Math II · S-CP.4

Using Two-Way Frequency Tables as Sample Spaces for Independence and Conditional Probability

Two-way tables turn messy categorical data into a probability map, letting students see totals, overlaps, conditions, and dependence clearly.

Concept Statistics and Probability
Domain Conditional Probability and the Rules of Probability
Read time 7 minutes

What this learning objective is really asking you to learn

This objective asks students to use two-way frequency tables as probability tools. A two-way frequency table organizes data according to two categorical variables. The rows represent categories of one variable, the columns represent categories of another, and the cells show how many observations fall into each combination.

For example, suppose a school surveys students about whether they participate in sports and whether they participate in music. A two-way table might show counts for students who do both, sports only, music only, or neither. The table becomes a sample space. Every surveyed student is one outcome, and the table organizes those outcomes into categories.

This objective builds directly on earlier work with two-way tables and probability language. Students should be able to interpret joint frequencies, marginal frequencies, and conditional frequencies.

A joint frequency is a count in an interior cell, representing both categories at once. For example, “plays sports and plays music.”

A marginal frequency is a row or column total, representing one category regardless of the other. For example, “plays sports” total.

A conditional frequency uses a row or column total as a restricted sample space. For example, “among students who play music, what fraction also play sports?”

The objective also asks students to use tables to reason about independence. If the conditional probability of one event is the same as its overall probability, the events are independent. If it changes, the events are dependent. Two-way tables make that comparison visible.

This objective matters because real data is often categorical. People are grouped by yes/no responses, grade level, preference, treatment/control group, outcome, region, device type, subscription status, and more. Two-way tables are one of the simplest tools for seeing relationships between such categories.

Why students should learn this math

Students should learn two-way tables because they are one of the most practical ways to analyze categorical data. A huge amount of real-world information comes in categories: voted or did not vote, passed or failed, subscribed or canceled, tested positive or negative, recovered or did not recover, uses app or does not use app, prefers A or B, bought product or did not buy product.

A one-way table can show how common one category is. A two-way table shows how two categories interact. That interaction is often the real question. Are students who attend tutoring more likely to pass? Are customers who use a feature less likely to cancel? Are people in one age group more likely to prefer a product? Are patients receiving a treatment more likely to recover? Are two survey responses related?

Two-way tables help students avoid vague claims. Instead of saying “sports and music seem related,” students can compute conditional percentages. What percentage of music students play sports? What percentage of non-music students play sports? If the percentages differ, the variables may be associated. If they are the same or very close, the data may suggest independence.

This is also a key skill for interpreting news and research. Many public claims are based on categorical comparisons: risk by group, outcome by treatment, preference by demographic, success by program participation. Without two-way-table fluency, students can be misled by raw counts. A group may have more cases simply because it is larger. Conditional percentages often tell the more meaningful story.

For example, if 80 students who did not attend tutoring passed and 20 tutoring students passed, raw counts make tutoring look worse. But if 100 non-tutoring students and 20 tutoring students were surveyed, the pass rates are 80% and 100%. The conditional rates tell a different story. Two-way tables teach students to choose the right denominator.

The “why” is that two-way tables are the bridge between data and probability reasoning. They make overlap, totals, conditions, and independence visible.

The historical machinery: categorical data and contingency tables

Two-way frequency tables are often called contingency tables in statistics. They became important because researchers needed a way to study whether categorical variables were associated. Are disease and exposure related? Are treatment and recovery related? Are gender and vote choice related? Are machine type and defect type related?

As statistics developed, contingency tables became a basic tool for organizing observed counts. Later methods such as chi-square tests built on them to determine whether observed differences were larger than might be expected by chance. Students in Math II are not doing formal chi-square inference yet, but they are learning the table structure that makes such inference possible.

The idea is simple but powerful: cross-classify observations by two variables. Once data is organized this way, you can compute joint, marginal, and conditional frequencies. You can compare conditional distributions. You can ask whether variables appear independent.

This history matters because students should not see two-way tables as a school formatting exercise. They are a foundational data-analysis tool used in medicine, social science, business analytics, education research, public policy, and quality control.

Where this fits in the big map of mathematics

This objective sits at the intersection of probability and statistics. It uses data counts to compute probabilities. It connects earlier descriptive statistics with later probability rules.

