Probability

15 Contingency Tables



[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.

Example 3.20

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
Table 3.2

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.

 

  1. Find P(Driver is a cell phone user).
  2. Find P(driver had no violation in the last year).
  3. Find P(Driver had no violation in the last year AND was a cell phone user).
  4. Find P(Driver is a cell phone user OR driver had no violation in the last year).
  5. Find P(Driver is a cell phone user GIVEN driver had a violation in the last year).
  6. Find P(Driver had no violation last year GIVEN driver was not a cell phone user)

Solutions:

  1. $\frac{\text{number of cell phone users}}{\text{total number in study}}= \frac{305}{755} = 0.404$
  2. $\frac{\text{number that had no violation}}{\text{total number in study}}= \frac{685}{755} = 0.9073$
  3. $\frac{280}{755} = 0.3709$
  4. $\left(\frac{305}{755}+\frac{685}{755}\right)-\frac{280}{755} = \frac{710}{755}=0.9404$
  5. $\frac{25}{70}=0.3571$
  6. $\frac{405}{450}=0.9$

Try It 3.20

Table 3.3 below 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
Table 3.3
  1. What is P(athlete stretches before exercising)?
  2. What is P(athlete stretches before exercising|no injury in the last year)?

Example 3.21

Table 3.4 below 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 ___ ___
Table 3.4 Hiking Area Preference
  1. Complete the table.
  2. 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.

  3. 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?
  4. 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).

Try It 3.21

Table 3.6 below 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
Table 3.6
  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?

Example 3.22

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 $\frac{1}{5}$ and the probability he is not caught is $\frac{4}{5}$ . 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 $\frac{1}{15}$ $\frac{1}{12}$ $\frac{1}{6}$ ____
Not Caught $\frac{4}{15}$ $\frac{3}{12}=\frac{1}{4}$ $\frac{1}{6}$ ____
Total ____ ____ ____ 1
Table 3.7 Door Choice
  • The first entry $\frac{1}{15}=\left( \frac{1}{5} \right) \left(  \frac{1}{3}\right)$ is $P(\text{Door One AND Caught})$
  • The entry $\frac{4}{15}=\left( \frac{4}{5} \right) \left(  \frac{1}{3}\right)$ is $P(\text{Door One AND Not Caught})$

Verify the remaining entries.

 

  1. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.
  2. What is the probability that Alissa does not catch Muddy?
  3. What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa?

Solution

  1. Caught or Not Door One Door Two Door Three Total
    Caught $\frac{1}{15}$ $\frac{1}{12}$ $\frac{1}{6}$ $\frac{19}{60}$
    Not Caught $\frac{4}{15}$ $\frac{3}{12}=\frac{1}{4}$ $\frac{1}{6}$ $\frac{41}{60}$
    Total $\frac{5}{15}=\frac{1}{3}$ $\frac{4}{12}=\frac{1}{3}$ $\frac{2}{6}=\frac{1}{3}$ 1
  2. $\frac{41}{60}=0.6833$
  3. $\frac{9}{19}=0.4737$

Example 3.23

Table 3.9 below 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
Table 3.9 United States Crime Index Rates Per 100,000 Inhabitants 2008–2011

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).

Solution

  1. 0.0294
  2. 0.1551
  3. 0.7165
  4. 0.2365
  5. 0.2575

Try It 3.23

Table 3.10 below 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
Table 3.10
  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?

License

Icon for the Creative Commons Attribution 4.0 International License

Introductory Statistics with Google Sheets by jkesler is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.