Contents

Contingency Tables

A two-way table provides a way of portraying data that can facilitate calculating probabilities. When used to calculate probabilities, a two-way table is often called a contingency table. 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. We used two-way tables in Chapters 1 and 2 to calculate marginal and conditional distributions. These tables organize data in a way that supports the calculation of relative frequency and, therefore, experimental (empirical) probability. 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 a cell phone while driving	25	280	305
Does not use a cell phone while driving	45	405	450
Total	70	685	755

Table 3.3

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.

Using the table, calculate the following probabilities:

Problem

Find P(Person uses a cell phone while driving).
Find P(Person had no violation in the last year).
Find P(Person had no violation in the last year and uses a cell phone while driving).
Find P(Person uses a cell phone while driving or person had no violation in the last year).
Find P(Person uses a cell phone while driving given person had a violation in the last year).
Find P(Person had no violation last year given person does not use a cell phone while driving).

Show solution

Solution

a. This is the same as the marginal distribution (Section 1.2).

P (Person uses a cell phone while driving) = \frac{number who use cell phones while driving}{number in study} = \frac{305}{755} \approx .4040 P (Person uses a cell phone while driving) = \frac{number who use cell phones while driving}{number in study} = \frac{305}{755} \approx .4040

3.5

b. The marginal distribution is

P (Person had no violation in the last year) = \frac{number who had no violation}{number in study} = \frac{685}{755} \approx .9073 . P (Person had no violation in the last year) = \frac{number who had no violation}{number in study} = \frac{685}{755} \approx .9073 .

3.6

c. Find the number of participants who satisfy both conditions.

\begin{matrix} P (P e r s o n h a d n o v i o l a t i o n i n t h e l a s t y e a r A N D u s e s a c e l l p h o n e w h i l e d r i v i n g) = \frac{n u m b e r w h o h a d n o v i o l a t i o n A N D u s e s c e l l p h o n e w h i l e d r i v i n g}{n u m b e r i n s t u d y} \\ = \frac{280}{755} \\ \approx .3709 \end{matrix} \begin{matrix} P (P e r s o n h a d n o v i o l a t i o n i n t h e l a s t y e a r A N D u s e s a c e l l p h o n e w h i l e d r i v i n g) = \frac{n u m b e r w h o h a d n o v i o l a t i o n A N D u s e s c e l l p h o n e w h i l e d r i v i n g}{n u m b e r i n s t u d y} \\ = \frac{280}{755} \\ \approx .3709 \end{matrix}

3.7

d. To find this probability, you need to identify how many participants use a cell phone while driving OR have no violation in the past year OR both.

$P (Person uses a cell phone while driving OR had no violation in the last year) = \frac{25 + 405 + 280}{755} P (Person uses a cell phone while driving) = \frac{25 + 405 + 280}{755}$

\begin{matrix} = \frac{710}{755} \\ \approx .9404 \end{matrix} \begin{matrix} = \frac{710}{755} \\ \approx .9404 \end{matrix}

3.8

e. This is a conditional probability. You are given that the person had no violation in the last year, so you need only consider the values in that column of data.

\begin{matrix} (P e r s o n u s e s a c e l l p h o n e w h i l e d r i v i n g G I V E N t h e p e r s o n h a d a v i o l a t i o n i n t h e l a s t y e a r) = \frac{n u m b e r w h o u s e d c e l l p h o n e A N D ​ h a d a v i o l a t i o n}{n u m b e r i n s t u d y w h o h a d a v i o l a t i o n i n t h e l a s t y e a r} \\ = \frac{25}{70} \\ \approx .3571 \end{matrix} \begin{matrix} (P e r s o n u s e s a c e l l p h o n e w h i l e d r i v i n g G I V E N t h e p e r s o n h a d a v i o l a t i o n i n t h e l a s t y e a r) = \frac{n u m b e r w h o u s e d c e l l p h o n e A N D ​ h a d a v i o l a t i o n}{n u m b e r i n s t u d y w h o h a d a v i o l a t i o n i n t h e l a s t y e a r} \\ = \frac{25}{70} \\ \approx .3571 \end{matrix}

3.9

f. For this conditional probability, consider only values in the row labeled “Does not use a cell phone while driving.”

P (Person had no violation last year GIVEN person does not use cell phone while driving) = \frac{405}{450} = .9 P (Person had no violation last year) = \frac{405}{450} = .9

Try It 3.20

Table 3.4 shows the number of athletes who stretch before exercising and how many had injuries within the past year.

	Injury in Past Year	No Injury in Past Year	Total
Stretches	55	295	350
Does not stretch	231	219	450
Total	286	514	800

Table 3.4

What is P(Athlete stretches before exercising)?
What is P(Athlete stretches before exercising|no injury in the last year)?

Example 3.21

Table 3.5 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 3.5 Hiking Area Preference

Problem

a. Complete the table.

Show solution

Solution

a. There are 45 females in the sample; 18 prefer the coastline and 16 prefer hiking near lakes and streams. So, we know there are 45 − 18 − 16 = 11 female students who prefer hiking on mountain peaks.

Continue reasoning in this way to complete the table.

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

Table 3.6 Hiking Area Preference

Problem

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.