It connects to Objective 053, where students used two-way tables for categorical data. Now the same table becomes a sample space for probability.

It connects to sample-space language. Each cell is an intersection of two events. Row and column totals represent broader events. The grand total represents the full sample space.

It connects to conditional probability. Conditional probabilities are computed by restricting to a row or column. The denominator changes depending on what is given.

It connects to independence. Tables make independence testable by comparing joint probability with product of marginals or by comparing conditional probabilities with overall probabilities.

It connects to inference. Later, students will evaluate reports based on data and distinguish surveys, experiments, and observational studies. Two-way tables are central to those tasks.

The big-map role is data structure. Students learn to turn categorical data into a probability map.

How to execute the skill technically

Start by identifying the two categorical variables. For example: tutoring status and pass/fail outcome. Create rows for one variable and columns for the other. Fill in counts carefully. Then compute row totals, column totals, and the grand total.

Example:

| | Passed | Did not pass | Total | |---|---:|---:|---:| | Tutoring | 18 | 2 | 20 | | No tutoring | 72 | 18 | 90 | | Total | 90 | 20 | 110 |

The grand total is 110 students.

A joint probability uses an interior cell divided by the grand total. For example:

\[P(tutoring and passed) = 18/110\].

A marginal probability uses a row or column total divided by the grand total. For example:

\[P(tutoring) = 20/110\].
\[P(passed) = 90/110\].

A conditional probability restricts the denominator. For example:

\[P(passed | tutoring) = 18/20 = 0.90\].
\[P(passed | no tutoring) = 72/90 = 0.80\].

These conditional probabilities compare pass rates within each tutoring group. In this table, students who attended tutoring passed at a higher rate.

To test independence, compare \(P(passed | tutoring)\) with \(P(passed)\). The overall pass rate is \(90/110 ≈ 0.818\). The tutoring pass rate is 0.90. Since these are not equal, tutoring status and passing are not independent in this data set.

Alternatively, compare \(P(tutoring and passed)\) with \(P(tutoring)P(passed)\). If equal, independence holds; if not, it does not.

Students should also learn to interpret in words. \(18/20\) means “among students who attended tutoring, 18 out of 20 passed.” That is different from \(18/110\), which means “18 out of all 110 surveyed students both attended tutoring and passed.” The denominator changes the meaning.

Another worked example: reading association from conditional distributions

Suppose a travel app surveys 300 users about whether they bought travel insurance and whether they traveled internationally.

| | Bought insurance | Did not buy insurance | Total | |---|---:|---:|---:| | International trip | 90 | 60 | 150 | | Domestic trip | 45 | 105 | 150 | | Total | 135 | 165 | 300 |

The overall probability of buying insurance is

\[P(insurance) = 135/300 = 0.45\].

Among international travelers,

\[P(insurance | international) = 90/150 = 0.60\].

Among domestic travelers,

\[P(insurance | domestic) = 45/150 = 0.30\].

These conditional probabilities are different from each other and from the overall probability. That suggests trip type and buying insurance are not independent in this table. International travelers bought insurance at a higher rate.

Now test with the multiplication rule:

\[P(international) = 150/300 = 0.50\].
\[P(insurance) = 135/300 = 0.45\].

If independent, \(P(international \cap insurance)\) would be \(0.50 \cdot 0.45 = 0.225\).

The actual joint probability is \(90/300 = 0.30\).

Since \(0.30 \ne 0.225\), the events are not independent.

Why two-way tables are better than memory for probability

Students often try to solve probability problems by remembering which formula applies. Two-way tables reduce that burden because they show the structure. The joint count is inside the table. The marginal totals are at the edges. Conditional denominators are row or column totals. Independence can be checked by comparing conditional percentages.

This is exactly how many professionals work. They do not keep everything in their heads. They structure data so the right comparisons are visible. A two-way table is a thinking tool. It prevents denominator mistakes, makes overlap visible, and turns a vague association question into a clear numerical comparison.

Common misconceptions and how to avoid them

One common mistake is using the grand total for every probability. Conditional probabilities require row or column totals.

Another mistake is confusing joint and conditional probabilities. \(P(A and B)\) and \(P(A | B)\) are different.

