# 3.5: Contingency Tables and Probability Trees

## Contingency Tables

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.

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 cell phone while driving 25 280 305
Does not use cell phone while driving 45 405 450
Total 70 685 755

Table3.2

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.

a. Find P(Driver is a cell phone user).

Solution 3.20

a. $$\frac{\text { number of cell phone users }}{\text { total number in study }}=\frac{305}{755}$$

b. Find P(Driver had no violation in the last year).

Solution 3.20

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

c. Find P(Driver had no violation in the last year $$\cap$$ was a cell phone user).

Solution 3.20

c. $$\frac{280}{755}$$

d. Find P(Driver is a cell phone user $$\cup$$ driver had no violation in the last year).

Solution 3.20

d. $$\left(\frac{305}{755}+\frac{685}{755}\right)-\frac{280}{755}=\frac{710}{755}$$

e. Find P(Driver is a cell phone user $$|$$ driver had a violation in the last year).

Solution 3.20

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

f. Find P(Driver had no violation last year $$|$$ driver was not a cell phone user)

Solution 3.20

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

Exercise 3.20

Table 3.3 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

Table3.3

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 3.4 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 ___ ___

Table3.4 Hiking Area Preference

a. Complete the table.

Solution 3.21

a.

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

Table3.5 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 $$P(F\cap C)$$.
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.

Solution 3.21

b.

1. $$P(F\cap C)=\frac{18}{100}$$ = 0.18
2. P(F)P(C) = $$\left(\frac{45}{100}\right)\left(\frac{34}{100}\right)$$ = (0.45)(0.34) = 0.153

$$P(F\cap C)$$ ≠ 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. 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?

Solution 3.21

c.

1.The word 'given' tells you that this is a conditional.

2.P(M||L) = $$\frac{25}{41}$$

3.No, the sample space for this problem is the 41 hikers who prefer lakes and streams.

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 $$P(F\cap P)$$.
4. Find $$P(F\cup P)$$.

Solution 3.21

d.

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

Exercise 3.21

Table 3.6 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

Table3.6

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 $$\frac{1}{15}$$ $$\frac{1}{12}$$ $$\frac{1}{6}$$ ____
Not caught $$\frac{4}{15}$$ $$\frac{3}{12}$$ $$\frac{1}{6}$$ ____
Total ____ ____ ____ 1

Table3.7 Door Choice

• The first entry $$\frac{1}{15}=\left(\frac{1}{5}\right)\left(\frac{1}{3}\right)$$ is $$P(Door One\cap Caught)$$
• The entry $$\frac{4}{15}=\left(\frac{4}{5}\right)\left(\frac{1}{3}\right)$$ is $$P(Door One\cap Not Caught)$$

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 $$\frac{1}{15}$$ $$\frac{1}{12}$$ $$\frac{1}{6}$$ $$\frac{19}{60}$$
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

Table3.8 Door Choice

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

Solution 3.22

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

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

Solution 3.22

c. $$\frac{9}{19}$$

Example 3.23

Table 3.9 contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the U.S.

Year Robbery Burglary Rape Vehicle 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

Table3.9 United States Crime Index Rates Per 100,000 Inhabitants 2008–2011

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

1. Find $$P(2009\cap Robbery)$$.
2. Find $$P(2010\cap Burglary)$$.
3. Find $$P(2010\cup Burglary)$$.
4. Find P(2011|Rape).
5. Find P(Vehicle|2008).

Solution 3.23

1. 0.0294
2. 0.1551
3. 0.7165
4. 0.2365
5. 0.2575

Exercise 3.23

Table 3.10 relates the weights and heights of a group of individuals participating in an observational study.

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

Table3.10

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

## Tree Diagrams

Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams can be used to visualize and solve conditional probabilities.

### Tree Diagrams

A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.

Example 3.24

In an urn, there are 11 balls. Three balls are red (R) and eight balls are blue (B). Draw two balls, one at a time, with replacement. "With replacement" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.

The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as R1, R2, and R3 and each blue ball as B1, B2, B3, B4, B5, B6, B7, and B8. Then the nine RR outcomes can be written as:

R1R1; R1R2; R1R3; R2R1; R2R2; R2R3; R3R1; R3R2; R3R3

The other outcomes are similar.

There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space.

a. List the 24 BR outcomes: B1R1, B1R2, B1R3, ...

Solution 3.24

a. B1R1; B1R2; B1R3; B2R1; B2R2; B2R3; B3R1; B3R2; B3R3; B4R1; B4R2; B4R3; B5R1; B5R2; B5R3; B6R1; B6R2; B6R3; B7R1; B7R2; B7R3; B8R1; B8R2; B8R3

b. Using the tree diagram, calculate P(RR).

