Probability Topics

18 Contingency Tables and Probability Trees

Contingency Tables

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
Uses cell phone while driving 25 280 305
Does not use cell phone while driving 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(Driver is a cell phone user).

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

b. Find P(Driver 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(Driver had no violation in the last year \cap was a cell phone user).

c. \frac{280}{755}

d. Find P(Driver is a cell phone user \cup driver 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(Driver is a cell phone user | driver had a violation in the last year).

e. \frac{25}{70} (The sample space is reduced to the number of drivers who had a violation.)

f. Find P(Driver had no violation last year | driver was not a cell phone user)

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

Try it

(Figure) 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

(Figure) 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\left(F\cap C\right).
  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.

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

P\left(F\cap C\right)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.

  1. The word ‘given’ tells you that this is a conditional.
  2. P(M|L) = \frac{25}{41}
  3. 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\left(F\cap P\right).
  4. Find P\left(F\cup P\right).

d.

  1. P(F) = \frac{45}{100}
  2. P(P) = \frac{25}{100}
  3. P\left(F\cap P\right) = \frac{11}{100}
  4. P\left(F\cup P\right) = \frac{45}{100} + \frac{25}{100}\frac{11}{100} = \frac{59}{100}
Try It

(Figure) 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\left(\text{Door One}\cap \text{Caught}\right)
  • The entry \frac{4}{15}=\left(\frac{4}{5}\right)\left(\frac{1}{3}\right) is P\left(\text{Door One}\cap \text{Not Caught}\right)

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 \cup Door Two given that Muddy is caught by Alissa?

c. \frac{9}{19}

(Figure) 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\left(2009\cap \text{Robbery}\right).
  2. Find P\left(2010\cap \text{Burglary}\right).
  3. Find P\left(2010\cup \text{Burglary}\right).
  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

(Figure) 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\left(\text{Obese}\cap \text{Tall}\right) = \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\left(\text{Tall}\cap \text{Underweight}\right) = \frac{12}{205} = 0.0585
  7. No. P(Tall) does not equal P(Tall|Obese).

Tree Diagrams

Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams can be used to visualize and solve conditional probabilities.

Tree Diagrams

A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of “branches” that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.

In an urn, there are 11 balls. Three balls are red (R) and eight balls are blue (B). Draw two balls, one at a time, with replacement. “With replacement” means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.

Total = 64 + 24 + 24 + 9 = 121

This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 8B and 3R. The second branch has a set of two lines (8B and 3R) for each line of the first branch. Multiply along each line to find 64BB, 24BR, 24RB, and 9RR.

The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the nine RR outcomes can be written as:

R1R1R1R2R1R3R2R1R2R2R2R3R3R1R3R2R3R3

The other outcomes are similar.

There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space.

a. List the 24 BR outcomes: B1R1, B1R2, B1R3, …

a. B1R1B1R2B1R3B2R1B2R2B2R3B3R1B3R2B3R3B4R1B4R2B4R3B5R1B5R2B5R3B6R1B6R2B6R3B7R1B7R2B7R3B8R1B8R2B8R3

b. Using the tree diagram, calculate P(RR).

b. P(RR) = \left(\frac{3}{11}\right)\left(\frac{3}{11}\right) = \frac{9}{121}

c. Using the tree diagram, calculate P\left(\mathrm{RB}\cup \mathrm{BR}\right).

c. P\left(\mathrm{RB}\cup \mathrm{BR}\right) = \left(\frac{3}{11}\right)\left(\frac{8}{11}\right) + \left(\frac{8}{11}\right)\left(\frac{3}{11}\right) = \frac{48}{121}

d. Using the tree diagram, calculate P\left(R\phantom{\rule{0.2em}{0ex}}\text{on 1st draw}\cap B\phantom{\rule{0.2em}{0ex}}\text{on 2nd draw}\right).