A third mistake is comparing raw counts instead of percentages. If groups are different sizes, raw counts can mislead.

A fourth mistake is reading table direction incorrectly. \(P(passed | tutoring)\) uses the tutoring row as the denominator. \(P(tutoring | passed)\) uses the passed column as the denominator. These are not the same.

A fifth mistake is claiming causation from a two-way table alone. A table can show association, but unless the data come from a well-designed experiment, it does not prove one variable caused the other.

The big takeaway

Two-way tables organize categorical data into a probability map. Interior cells show joint outcomes. Margins show totals. Rows and columns can become restricted sample spaces for conditional probability. Comparing conditional probabilities helps students judge independence or association. This is one of the most practical probability tools students learn because it connects directly to real data and real claims.

Problem Library

Problems in the App From This Objective

207 problems across 15 archetypes in the app.

organize counts by two variables.
15 problems Warmup Practice Mixed Review Assessment
Problem 1

Construct a two-way table from raw categorical data 12 ninth graders like tea, 8 ninth graders do not, 10 tenth graders like tea, 20 tenth graders do not.

Problem 2

Construct a two-way table from raw categorical data male yes 15, male no 5, female yes 18, female no 12.

Problem 3

Construct a two-way table from raw categorical data red small 6, red large 4, blue small 5, blue large 7.

Problem 4

Construct a two-way table from raw categorical data freshman bus 9, freshman car 6, sophomore bus 8, sophomore car 12.

Problem 5

Construct a two-way table from raw categorical data 10 indoor cats, 5 outdoor cats, 12 indoor dogs, 8 outdoor dogs.

Problem 6

Construct a two-way table from raw categorical data 15 males play soccer, 10 males play basketball, 18 females play soccer, 12 females play basketball.

Problem 7

Construct a two-way table from raw categorical data 20 people liked action movies, 5 disliked action movies, 18 people liked comedy movies, 7 disliked comedy movies.

Problem 8

Construct a two-way table from raw categorical data math A 10, math B 8, science A 12, science B 7.

Problem 9

Construct a two-way table from raw categorical data 10 people drink coffee in the morning, 5 drink coffee in the afternoon, 8 people drink juice in the morning, 12 drink juice in the afternoon.

Problem 10

Construct a two-way table from raw categorical data 15 ripe apples, 5 unripe apples, 10 ripe bananas, 8 unripe bananas.

Problem 11

Construct a two-way table from raw categorical data white cars 20, black cars 15, white trucks 10, black trucks 8.

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Problem 12

Construct a two-way table from raw categorical data 12 people swim in summer, 3 people ski in summer, 2 people swim in winter, 18 people ski in winter.

Problem 13

Construct a two-way table from raw categorical data 10 teens like pop, 5 teens like rock, 15 adults like pop, 12 adults like rock.

Problem 14

Construct a two-way table from raw categorical data 25 people like pizza, 5 dislike pizza, 18 people like pasta, 7 dislike pasta.

Problem 15

Construct a two-way table from raw categorical data sunny happy 15, sunny sad 3, rainy happy 5, rainy sad 10.

use row, column, and grand totals.
12 problems Warmup Practice Mixed Review Assessment
Problem 16

Complete missing totals in a two-way table from row entries 12 and 8.

Problem 17

Complete missing totals in a two-way table from column entries 15 and 10.

Problem 18

Complete missing totals in a two-way table from grand total 60 and first row total 25.

Problem 19

Complete missing totals in a two-way table from row total 30 with one cell 18.

Problem 20

Complete missing totals in a two-way table from column total 45 with one cell 20.

Problem 21

Complete missing totals in a two-way table from grand total 100, first row total 30, and second row total 40.

Problem 22

Complete missing totals in a two-way table from grand total 90, first column total 25, and second column total 35.

Problem 23

Complete missing totals in a two-way table from all data cells 10, 15, 20, 25.

Problem 24

Complete missing totals in a two-way table from first row total 30, second row total 40, and first cell of first row 12.

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Problem 25

Complete missing totals in a two-way table from first column total 25, second column total 35, and first cell of first column 10.

Problem 26

Complete missing totals in a two-way table from row total 50 with two cells 15 and 20.