Solution 3.24

b. P(RR) = $$\left(\frac{3}{11}\right)\left(\frac{3}{11}\right) = \frac{9}{121}$$

c. Using the tree diagram, calculate P(RB\cup BR)P(RB\cup BR).

Solution 3.24

c. $$P(RB\cup BR)$$ = $$\left(\frac{3}{11}\right)\left(\frac{8}{11}\right)+\left(\frac{8}{11}\right)\left(\frac{3}{11}\right)=\frac{48}{121}$$

d. Using the tree diagram, calculate $$P(Ron 1st draw\cap Bon 2nd draw)$$.

Solution 3.24

d. $$P(Ron 1st draw\cap Bon 2nd draw) = \left(\frac{3}{11}\right)\left(\frac{8}{11}\right)=\frac{24}{121}$$

e. Using the tree diagram, calculate P(R on 2nd draw|B on 1st draw).

Solution 3.24

e. P(R on 2nd draw|B on 1st draw) = P(R on 2nd|B on 1st) = $$\frac{24}{88} = \frac{3}{11}$$

This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 BR and 64 BB). Twenty-four of the 88 possible outcomes are BR. $$\frac{24}{88} = \frac{3}{11}$$.

f. Using the tree diagram, calculate P(BB).

Solution 3.24

f. P(BB) = $$\frac{64}{121}$$

g. Using the tree diagram, calculate P(B on the 2nd draw|R on the first draw).

Solution 3.24

g. P(B on 2nd draw|R on 1st draw) = $$\frac{8}{11}$$

There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have B on the second draw. The probability is then $$\frac{24}{33}$$.

Exercise 3.24

In a standard deck, there are 52 cards. 12 cards are face cards (event F) and 40 cards are not face cards (event N). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate P(FF).

Example 3.25

An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. "Without replacement" means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, $$\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}$$.

NOTE

If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement, so that on the second draw there are ten marbles left in the urn.

Calculate the following probabilities using the tree diagram.

a. P(RR) = ________

Solution 3.25

a. P(RR) = $$\left(\frac{3}{11}\right)\left(\frac{2}{10}\right)=\frac{6}{110}$$

b. Fill in the blanks:

$$P(RB\cup BR) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right)$$ + (___)(___) = $$\frac{48}{110}$$

Solution 3.25

b. $$P(RB\cup BR) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right)+\left(\frac{8}{11}\right)\left(\frac{3}{10}\right)=\frac{48}{110}$$

c. P(R on 2nd|B on 1st) =

Solution 3.25

c. P(R on 2nd|B on 1st) = $$\frac{3}{10}$$

d. Fill in the blanks.

$$P(Ron 1st\cap Bon 2nd)$$ = (___)(___) = $$\frac{24}{100}$$

Solution 3.25

d. $$P(R \text{ on 1st }\cap B \text{ on 2nd}) = \left(\frac{3}{11}\right)\left(\frac{8}{10}\right)=\frac{24}{110}$$

e. Find P(BB).

Solution 3.25

e. P(BB) = $$\left(\frac{8}{11}\right)\left(\frac{7}{10}\right)$$

f. Find P(B on 2nd|R on 1st).

Solution 3.25

f. Using the tree diagram, P(B on 2nd|R on 1st) = P(R|B) = $$\frac{8}{10}$$.

If we are using probabilities, we can label the tree in the following general way.

• P(R|R) here means P(R on 2nd|R on 1st)
• P(B|R) here means P(B on 2nd|R on 1st)
• P(R|B) here means P(R on 2nd|B on 1st)
• P(B|B) here means P(B on 2nd|B on 1st)

Exercise 3.25

In a standard deck, there are 52 cards. Twelve cards are face cards (F) and 40 cards are not face cards (N). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities.

1. Find $$P(FN\cup NF)$$.
2. Find P(N|F).
3. Find P(at most one face card).
Hint: "At most one face card" means zero or one face card.
4. Find P(at least on face card).
Hint: "At least one face card" means one or two face cards.

Example 3.26

A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption.

1. What is the probability that both kittens are tabby?

$$a \cdot\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) b \cdot\left(\frac{4}{9}\right)\left(\frac{4}{9}\right) c \cdot\left(\frac{4}{9}\right)\left(\frac{3}{8}\right) d \cdot\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)$$
2. What is the probability that one kitten of each coloring is selected?

a.$$\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)$$ b.$$\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)$$ c.$$\left(\frac{4}{9}\right)\left(\frac{5}{9}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{9}\right)$$ d.$$\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)+\left(\frac{5}{9}\right)\left(\frac{4}{8}\right)$$
3. What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?
4. What is the probability of choosing two kittens of the same color?
a. c, b. d, c. $$\frac{4}{8}$$, d. $$\frac{32}{72}$$