Show solution

Solution

P(F AND C) = $\frac{18}{100} \frac{18}{100}$ = .18
P(F)P(C) = $(\frac{45}{100}) (\frac{34}{100}) (\frac{45}{100}) (\frac{34}{100})$ = (.45)(.34) = .153

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

Problem

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?
Is the sample space for this problem all 100 hikers? If not, what is it?
Fill in the blanks and calculate the probability: P(_____|_____) = _____.

Show solution

Solution

The word given tells you that this is a conditional.
No, the sample space for this problem is the 41 hikers who prefer lakes and streams.
Find the conditional probability P(M|L). Because it is given that the person prefers hiking near lakes and streams, you need only consider the values in the column labeled "Near Lakes and Streams." P(M|L) = $\frac{25}{41} \frac{25}{41}$

Problem

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).

Show solution

Solution

P(F) = $\frac{45}{100} \frac{45}{100}$
P(P) = $\frac{25}{100} \frac{25}{100}$
P(F AND P) = $\frac{number of hikers that are both female AND prefers mountain peaks}{number of hikers in study} \frac{number of hikers that are both female AND prefers mountain peaks}{number of hikers in study}$ = $\frac{11}{100} \frac{11}{100}$
P(F OR P) = P(F) + P(P) − P(F AND P) = $\frac{45}{100} \frac{45}{100}$ + $\frac{25}{100} \frac{25}{100}$ - $\frac{11}{100} \frac{11}{100}$ = $\frac{59}{100} \frac{59}{100}$

Try It 3.21

Table 3.7 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 3.7

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?

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

Caught or Not	Door One	Door Two	Door Three	Total
Caught	$\frac{1}{15} \frac{1}{15}$	$\frac{1}{12} \frac{1}{12}$	$\frac{1}{6} \frac{1}{6}$	____
Not Caught	$\frac{4}{15} \frac{4}{15}$	$\frac{3}{12} \frac{3}{12}$	$\frac{1}{6} \frac{1}{6}$	____
Total	____	____	____	1

Table 3.8 Door Choice

The first entry $\frac{1}{15} = (\frac{1}{5}) (\frac{1}{3}) \frac{1}{15} = (\frac{1}{5}) (\frac{1}{3})$ is P(Door One AND Caught).
The entry $\frac{4}{15} = (\frac{4}{5}) (\frac{1}{3}) \frac{4}{15} = (\frac{4}{5}) (\frac{1}{3})$ is P(Door One AND Not Caught).

Verify the remaining entries.

Problem

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

Show solution

Solution

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

Table 3.9 Door Choice

Problem

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

Show solution

Solution

b. $\frac{41}{60} \frac{41}{60}$

Problem

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

Show solution

Solution

c. This is a conditional probability, so consider only probabilities in the row labeled "Caught." Choosing Door One and choosing Door Two are mutually exclusive, so

P (Choosing Door One OR Choosing Door Two AND Caught) = \frac{1}{15} + \frac{1}{12} = \frac{9}{60} . P (Choosing Door One OR Choosing Door Two AND Caught) = \frac{1}{15} + \frac{1}{12} = \frac{9}{60} .

Use the formula for conditional probability $P (A | B) = \frac{P (A AND B)}{P (B)} . P (A | B) = \frac{P (A AND B)}{P (B)} .$

P (Door One OR Door Two|Caught) = \frac{P (Door One OR Door Two AND Caught)}{P (Caught)} = \frac{\frac{9}{60}}{\frac{19}{60}} = \frac{9}{19.} . P (Door One OR Door Two|Caught) = \frac{P (Door One OR Door Two AND Caught)}{P (Caught)} = \frac{\frac{9}{60}}{\frac{19}{60}} = \frac{9}{19.} .

Example 3.23

Table 3.10 contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the United States.

Year	Crime A	Crime B	Crime C	Crime D
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 3.10 U.S. Crime Index Rates Per 100,000 Inhabitants 2008–2011

Problem

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

Find P(2009 AND Crime A).
Find P(2010 AND Crime B).
Find P(2010 OR Crime B).
Find P(2011|Crime A).
Find P(Crime D|2008).

Show solution

Solution

a. $\frac{133 .1}{4,520 .7} \frac{133 .1}{4,520 .7}$ = .0294, b. $\frac{701}{4,520 .7} \frac{701}{4,520 .7}$ = .1551, c. P(2010 OR Crime B) = P(2010) + P(Crime B) – P(2010 AND Crime B) = $\frac{1,087 .1}{4,520 .7} \frac{1,087 .1}{4,520 .7}$ + $\frac{2,852 .9}{4,520 .7} \frac{2,852 .9}{4,520 .7}$ − $\frac{701}{4,520 .7} \frac{701}{4,520 .7}$ = .7165, d. $\frac{113 .7}{511 .8} \frac{113 .7}{511 .8}$ = .2222, e. $\frac{314 .7}{1,222 .2} \frac{314 .7}{1,222 .2}$ = .2575

Try It 3.23

Table 3.11 relates the weights and heights of a group of individuals participating in an observational study.

Ages	Tall	Medium	Short
Under 18	18	28	14
18–50	20	51	28
51+	12	25	9
Totals

Table 3.11

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 Under 18 and tall.
Find the probability that a randomly chosen individual from this group is tall given that the individual is Under 18.
Find the probability that a randomly chosen individual from this group is Under 18 given that the individual is tall.
Find the probability a randomly chosen individual from this group is tall and age 51+.
Are the events under 18 and tall independent?