Problem 27

Complete missing totals in a two-way table from column total 60 with two cells 20 and 25.

divide cell count by grand total.
15 problems Warmup Practice Mixed Review Assessment
Problem 28

Compute the joint probability from table data cell A and B is 18, grand total 100.

Problem 29

Compute the joint probability from table data students who are seniors and in band: 12 of 80.

Problem 30

Compute the joint probability from table data cell count 25 in a total of 200.

Problem 31

Compute the joint probability from table data cell count c and grand total n.

Problem 32

Compute the joint probability from table data event X and Y occurred 7 times out of a total of 50 trials.

Problem 33

Compute the joint probability from table data people who prefer coffee and read books: 30 out of 150 surveyed.

Problem 34

Compute the joint probability from table data 60 items are defective and red, from a grand total of 400 items.

Problem 35

Compute the joint probability from table data employees who are full-time and in department Z: 45 out of 300 employees.

Problem 36

Compute the joint probability from table data the intersection of set P and set Q has 10 elements, and the universal set has 25 elements.

Problem 37

Compute the joint probability from table data students who passed both math and science: 80 out of 240 students.

Problem 38

Compute the joint probability from table data cell (row 1, col 1) has 13 entries, grand total is 52.

Problem 39

Compute the joint probability from table data respondents who chose option A and option B: 22 from a total of 110 participants.

Problem 40

Compute the joint probability from table data the number of successful trials for condition Alpha and Beta was 35, out of 175 total trials.

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Problem 41

Compute the joint probability from table data vehicles that are trucks and red: 50 out of 250 vehicles.

Problem 42

Compute the joint probability from table data cell value p and total q.

divide row/column total by grand total.
15 problems Warmup Practice Mixed Review Assessment
Problem 43

Compute the marginal probability from table data row total 40, grand total 100.

Problem 44

Compute the marginal probability from table data column total 35, grand total 80.

Problem 45

Compute the marginal probability from table data total who prefer bus is 18 out of 60.

Problem 46

Compute the marginal probability from table data marginal total m and grand total n.

Problem 47

Compute the marginal probability from table data row total 25, grand total 50.

Problem 48

Compute the marginal probability from table data column total 12, grand total 36.

Problem 49

Compute the marginal probability from table data total students who chose art is 45 out of 150.

Problem 50

Compute the marginal probability from table data the number of people who prefer coffee is 70 from a survey of 200.

Problem 51

Compute the marginal probability from table data row sum 120, overall total 300.

Problem 52

Compute the marginal probability from table data column sum 64, grand total 256.

Problem 53

Compute the marginal probability from table data 27 people selected option A from a total of 90 participants.

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Problem 54

Compute the marginal probability from table data the total for category X is 55, and the grand total is 110.

Problem 55

Compute the marginal probability from table data total males in a study is 75 out of 125 subjects.

Problem 56

Compute the marginal probability from table data the sum of the 'Yes' responses is 88 from 400 responses.

Problem 57

Compute the marginal probability from table data a specific group has a total of 15 members, and the entire population is 75.

divide cell by row total.
15 problems Warmup Practice Mixed Review Assessment
Problem 58

Compute the row conditional probability from row total 30 and cell in that row 12.

Problem 59

Compute the row conditional probability from grade 9 row has 20 students; 8 prefer bus.

Problem 60

Compute the row conditional probability from row total r and cell c.

Problem 61

Compute the row conditional probability from sports row total 50 and yes cell 35.

Problem 62

Compute the row conditional probability from row total 40, cell value 10.

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Problem 63

Compute the row conditional probability from row total 25, cell value 5.

Problem 64

Compute the row conditional probability from In the 'Fiction' row, there are 60 books total, and 15 are sci-fi.

Problem 65

Compute the row conditional probability from Out of 80 employees in the 'Marketing' department, 20 are managers.

Problem 66

Compute the row conditional probability from The 'Adults' row has 120 people, 48 of whom chose option A.

Problem 67

Compute the row conditional probability from A survey found that 200 people are in the 'Rural' category, with 70 owning pets.

Problem 68

Compute the row conditional probability from row total 15, cell value 7.

Problem 69

Compute the row conditional probability from Among the 35 students in 'Class B', 13 passed the exam.

