4.4 Probability topics: contingency tables (Page 25/2)

Collaborative statistics using Page 1 / 2

This module introduces the contingency table as a way of determining conditional probabilities.

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. Contingincy tables provide a way of portraying data that can facilitate calculating probabilities.

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

	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 $305 + 450 = 755$ and $70 + 685 = 755$ .

Calculate the following probabilities using the table

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

$\frac{number of car phone users}{total number in study} = \frac{305}{755}$

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

$\frac{number that had no violation}{total number in study} = \frac{685}{755}$

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

$\frac{280}{755}$

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

$(\frac{305}{755} + \frac{685}{755}) - \frac{280}{755} = \frac{710}{755}$

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

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

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

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

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

Hiking area preference
Sex	The Coastline	Near Lakes and Streams	On Mountain Peaks	Total
Female	18	16	___	45
Male	___	___	14	55
Total	___	41	___	___

Complete the table.

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

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

Let $F$ = being female and let $C$ = preferring the coastline.

a $P(F AND C)$ =
b $P(F) \cdot P(C)$ =

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.

a $P(F AND C) = \frac{18}{100} = 0.18$
b $P(F) \cdot P(C) = \frac{45}{100} \cdot \frac{34}{100} = 0.45 \cdot 0.34 = 0.153$

$P(F AND C) \neq P(F) \cdot P(C)$ , so the events $F$ and $C$ are not independent.

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.

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

a The word 'given' tells you that this is a conditional.
b $P(M|L) = \frac{25}{41}$
c No, the sample space for this problem is 41.

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

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

a $P(F) = \frac{45}{100}$
b $P(P) = \frac{25}{100}$
c $P(F AND P) = \frac{11}{100}$
d $P(F OR P) = \frac{45}{100} + \frac{25}{100} - \frac{11}{100} = \frac{59}{100}$

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 $\frac{1}{5}$ 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 $\frac{1}{3}$ .

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

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

Verify the remaining entries.

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

Door choice
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	$\frac{4}{15}$	$\frac{3}{12}$	$\frac{1}{6}$	$\frac{41}{60}$
Total	$\frac{5}{15}$	$\frac{4}{12}$	$\frac{2}{6}$	1

What is the probability that Alissa does not catch Muddy?

$\frac{41}{60}$

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

$\frac{9}{19}$

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

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Source: OpenStax, Collaborative statistics using spreadsheets. OpenStax CNX. Jan 05, 2016 Download for free at http://legacy.cnx.org/content/col11521/1.23

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