text{2,000}}{\text{29,760}}\)
 \(\frac{\text{23,720}}{\text{29,760}}\)
 \(\frac{\text{5,010}}{\text{6,020}}\)
Use the following information to answer the next two exercises. The table of data obtained from www.baseballalmanac.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).
 \(\frac{1518}{2873}\)
 \(\frac{2873}{12351}\)
 \(\frac{583}{12351}\)
 \(\frac{4189}{12351}\)
Find P(hit was made by Ty CobbThe hit was a Home Run).
 \(\frac{4189}{12351}\)
 \(\frac{114}{1720}\)
 \(\frac{1720}{4189}\)
 \(\frac{114}{12351}\)
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 
 Complete the table.
 What is the probability that a randomly selected child will have wavy hair?
 What is the probability that a randomly selected child will have either brown or blond hair?
 What is the probability that a randomly selected child will have wavy brown hair?
 What is the probability that a randomly selected child will have red hair, given that he or she has straight hair?
 If B is the event of a child having brown hair, find the probability of the complement of B.
 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.
 \(\frac{26}{106}\)
 \(\frac{33}{106}\)
 \(\frac{21}{106}\)
 \(\left(\frac{26}{106}\right)\) + \(\left(\frac{33}{106}\right)\) – \(\left(\frac{21}{106}\right)\) = \(\left(\frac{38}{106}\right)\)
 \(\frac{21}{33}\)
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.