Problem 70

Compute the row conditional probability from row total 100, cell value 50.

Problem 71

Compute the row conditional probability from row total 50, cell value 0.

Problem 72

Compute the row conditional probability from row total 75, cell value 75.

divide cell by column total.
15 problems Warmup Practice Mixed Review Assessment
Problem 73

Compute the column conditional probability from column total 40 and cell in that column 10.

Problem 74

Compute the column conditional probability from bus column has 25 students; 15 are freshmen.

Problem 75

Compute the column conditional probability from column total k and cell c.

Problem 76

Compute the column conditional probability from yes column total 80 and senior yes cell 20.

Problem 77

Compute the column conditional probability from The 'red' column has a total of 50 items, and 10 of them are 'small'.

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Problem 78

Compute the column conditional probability from In the 'fiction' column, there are 120 books. 30 of these are 'mystery'.

Problem 79

Compute the column conditional probability from A survey's 'agree' column shows 200 responses. 40 of these are from 'women'.

Problem 80

Compute the column conditional probability from For the 'morning shift' column, 300 employees were present. 75 of them were 'part-time'.

Problem 81

Compute the column conditional probability from The 'apples' column has 30 fruits. 10 of them are 'green'.

Problem 82

Compute the column conditional probability from Out of 60 'dogs' in a column, 15 are 'poodles'.

Problem 83

Compute the column conditional probability from Given a column with total X, and a specific cell value Y within that column.

Problem 84

Compute the column conditional probability from In the 'pass' column, there are 80 students. 64 of them scored 'above average'.

Problem 85

Compute the column conditional probability from A 'Type A' column contains 20 samples. 5 of these samples tested 'positive'.

Problem 86

Compute the column conditional probability from The 'online' column has 150 transactions. 45 of these were 'returns'.

Problem 87

Compute the column conditional probability from In the 'blue car' column, there are 75 cars. 25 of them are 'sedans'.

state denominator and event meaning.
15 problems Warmup Practice Mixed Review Assessment
Problem 88

Interpret joint, marginal, or conditional probability in context: P(senior and band)=12/80.

Problem 89

Interpret joint, marginal, or conditional probability in context: P(bus)=18/60.

Problem 90

Interpret joint, marginal, or conditional probability in context: P(bus|freshman)=8/20.

Problem 91

Interpret joint, marginal, or conditional probability in context: P(freshman|bus)=8/18.

Problem 92

Interpret joint, marginal, or conditional probability in context: P(male and likes_math)=25/100.

Problem 93

Interpret joint, marginal, or conditional probability in context: P(likes_science)=40/100.

Problem 94

Interpret joint, marginal, or conditional probability in context: P(likes_science|female)=15/50.

Problem 95

Interpret joint, marginal, or conditional probability in context: P(female|likes_science)=15/40.

Problem 96

Interpret joint, marginal, or conditional probability in context: P(dog and short_haired)=10/30.

Problem 97

Interpret joint, marginal, or conditional probability in context: P(cat)=12/30.

Problem 98

Interpret joint, marginal, or conditional probability in context: P(short_haired|cat)=5/12.

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Problem 99

Interpret joint, marginal, or conditional probability in context: P(cat|short_haired)=5/15.

Problem 100

Interpret joint, marginal, or conditional probability in context: P(faulty and car)=7/200.

Problem 101

Interpret joint, marginal, or conditional probability in context: P(truck)=80/200.

Problem 102

Interpret joint, marginal, or conditional probability in context: P(faulty|truck)=3/80.

compare conditional to marginal probabilities.
15 problems Warmup Practice Mixed Review Assessment
Problem 103

Test independence using conditional probabilities from table data P(A)=0.40 and P(A|B)=0.40.

Problem 104

Test independence using conditional probabilities from table data P(prefers bus)=0.30 and P(prefers bus|freshman)=0.50.

Problem 105

Test independence using conditional probabilities from table data P(B)=2/5 and P(B|A)=3/5.

Problem 106

Test independence using conditional probabilities from table data P(A)=a and P(A|B)=a.

Problem 107

Test independence using conditional probabilities from table data P(C)=0.75 and P(C|D)=0.75.

