Probabilities and Contingency Tables

Learning Outcomes

Calculate probabilities for events that are mutually exclusive and not mutually exclusive for a given contingency table
Calculate probabilities for independent events for a given contingency table
Calculate conditional probabilities for a given contingency table
Determine if two events are independent for a given contingency table

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.

The following video shows and example of finding the probability of an event from a table.

Example

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
Cell phone user	$25$	$280$	$305$
Not a cell 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 $305 + 450 = 755 \text{ and } 70 + 685 = 755$ .

Calculate the following probabilities using the table.

Find $P$ (Person is a car phone user).
Find $P$ (Person had no violation in the last year).
Find $P$ (Person had no violation in the last year AND was a car phone user).
Find $P$ (Person is a car phone user OR person had no violation in the last year).
Find $P$ (Person is a car phone user GIVEN person had a violation in the last year).
Find $P$ (Person had no violation last year GIVEN person was not a car phone user).

Show Solution

Solution:

$\displaystyle\frac{{\text{number of car phone users}}}{{\text{total number in study}}}=\frac{{305}}{{755}}$
$\displaystyle\frac{{\text{number that had no violation}}}{{\text{total number in study}}}=\frac{{685}}{{755}}$
$\displaystyle\frac{{280}}{{755}}$
$\displaystyle{(\frac{{305}}{{755}}+\frac{{685}}{{755}})}-\frac{{280}}{{755}}=\frac{{710}}{{755}}$
$\displaystyle\frac{{25}}{{70}}$ (The sample space is reduced to the number of persons who had a violation.)
$\displaystyle\frac{{405}}{{450}}$ (The sample space is reduced to the number of persons who were not car phone users.)

This video shows an example of how to determine the probability of an AND event using a contingency table.

try it

This table 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	$286$	$514$	$800$

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

Show Solution

$P$ (athlete stretches before exercising) = $\displaystyle\frac{{350}}{{800}} = 0.4375$
$P$ (athlete stretches before exercising|no injury in the last year) = $\displaystyle\frac{{295}}{{514}} = 0.5739$

Example

This table shows a random sample of $100$ hikers and the areas of hiking they prefer.

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.
Are the events “being female” and “preferring the coastline” independent events? Let $F$ = being female and let $C$ = preferring the coastline.
1. Find $P(F \text{ AND } C)$ .
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.
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?
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 $P(F \text{ AND } P)$ .
4. Find $P(F \text{ OR } P)$ .

Show Solution

Solution:

1. 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 $34$ $41$ $25$ $100$
2. $P(F \text{ AND }C) = \displaystyle\frac{{18}}{{100}} = 0.18$
  $P(F)P(C) = \displaystyle(\frac{{45}}{{100}})(\frac{{34}}{{100}}) = (0.45)(0.34) = 0.153$
  $P(F \text{ AND } C) \neq P(F)P(C)$ , so the events $F$ and $C$ are not independent.
3. The word given tells you that this is a conditional.
  $\displaystyle{P}{({M}|{L})}=\frac{{25}}{{41}}$
  No, the sample space for this problem is the 41 hikers who prefer lakes and streams.
4. $P(F) = \displaystyle\frac{{45}}{{100}}$
  $P(P) = \displaystyle\frac{{25}}{{100}}$
  $P(F \text{ AND } P) = \displaystyle\frac{{11}}{{100}}$
  $P(F \text{ OR } P) = \displaystyle\frac{{45}}{{100}} + \frac{{25}}{{100}} - \frac{{11}}{{100}} = \frac{{59}}{{100}}$

Sex	The Coastline	Near Lakes and Streams	On Mountain Peaks	Total
Female	$18$	$16$	$11$	$45$
Male	$16$	$25$	$14$	$55$
Total	$34$	$41$	$25$	$100$

try it

This table 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	$71$	$90$	$39$	$200$

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?

Show Solution

$P(H|M) = \displaystyle\frac{{52}}{{90}} = 0.5778$
For $M$ and $H$ to be independent, show $P(H|M) =P(H)$
$P(H|M) = 0.5778, P(H) =\displaystyle\frac{{90}}{{200}} = 0.45$
$P(H|M)$ does not equal $P(H)$ , so $M$ and $H$ are not independent.

Example

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

Door Choice

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

The first entry $\displaystyle\frac{{1}}{{15}}={(\frac{{1}}{{5}})}{(\frac{{1}}{{3}})}$ is $P$ (Door One AND Caught)
The entry $\displaystyle\frac{{4}}{{15}}={(\frac{{4}}{{5}})}{(\frac{{1}}{{3}})}$ 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 $1$ .
What is the probability that Alissa does not catch Muddy?
What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa?

Show Solution

Solution:

Door Choice

Caught or Not	Door One	Door Two	Door Three	Total
Caught	$\displaystyle\frac{{1}}{{15}}$	$\displaystyle\frac{{1}}{{12}}$	$\displaystyle\frac{{1}}{{6}}$	$\displaystyle\frac{{19}}{{60}}$
Not Caught	$\displaystyle\frac{{4}}{{15}}$	$\displaystyle\frac{{3}}{{12}}$	$\displaystyle\frac{{1}}{{6}}$	$\displaystyle\frac{{41}}{{60}}$
Total	$\displaystyle\frac{{5}}{{15}}$	$\displaystyle\frac{{4}}{{12}}$	$\displaystyle\frac{{2}}{{16}}$	$1$

$\displaystyle\frac{{41}}{{60}}$
$\displaystyle\frac{{9}}{{19}}$

example

This table contains the number of crimes per $100,000$ inhabitants from 2008 to 2011 in the U.S.

Year	Robbery	Burglary	Rape	Vehicle
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

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

Find $P$ (2009 AND Robbery).
Find $P$ (2010 AND Burglary).
Find $P$ (2010 OR Burglary).
Find $P$ (2011|Rape).
Find $P$ (Vehicle|2008).

Show Solution

Solution:

$0.0294$
$0.1551$
$0.7165$
$0.2365$
$0.2575$

This video gives and example of determining an “OR” probability given a table.

try it

This table relates the weights and heights of a group of individuals participating in an observational study.

Weight/Height	Tall	Medium	Short
Obese	$18$	$28$	$14$
Normal	$20$	$51$	$28$
Underweight	$12$	$25$	$9$
Totals

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 Obese and Tall.
Find the probability that a randomly chosen individual from this group is Tall given that the idividual is Obese.
Find the probability that a randomly chosen individual from this group is Obese given that the individual is Tall.
Find the probability a randomly chosen individual from this group is Tall and Underweight.
Are the events Obese and Tall independent?

Show Solution

Weight/Height	Tall	Medium	Short	Total
Obese	$18$	$28$	$14$	$60$
Normal	$20$	$51$	$28$	$99$
Underweight	$12$	$25$	$9$	$46$
Totals	$50$	$104$	$51$	$205$

Row Totals: $60, 99, 46$ . Column totals: $50, 104, 51$ .
$P$ (Tall) = $\displaystyle\frac{{50}}{{205}} = 0.244$
$P$ (Obese AND Tall) = $\displaystyle\frac{{18}}{{205}} = 0.088$
$P$ (Tall|Obese) = $\displaystyle\frac{{18}}{{60}} = 0.3$
$P$ (Obese|Tall) = $\displaystyle\frac{{18}}{{50}} = 0.36$
$P$ (Tall AND Underweight = $\displaystyle\frac{{12}}{{205}} = 0.0585$
No. $P$ (Tall) does not equal $P$ (Tall|Obese).

Module 3: Probability

Learning Outcomes

Example

try it

Example

try it

Example

example

try it

Candela Citations