d. P\left(R\phantom{\rule{0.2em}{0ex}}\text{on 1st draw}\cap B\phantom{\rule{0.2em}{0ex}}\text{on 2nd draw}\right) = \left(\frac{3}{11}\right)\left(\frac{8}{11}\right) = \frac{24}{121}

e. Using the tree diagram, calculate P(R on 2nd draw|B on 1st draw).

e. P(R on 2nd draw|B on 1st draw) = P(R on 2nd|B on 1st) = \frac{24}{88} = \frac{3}{11}

This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible outcomes are BR. \frac{24}{88} = \frac{3}{11}.

f. Using the tree diagram, calculate P(BB).

f. P(BB) = \frac{64}{121}

g. Using the tree diagram, calculate P(B on the 2nd draw|R on the first draw).

g. P(B on 2nd draw|R on 1st draw) = \frac{8}{11}

There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw. The probability is then \frac{24}{33}.

Try It

In a standard deck, there are 52 cards. 12 cards are face cards (event F) and 40 cards are not face cards (event N). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate P(FF).

This is a tree diagram with branches showing frequencies of each draw. The first branch shows two lines: 12F and 40N. The second branch has a set of two lines (12F and 40N) for each line of the first branch. Multiply along each line to find 144FF, 480FN, 480NF, and 1,600NN.

Total number of outcomes is 144 + 480 + 480 + 1600 = 2,704.

P(FF) = \frac{144}{\text{144 + 480 + 480 + 1,600}}=\frac{144}{2,704}=\frac{9}{169}

An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. “Without replacement” means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, \left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}.

Total = \frac{56+24+24+6}{110}=\frac{110}{110}=1

This is a tree diagram with branches showing probabilities of each draw. The first branch shows 2 lines: B 8/11 and R 3/11. The second branch has a set of 2 lines for each first branch line. Below B 8/11 are B 7/10 and R 3/10. Below R 3/11 are B 8/10 and R 2/10. Multiply along each line to find BB 56/110, BR 24/110, RB 24/110, and RR 6/110.

NOTE

If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement, so that on the second draw there are ten marbles left in the urn.


Calculate the following probabilities using the tree diagram.

a. P(RR) = ________

a. P(RR) = \left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}

b. Fill in the blanks:

P\left(\mathrm{RB}\cup \mathrm{BR}\right) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right)\text{ }+\text{ (___)(___) }=\text{ }\frac{48}{110}

b. P\left(\mathrm{RB}\cup \mathrm{BR}\right) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) + \left(\frac{8}{11}\right)\left(\frac{3}{10}\right) = \frac{48}{110}

c. P(R on 2nd|B on 1st) =

c. P(R on 2nd|B on 1st) = \frac{3}{10}

d. Fill in the blanks.

P\left(R\phantom{\rule{0.2em}{0ex}}\text{on 1st}\cap B\phantom{\rule{0.2em}{0ex}}\text{on 2nd}\right) = (___)(___) = \frac{24}{100}

d. P\left(R\phantom{\rule{0.2em}{0ex}}\text{on 1st}\cap B\phantom{\rule{0.2em}{0ex}}\text{on 2nd}\right) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right) = \frac{24}{100}

e. Find P(BB).

e. P(BB) = \left(\frac{8}{11}\right)\left(\frac{7}{10}\right)

f. Find P(B on 2nd|R on 1st).

f. Using the tree diagram, P(B on 2nd|R on 1st) = P(R|B) = \frac{8}{10}.

If we are using probabilities, we can label the tree in the following general way.

This is a tree diagram for a two-step experiment. The first branch shows first outcome: P(B) and P(R). The second branch has a set of 2 lines for each line of the first branch: the probability of B given B = P(BB), the probability of R given B = P(RB), the probability of B given R = P(BR), and the probability of R given R = P(RR).