Problem 108

Test independence using conditional probabilities from table data P(E)=0.6 and P(E|F)=0.5.

Problem 109

Test independence using conditional probabilities from table data P(X)=1/3 and P(X|Y)=1/3.

Problem 110

Test independence using conditional probabilities from table data P(M)=3/4 and P(M|N)=1/2.

Problem 111

Test independence using conditional probabilities from table data P(R)=x and P(R|S)=y.

Problem 112

Test independence using conditional probabilities from table data P(pass exam)=0.85 and P(pass exam|studied)=0.85.

Problem 113

Test independence using conditional probabilities from table data P(has flu)=0.10 and P(has flu|vaccinated)=0.05.

Problem 114

Test independence using conditional probabilities from table data P(rain|cloudy)=0.60 and P(rain)=0.60.

Problem 115

Test independence using conditional probabilities from table data P(win game|home field)=0.70 and P(win game)=0.50.

Problem 116

Test independence using conditional probabilities from table data P(G)=k and P(G|H)=k.

Open in simulator
Problem 117

Test independence using conditional probabilities from table data P(J)=p and P(J|K)=q.

compare joint to product of marginals.
15 problems Warmup Practice Mixed Review Assessment
Problem 118

Test independence using joint probability from table data P(A)=0.50, P(B)=0.40, P(A and B)=0.20.

Problem 119

Test independence using joint probability from table data P(A)=0.60, P(B)=0.25, P(A and B)=0.10.

Problem 120

Test independence using joint probability from table data A total 20, B total 30, overlap 12, grand total 100.

Problem 121

Test independence using joint probability from table data P(A)=p, P(B)=q, P(A and B)=pq.

Problem 122

Test independence using joint probability from table data P(A)=0.25, P(B)=0.80, P(A and B)=0.20.

Problem 123

Test independence using joint probability from table data P(A)=0.70, P(B)=0.30, P(A and B)=0.25.

Problem 124

Test independence using joint probability from table data P(A)=1/2, P(B)=1/3, P(A and B)=1/6.

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Problem 125

Test independence using joint probability from table data P(A)=2/5, P(B)=1/4, P(A and B)=1/8.

Problem 126

Test independence using joint probability from table data P(A)=1.00, P(B)=0.75, P(A and B)=0.75.

Problem 127

Test independence using joint probability from table data P(A)=0.50, P(B)=0.00, P(A and B)=0.10.

Problem 128

Test independence using joint probability from table data A total 40, B total 50, overlap 20, grand total 100.

Problem 129

Test independence using joint probability from table data A total 25, B total 60, overlap 10, grand total 100.

Problem 130

Test independence using joint probability from table data A total 60, B total 70, overlap 42, grand total 100.

Problem 131

Test independence using joint probability from table data A total 150, B total 200, overlap 40, grand total 500.

Problem 132

Test independence using joint probability from table data Number of A is 30, Number of B is 40, Number of A and B is 12, Total observations is 100.

convert each row to conditional distribution.
12 problems Warmup Practice Mixed Review Assessment
Problem 133

Create a row-relative frequency table from row 1 cells 12,8; row 2 cells 10,20.

Problem 134

Create a row-relative frequency table from freshman bus 8 car 12; sophomore bus 15 car 15.

Problem 135

Create a row-relative frequency table from row cells a,b.

Problem 136

Create a row-relative frequency table from yes/no row counts 18,2.

Problem 137

Create a row-relative frequency table from classA boys 10 girls 10; classB boys 15 girls 5; classC boys 8 girls 12.

Problem 138

Create a row-relative frequency table from fruit apples 20 bananas 30 oranges 50; vegetables carrots 10 potatoes 40 broccoli 50.

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Problem 139

Create a row-relative frequency table from survey agree 40 strongly_agree 20 disagree 30 strongly_disagree 10.

Problem 140

Create a row-relative frequency table from groupX male 7 female 3; groupY male 11 female 4.

Problem 141

Create a row-relative frequency table from departmentA full_time 25 part_time 5; departmentB full_time 15 part_time 0.

Problem 142

Create a row-relative frequency table from region1 rural 100 urban 200 suburban 300; region2 rural 50 urban 150 suburban 100.

Problem 143

Create a row-relative frequency table from productX good 18 bad 2; productY good 10 bad 10.

