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4.5: Contingency Tables and Probability Trees

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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.

Exercise $\PageIndex{1}$

Suppose a public health study on the effectiveness of a campus-wide vaccination clinic produced the following data:

	Contracted the Flu	Did NOT contract the Flu	Total
Received Flu Vaccine	25	280	305
Did NOT receive Flu Vaccine	45	405	450
Total	70	685	755

Table $\PageIndex{1}$

The total number of people in the sample is 755. The row totals (vaccination status) 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.

a. Find P(Person was vaccinated).

b. Find P(Person did not contract the flu).

c. Find P(Person did not contract the flu AND was vaccinated).

d. Find P(Person was vaccinated OR did not contract the flu).

e. Find P(Person was vaccinated GIVEN they contracted the flu).

f. Find P(Person did not contract the flu GIVEN they were NOT vaccinated).

Answer

a. Find P(Person was vaccinated): 305/755 = 0.4040

b. Find P(Person did not contract the flu): 685/755 = 0.9073

c. Find P(Person did not contract the flu AND was vaccinated): 280/755 = 0.3709

d. Find P(Person was vaccinated OR did not contract the flu): (305/755) + (685/755) - (280/755) = 710/755 = 0.9404

e. Find P(Person was vaccinated GIVEN they contracted the flu): 25/70 = 0.3571

f. Find P(Person did not contract the flu GIVEN they were NOT vaccinated): 405/450 = 0.9000

Solution

a. 0.0294, b. 0.1551, c. 0.7165, d. 0.2365, e. 0.2575

Exercise $\PageIndex{2}$

Table (\PageIndex{3}\) 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
Women	18	16		45
Men			14	55
Total		41

Table $\PageIndex{3}$ Hiking Area Preference

a. Complete the table.

b. Are the events "being a woman" and "preferring the coastline" independent events?

Let $F=$ being a woman and let $C=$ preferring the coastline.
1. Find $P(F \cap C)$.
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.

c. Find the probability that a person is a man given that the person prefers hiking near lakes and streams. Let $M=$ being a man, 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: $P$ ( $\qquad$ | $\qquad$ ) = $\qquad$.
3. Is the sample space for this problem all 100 hikers? If not, what is it?

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

1. Find $P(F)$.
2. Find $P(P)$.
3. Find $P(F \cap P)$.
4. Find $P(F \cup P)$.

Answer

Sex	The coastline	Near lakes and streams	On mountain peaks	Total
Women	18	16	11	45
Men	16	25	14	55
Total	34	41	25	100

Table $\PageIndex{4}$ Hiking Area Preference

1. $P(F \cap C)=\dfrac{18}{100}=0.18$
2. $P(F) P(C)=\left(\dfrac{45}{100}\right)\left(\dfrac{34}{100}\right)=(0.45)(0.34)=0.153$
$P(F \cap C) \neq P(F) P(C)$, so the events $F$ and $C$ are not independent.

1. The word 'given' tells you that this is a conditional.
2. $P(M \mid L)=\dfrac{25}{41}$
3. No, the sample space for this problem is the 41 hikers who prefer lakes and streams.

1. $P(F)=\dfrac{45}{100}$
2. $P(P)=\dfrac{25}{100}$
3. $P(F \cap P)=\dfrac{11}{100}$
4. $P(F \cup P)=\dfrac{45}{100}+\dfrac{25}{100}-\dfrac{11}{100}=\dfrac{59}{100}$

$P (A O R B)$

Exercise $\PageIndex{4}$

Table $\PageIndex{9}$ contains the number of Hospitalization incidents per 100,000 individuals in the United States over the span of several years.

Year	Flu	COVID	Injury	Cardiovascular
1	145.7	732.1	29.7	314.7
2	133.1	717.7	29.1	259.2
3	119.3	701	27.7	239.1
4	113.7	702.2	26.8	229.6
Total

Table $\PageIndex{9}$ United States Hospitalization rates Per 100,000 individuals

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

a. P(Year 2 AND Flu)

b. P(Year 3 AND COVID)

c. P(Year 4 AND Injury)

d. P(Cardiovascular AND Year 1)

Answer: a. 0.0294, b. 0.1551, c. 0.2365, d. 0.2575

Solution

a. 0.0294, b. 0.1551, c. 0.7165, d. 0.2365, e. 0.2575

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