## Introduction

### Introduction

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.

Calculate the following probabilities using the table:

1. Find P(Person uses a cell phone while driving).
2. Find P(Person had no violation in the last year).
3. Find P(Person had no violation in the last year and uses a cell phone while driving).
4. Find P(Person uses a cell phone while driving or person had no violation in the last year).
5. Find P(Person uses a cell phone while driving given person had a violation in the last year).
6. Find P(Person had no violation last year given person does not use a cell phone while driving).
Solution 3.20

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

$P(Person uses a cell phone while driving)=number who use cell phones while drivingnumber in study=305755≈.4040P(Person uses a cell phone while driving)=number who use cell phones while drivingnumber in study=305755≈.4040$

b. The marginal distribution is

$P(Person had no violation in the last year)=number who had no violationnumber in study=685755≈.9073.P(Person had no violation in the last year)=number who had no violationnumber in study=685755≈.9073.$

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

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) $=25+405+280755=25+405+280755$

$=710755≈.9404=710755≈.9404$

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.

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

$P(Person had no violation last yearGIVENperson does not use cell phone while driving)=405450=.9P(Person had no violation last yeargivenperson does not use cell phone while driving)=405450=.9$
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
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 3.5 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

a. Complete the table.

Solution 3.21

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

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(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.

Solution 3.21

b.

1. P(F AND C) = $1810018100$ = .18
2. P(F)P(C) = $(45100)(34100)(45100)(34100)$ = (.45)(.34) = .153

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.

1. What word tells you this is a conditional?
2. Is the sample space for this problem all 100 hikers? If not, what is it?
3. Fill in the blanks and calculate the probability: P(_____|_____) = _____.
Solution 3.21

c.

1. The word given tells you that this is a conditional.
2. No, the sample space for this problem is the 41 hikers who prefer lakes and streams.
3. 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) = $25412541$

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(F AND P).
4. Find P(F OR P).
Solution 3.21

d.

1. P(F) = $4510045100$
2. P(P) = $2510025100$
3. P(F AND P) = $number of hikers that are both female AND prefers mountain peaksnumber of hikers in studynumber of hikers that are both female AND prefers mountain peaksnumber of hikers in study$ = $1110011100$
4. P(F OR P) = P(F) + P(P) − P(F AND P) = $4510045100$ + $2510025100$ - $1110011100$ = $5910059100$
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
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 $1515$ and the probability he is not caught is $4545$. If he goes out the second door, the probability he gets caught by Alissa is $1414$ and the probability he is not caught is $3434$. The probability that Alissa catches Muddy coming out of the third door is $1212$ and the probability she does not catch Muddy is $1212$. It is equally likely that Muddy will choose any of the three doors, so the probability of choosing each door is $1313$.

Caught or Not Door One Door Two Door Three Total
Caught $115115$ $112112$ $1616$ ____
Not Caught $415415$ $312312$ $1616$ ____
Total ____ ____ ____ 1
Table 3.8 Door Choice
• The first entry $115 = (15)(13)115 = (15)(13)$ is P(Door One AND Caught).
• The entry $415=(45)(13)415=(45)(13)$ 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 lower-right corner entry is 1.

Solution 3.22

a.

Caught or Not Door One Door Two Door Three Total
Caught $115115$ $112112$ $1616$ $19601960$
Not Caught $415415$ $312312$ $1616$ $41604160$
Total $515515$ $412412$ $2626$ 1
Table 3.9 Door Choice

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

Solution 3.22

b. $41604160$

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

Solution 3.22

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

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

$P(Door One OR Door Two|Caught)=P(Door One OR Door Two AND Caught)P(Caught)=9601960=919.$

### 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 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 3.10 U.S. 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 Crime A).
2. Find P(2010 AND Crime B).
3. Find P(2010 OR Crime B).
4. Find P(2011|Crime A).
5. Find P(Crime D|2008).
Solution 3.23

a. $133.14,520.7133.14,520.7$ = .0294, b. $7014,520.77014,520.7$ = .1551, c. P(2010 OR Crime B) = P(2010) + P(Crime B) − P(2010 AND Crime B) = $1,087.14,520.71,087.14,520.7$ + $2,852.94,520.72,852.94,520.7$$7014,520.77014,520.7$ = .7165, d. $113.7511.8113.7511.8$ = .2222, e. $314.71,222.2314.71,222.2$ = .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 Totals
Under 18 18 28 14
18–50 20 51 28
51+ 12 25 9
Totals
Table 3.11
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 Under 18 and Tall.
4. Find the probability that a randomly chosen individual from this group is tall given that the individual is Under 18.
5. Find the probability that a randomly chosen individual from this group is Under 18 given that the individual is tall.
6. Find the probability a randomly chosen individual from this group is tall and age 51+.
7. Are the events under 18 and tall independent?