Problem 144

Create a row-relative frequency table from colors red 10 blue 20 green 30.

convert each column to conditional distribution.
12 problems Warmup Practice Mixed Review Assessment
Problem 145

Create a column-relative frequency table from column 1 cells 12,8; column 2 cells 10,20.

Problem 146

Create a column-relative frequency table from bus column freshman 8 sophomore 12; car column freshman 6 sophomore 14.

Problem 147

Create a column-relative frequency table from column cells a,b.

Problem 148

Create a column-relative frequency table from yes column counts 21,9.

Problem 149

Create a column-relative frequency table from apples column red 15 green 5; oranges column red 10 green 20.

Problem 150

Create a column-relative frequency table from gender column male 40 female 60; age column under_30 25 over_30 75.

Problem 151

Create a column-relative frequency table from day column sunny 10 cloudy 20 rainy 30; night column clear 15 overcast 5.

Problem 152

Create a column-relative frequency table from dogs column small 10 medium 20 large 30; cats column small 5 medium 10 large 15.

Problem 153

Create a column-relative frequency table from north column sales 100 expenses 50; south column sales 150 expenses 75.

Problem 154

Create a column-relative frequency table from vote column yes 70 no 30; abstain column yes 10 no 90.

Problem 155

Create a column-relative frequency table from product_A column good 2 bad 8; product_B column good 3 bad 7.

Problem 156

Create a column-relative frequency table from morning column coffee 25 tea 15; afternoon column coffee 10 tea 30.

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compare conditional distributions.
12 problems Warmup Practice Mixed Review Assessment
Problem 157

Use the two-way table to evaluate association: P(bus|freshman)=0.40 and P(bus|sophomore)=0.42.

Problem 158

Use the two-way table to evaluate association: P(prefers math|club)=0.75 and P(prefers math|not club)=0.35.

Problem 159

Use the two-way table to evaluate association: row conditional distributions are identical.

Problem 160

Use the two-way table to evaluate association: one group has twice the conditional rate of another.

Problem 161

Use the two-way table to evaluate association: P(success|treatment)=0.61 and P(success|control)=0.59.

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Problem 162

Use the two-way table to evaluate association: P(pass|studied)=0.85 and P(pass|didn't study)=0.30.

Problem 163

Use the two-way table to evaluate association: P(likes coffee|male)=0.55 and P(likes coffee|female)=0.55.

Problem 164

Use the two-way table to evaluate association: P(owns pet|children)=0.70 and P(owns pet|no children)=0.35.

Problem 165

Use the two-way table to evaluate association: P(reads daily|under 30)=0.20 and P(reads daily|over 30)=0.60.

Problem 166

Use the two-way table to evaluate association: P(satisfied|new product)=0.78 and P(satisfied|old product)=0.80.

Problem 167

Use the two-way table to evaluate association: P(travels abroad|college degree)=0.65 and P(travels abroad|no college degree)=0.25.

Problem 168

Use the two-way table to evaluate association: P(buys online|urban)=0.60 and P(buys online|rural)=0.20.

distinguish total, row, and column denominators.
12 problems Warmup Practice Mixed Review Assessment
Problem 169

Choose the correct denominator for the table-based question: What percent of freshmen ride the bus?.

Problem 170

Choose the correct denominator for the table-based question: What percent of bus riders are freshmen?.

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Problem 171

Choose the correct denominator for the table-based question: What percent of all students are freshmen bus riders?.

Problem 172

Choose the correct denominator for the table-based question: What percent of students who prefer math are seniors?.

Problem 173

Choose the correct denominator for the table-based question: What percent of students who play soccer are also in the band?.

Problem 174

Choose the correct denominator for the table-based question: What percent of students in the band also play soccer?.

Problem 175

Choose the correct denominator for the table-based question: What percent of all students play soccer and are in the band?.

Problem 176

Choose the correct denominator for the table-based question: Among the employees who work full-time, what percentage are salaried?.

Problem 177

Choose the correct denominator for the table-based question: Of the salaried employees, what percentage work full-time?.

Problem 178

Choose the correct denominator for the table-based question: Given that a customer bought item X, what is the probability they also bought item Y?.

