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Probability Topics: Contingency Tables

Module by: Barbara Illowsky, Ph.D., Susan Dean. E-mail the authors

Summary: This module introduces the contingency table as a way of determining conditional probabilities. Note: This module is currently under revision, and its content is subject to change. This module is being prepared as part of a statistics textbook that will be available for the Fall 2008 semester.

Note: You are viewing an old version of this document. The latest version is available here.

A contingency table provides a different way of 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.

Example 1

Suppose a study of speeding violations and drivers who use car phones produced the following fictional data:

Table 1
  Speeding violation in the last year No speeding violation in the last year Total
Car phone user 25 280 305
Not a car 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 = 755305 + 450 = 755 and 70 + 685 = 75570 + 685 = 755.

Calculate the following probabilities using the table

Problem 1

P(person is a car phone user) = P(person is a car phone user) =

Solution

number of car phone users total number in study = 305 755 number of car phone users total number in study = 305 755

Problem 2

P(person had no violation in the last year) = P(person had no violation in the last year) =

Solution

number that had no violation total number in study = 685 755 number that had no violation total number in study = 685 755

Problem 3

P(person had no violation in the last year AND was a car phone user) =P(person had no violation in the last year AND was a car phone user) =

Solution

280 755 280 755

Problem 4

P(person is a car phone user OR person had no violation in the last year) =P(person is a car phone user OR person had no violation in the last year) =

Solution

( 305 755 + 685 755 ) - 280 755 = 710 755 ( 305 755 + 685 755 ) - 280 755 = 710 755

Problem 5

P(person is a car phone user GIVEN person had a violation in the last year) = P(person is a car phone user GIVEN person had a violation in the last year) =

Solution

25 70 25 70 (The sample space is reduced to the number of persons who had a violation.)

Problem 6

P(person had no violation last year GIVEN person was not a car phone user) = P(person had no violation last year GIVEN person was not a car phone user) =

Solution

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

Example 2

The following table shows a random sample of 100 hikers and the areas of hiking preferred:

Table 2: Hiking Area Preference
Sex The Coastline Near Lakes and Streams On Mountain Peaks Total
Female 18 16 ___ 45
Male ___ ___ 14 55
Total ___ 41 ___ ___

Problem 1

Complete the table.

Solution

Table 3: 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

Problem 2

Are the events "being female" and "preferring the coastline" independent events?

Let FF = being female and let CC = preferring the coastline.

  • a. P(F AND C)P(F AND C) =
  • b. P(F)P(C)P(F)P(C) =

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

Solution

  • a. P(F AND C) = 18 100 = 0.18 P(F AND C) = 18 100 = 0.18
  • b. P(F)P(C) = 45 100 45 100 = 0.45 0.45 = 0.153 P(F)P(C) = 45 100 45 100 = 0.450.45 = 0.153

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

Problem 3

Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let MM = being male and let LL = prefers hiking near lakes and streams.

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

Solution

  • a. The word 'given' tells you that this is a conditional.
  • b. P(M|L) = 2541P(M|L) = 2541
  • c. No, the sample space for this problem is 41.

Problem 4

Find the probability that a person is female or prefers hiking on mountain peaks. Let FF = being female and let PP = prefers mountain peaks.

  • a. P(F) = P(F) =
  • b. P(P) = P(P) =
  • c. P(F AND P) = P(F AND P) =
  • d. Therefore, P(F OR P) = P(F OR P) =

Solution

  • a. P(F) = 45100P(F) = 45100
  • b. P(P) = 25100P(P) = 25100
  • c. P(F AND P) = 11100P(F AND P) = 11100
  • d. P(F OR P) = 45100 + 25100 - 11100 = 59100P(F OR P) = 45100 + 25100 - 11100 = 59100

Example 3

Muddy Mouse lives in a cage with 3 doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is 1 5 1 5 and the probability he is not caught is 4 5 4 5 . If he goes out the second door, the probability he gets caught by Alissa is 1 4 1 4 and the probability he is not caught is 3 4 3 4 . The probability that Alissa catches Muddy coming out of the third door is 1 2 1 2 and the probability she does not catch Muddy is 1 2 1 2 . It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is 1 3 1 3 .

Table 4: Door Choice
Caught or Not Door One Door Two Door Three Total
Caught 1 15 1 15 1 12 1 12 1 6 1 6 ____
Not Caught 4 15 4 15 3 12 3 12 1 6 1 6 ____
Total ____ ____ ____ 1
  • The first entry 1 15 = ( 1 4 ) ( 1 3 ) 1 15 =( 1 4 )( 1 3 ) is P(Door One AND Caught)P(Door One AND Caught).
  • The entry 4 15 = ( 4 5 )( 1 3 ) 4 15 =( 4 5 )( 1 3 ) is P(Door One AND Not Caught)P(Door One AND Not Caught).

Verify the remaining entries.

Problem 1

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

Solution

Table 5: Door Choice
Caught or Not Door One Door Two Door Three Total
Caught 1 15 1 15 1 12 1 12 1 6 1 6 19601960
Not Caught 4 15 4 15 3 12 3 12 1 6 1 6 31603160
Total 515515 412412 2626 1

Problem 2

What is the probability that Alissa does not catch Muddy?

Solution

41604160

Problem 3

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

Solution

919919

Note:

You could also do this problem by using a probability tree. See the Tree Diagrams (Optional) section of this chapter for examples.

Glossary

Contingency Table:
The method of displaying a frequency distribution in case of dependable (contingent) variables; the table provides the easy way to calculate conditional probabilities.

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