  • P(R|R) here means P(R on 2nd|R on 1st)
  • P(B|R) here means P(B on 2nd|R on 1st)
  • P(R|B) here means P(R on 2nd|B on 1st)
  • P(B|B) here means P(B on 2nd|B on 1st)
Try It

In a standard deck, there are 52 cards. Twelve cards are face cards (F) and 40 cards are not face cards (N). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities.

This is a tree diagram with branches showing frequencies of each draw. The first branch shows 2 lines: F 12/52 and N 40/52. The second branch has a set of 2 lines (F 11/52 and N 40/51) for each line of the first branch. Multiply along each line to find FF 121/2652, FN 480/2652, NF 480/2652, and NN 1560/2652.
  1. Find P\left(\mathrm{FN}\cup \mathrm{NF}\right).
  2. Find P(N|F).
  3. Find P(at most one face card).
    Hint: “At most one face card” means zero or one face card.
  4. Find P(at least on face card).
    Hint: “At least one face card” means one or two face cards.
  1. P\left(\mathrm{FN}\cup \mathrm{NF}\right) = \frac{\text{480}}{\text{2,652}}\text{ + }\frac{\text{480}}{\text{2,652}}\text{ = }\frac{\text{960}}{\text{2,652}}\text{ = }\frac{\text{80}}{\text{221}}
  2. P(N|F) = \frac{40}{51}
  3. P(at most one face card) = \frac{\text{(480  +  480  +  1,560)}}{\text{2,652}} = \frac{2,520}{2,652}
  4. P(at least one face card) = \frac{\text{(132 + 480 + 480)}}{\text{2,652}} = \frac{\text{1,092}}{\text{2,652}}

A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption.

This is a tree diagram with branches showing probabilities of kitten choices. The first branch shows two lines: T 4/9 and B 5/9. The second branch has a set of 2 lines for each first branch line. Below T 4/9 are T 3/8 and B 5/8. Below B 5/9 are T 4/8 and B 4/8. Multiply along each line to find probabilities of possible combinations.

  1. What is the probability that both kittens are tabby?

    a.\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) b.\left(\frac{4}{9}\right)\left(\frac{4}{9}\right) c.\left(\frac{4}{9}\right)\left(\frac{3}{8}\right) d.\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)

  2. What is the probability that one kitten of each coloring is selected?

    a.\left(\frac{4}{9}\right)\left(\frac{5}{9}\right) b.\left(\frac{4}{9}\right)\left(\frac{5}{8}\right) c.\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{9}\right) d.\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{8}\right)

  3. What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?
  4. What is the probability of choosing two kittens of the same color?

a. c, b. d, c. \frac{4}{8}, d. \frac{32}{72}

Try It

Suppose there are four red balls and three yellow balls in a box. Two balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected?

\left(\frac{4}{7}\right)\left(\frac{3}{6}\right) + \left(\frac{3}{7}\right)\left(\frac{4}{6}\right)

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).

“Human Blood Types.” Unite Blood Services, 2011. Available online at http://www.unitedbloodservices.org/learnMore.aspx (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).

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).

Data from Clara County Public H.D.

Data from the American Cancer Society.

Data from The Data and Story Library, 1996. Available online at http://lib.stat.cmu.edu/DASL/ (accessed May 2, 2013).

Data from the Federal Highway Administration, part of the United States Department of Transportation.

Data from the United States Census Bureau, part of the United States Department of Commerce.

Data from USA Today.

“Environment.” The World Bank, 2013. Available online at http://data.worldbank.org/topic/environment (accessed May 2, 2013).

“Search for Datasets.” Roper Center: Public Opinion Archives, University of Connecticut., 2013. Available online at https://ropercenter.cornell.edu/?s=Search+for+Datasets (accessed February 6, 2019).

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.

A tree diagram use branches to show the different outcomes of experiments and makes complex probability questions easy to visualize.

Key Terms

Tree Diagram
the useful visual representation of a sample space and events in the form of a “tree” with branches marked by possible outcomes together with associated probabilities (frequencies, relative frequencies)
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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