Problem 179

Choose the correct denominator for the table-based question: Among the voters who are registered as independent, what percentage are under 40?.

Problem 180

Choose the correct denominator for the table-based question: What percent of all surveyed individuals are under 40 and registered as independent?.

cite appropriate probabilities.
12 problems Warmup Practice Mixed Review Assessment
Problem 181

Write a supported conclusion from two-way table evidence 60% of club members prefer math versus 35% of nonmembers.

Problem 182

Write a supported conclusion from two-way table evidence P(bus|freshman)=0.40 and P(bus|sophomore)=0.40.

Problem 183

Write a supported conclusion from two-way table evidence raw count is higher for group A but group A is much larger overall.

Problem 184

Write a supported conclusion from two-way table evidence P(A|B)=0.70 and P(A|not B)=0.72.

Problem 185

Write a supported conclusion from two-way table evidence 70% of students who attended tutoring passed the exam, compared to 40% of those who did not attend tutoring.

Problem 186

Write a supported conclusion from two-way table evidence P(owns a tablet|under 30)=0.62 and P(owns a tablet|over 30)=0.65.

Problem 187

Write a supported conclusion from two-way table evidence Among surveyed customers, 3 out of 4 who used the new feature rated it excellent, while only 1 out of 4 who didn't use it rated it excellent.

Problem 188

Write a supported conclusion from two-way table evidence P(prefers online shopping|urban)=0.75 and P(prefers online shopping|rural)=0.75.

Open in simulator
Problem 189

Write a supported conclusion from two-way table evidence The rate of successful project completion was 80% for teams with a dedicated project manager, versus 55% for teams without one.

Problem 190

Write a supported conclusion from two-way table evidence P(satisfied|product A)=0.88 and P(satisfied|product B)=0.85.

Problem 191

Write a supported conclusion from two-way table evidence Among surveyed voters, 90% of those registered with Party X voted, while 45% of those not registered with Party X voted.

Problem 192

Write a supported conclusion from two-way table evidence P(read daily|high income)=0.60 and P(read daily|low income)=0.30.

catch wrong denominator, reversed condition, and count/percent confusion.
15 problems Warmup Practice Mixed Review Assessment
Problem 193

Correct the two-way-table probability error: A student divides a conditional cell by the grand total.

Problem 194

Correct the two-way-table probability error: A student computes P(A|B) when the question asks P(B|A).

Open in simulator
Problem 195

Correct the two-way-table probability error: A student compares raw counts from unequal group sizes.

Problem 196

Correct the two-way-table probability error: A student treats 0.35 as 35 students in a relative-frequency table.

Problem 197

Correct the two-way-table probability error: When finding the probability of 'likes dogs' given 'is male', a student uses the total number of students as the denominator.

Problem 198

Correct the two-way-table probability error: A student calculates P(A and B) by dividing the cell count by the total of row A.

Problem 199

Correct the two-way-table probability error: A student is asked for the probability of 'passing the test given they studied', but they calculate the probability of 'studying given they passed the test'.

Problem 200

Correct the two-way-table probability error: A student states that 'more people prefer coffee than tea' based on comparing the raw counts of coffee drinkers and tea drinkers, even though the groups surveyed were different sizes.

Problem 201

Correct the two-way-table probability error: In a table of relative frequencies, a student interprets the value 0.25 in a cell as '25 individuals'.

Problem 202

Correct the two-way-table probability error: A student states that the probability of being female is simply the total number of females in the table.

Problem 203

Correct the two-way-table probability error: The problem asks for P(event A | event B), but the student calculates (Number of B and A) / (Number of A).

Problem 204

Correct the two-way-table probability error: A student concludes that 'females are more likely to vote for candidate X' because more females voted for X than males, without considering the total number of female vs. male voters.

Problem 205

Correct the two-way-table probability error: In a table showing joint relative frequencies, a student treats a cell value like P(A) or P(B).

Problem 206

Correct the two-way-table probability error: A student calculates the probability of 'being left-handed and male' by dividing the count of left-handed males by the total number of males.

Problem 207

Correct the two-way-table probability error: When asked for the probability that a randomly selected person *who owns a cat* also *owns a dog*, a student uses the total number of people who own dogs as the denominator.