Math II · S-CP.2

Determining Event Independence with `P(A and B) = P(A)P(B)`

Independence helps students distinguish events that genuinely do not affect each other from events that only appear unrelated.

Concept Statistics and Probability
Domain Conditional Probability and the Rules of Probability
Read time 8 minutes

What this learning objective is really asking you to learn

This objective asks students to determine whether two events are independent. In probability, two events are independent if knowing that one event occurred does not change the probability of the other event. The formal test in this standard is

\[P(A and B) = P(A)P(B)\].

Using set notation, this is

\[P(A \cap B) = P(A)P(B)\].

The event \(A \cap B\) means both A and B occur. If the probability of both events occurring equals the product of their separate probabilities, the events are independent. If not, they are dependent.

For example, suppose you flip a fair coin and roll a fair die. Let A be “the coin lands heads,” and B be “the die shows 6.” The probability of heads is \(1/2\). The probability of rolling a 6 is \(1/6\). The probability of both heads and 6 is \(1/12\). Since

\[1/12 = (1/2)(1/6)\],

the events are independent. The coin result does not affect the die result.

Now consider drawing one card from a deck. Let A be “the card is a king,” and B be “the card is a face card.” The probability of a king is \(4/52\). The probability of a face card is \(12/52\). The probability of both a king and a face card is \(4/52\), because every king is a face card. But \((4/52)(12/52)\) is not equal to \(4/52\). So the events are not independent. Knowing the card is a king changes the probability that it is a face card.

This objective is asking students to do more than use a formula. They must understand the meaning: independence means no probability change. The multiplication equation is a test for that meaning.

Why students should learn this math

Students should learn independence because the world is full of claims that one thing affects another. Does studying affect test performance? Does a medication affect recovery? Does weather affect traffic? Does an ad campaign affect sales? Does one machine defect predict another? Does one genetic trait affect another? Does one event truly change the chance of another, or are they unrelated?

Probability independence gives students a disciplined way to reason about such questions. It is not enough to say two things “seem unrelated.” We need to compare probabilities. If the chance of B is the same whether or not A occurred, the events are independent. If the chance changes, they are dependent.

This matters in everyday reasoning. People often see patterns where none exist. If someone wins two coin tosses in a row, they may think they are “due” for tails. But independent events do not remember the past. A fair coin flip remains \(1/2\) heads each time. This misconception, sometimes related to the gambler's fallacy, causes bad decisions in gambling, investing, sports predictions, and risk judgment.

Independence also matters when events are not independent but people assume they are. Medical symptoms are not independent of disease status. Test results are not independent of whether a person has the condition being tested. Product failures may not be independent if they come from the same manufacturing defect. Survey responses may not be independent if respondents influence one another. Assuming independence when it is false can produce dangerously wrong conclusions.

In statistics, independence is a foundation for inference. Many statistical methods depend on independent samples or independent trials. If data points influence each other, calculations can become misleading. In science, experiments are designed to isolate effects and reduce dependence. In data science, correlated or dependent variables can distort models.

The “why” is that independence is a reality check. It helps students ask: does this event actually change the probability of that event? That question is central to reasoning under uncertainty.

The historical machinery: from games of chance to statistical assumptions

Independence emerged naturally from games of chance. If a die is fair and well-made, one roll should not affect the next roll. If a coin is flipped honestly, previous flips should not determine future flips. These simple settings made independence intuitive and calculable.

As probability developed into a broader mathematical field, independence became a formal concept. It was no longer enough to say events “do not influence each other” in ordinary language. Mathematics needed a precise test. The equation \(P(A \cap B) = P(A)P(B)\) became that test.

This formalization mattered because probability moved into science, insurance, economics, and statistics. In these fields, independence is often an assumption that must be examined. Insurance models may assume certain risks are independent, but natural disasters can make many claims happen together. Scientific experiments may assume observations are independent, but shared conditions can create dependence. Statistical models may assume independent errors, but real data can violate that assumption.

The history shows why independence is not just a classroom formula. It is one of the major assumptions that allows probability models to work. When independence is true, multiplication rules become simple. When it is false, more careful conditional reasoning is needed.

Where this fits in the big map of mathematics

This objective comes after students learn events as subsets of sample spaces. To test independence, students need to know what \(A\), \(B\), and \(A \cap B\) mean. It also prepares directly for conditional probability. In fact, another way to say independence is \(P(A | B) = P(A)\), provided \(P(B) \ne 0\). That means the probability of A given B is the same as the original probability of A.

It connects to two-way tables. Tables can show joint frequencies and marginal frequencies. Students can compute \(P(A)\), \(P(B)\), and \(P(A \cap B)\) from a table and test independence.

It connects to multiplication rules. If events are independent, the probability of both is the product of the individual probabilities. If they are not independent, the general multiplication rule uses conditional probability: \(P(A \cap B) = P(A)P(B | A)\).

It connects to statistics and inference. Independent trials are central in binomial probability, simulations, randomized experiments, and many models.

It connects to real-world decision-making because independence assumptions are often hidden. A model may multiply probabilities as if events are independent. Students should learn to ask whether that assumption is justified.

The big-map role is probability structure. Independence tells when event relationships are simple and when conditional reasoning is required.

How to execute the skill technically

To test independence using this standard, follow four steps.

First, identify the events A and B clearly.

Second, compute \(P(A)\), \(P(B)\), and \(P(A \cap B)\).

Third, compute the product \(P(A)P(B)\).

Fourth, compare. If \(P(A \cap B) = P(A)P(B)\), the events are independent. If not, they are dependent.

Example: roll one fair die. Let A be “roll an even number,” and B be “roll a number greater than 4.” Then \(A = {2, 4, 6}\) and \(B = {5, 6}\). So \(P(A) = 3/6 = 1/2\), \(P(B) = 2/6 = 1/3\), and \(A \cap B = {6}\), so \(P(A \cap B) = 1/6\).

Now multiply:

\[P(A)P(B) = (1/2)(1/3) = 1/6\].

Since this equals \(P(A \cap B)\), A and B are independent in this die-roll example.

Another example: draw one card. Let A be “heart,” and B be “red.” \(P(A) = 13/52 = 1/4\). \(P(B) = 26/52 = 1/2\). \(A \cap B\) is hearts, because every heart is red, so \(P(A \cap B) = 1/4\). But \(P(A)P(B) = 1/8\). Since \(1/4 \ne 1/8\), the events are dependent.

From a two-way table, use counts. Suppose 100 students are surveyed. 40 play sports, 30 play music, and 12 do both. Let A be sports and B be music. Then \(P(A) = 40/100\), \(P(B) = 30/100\), and \(P(A \cap B) = 12/100\). The product is \((40/100)(30/100) = 0.12\), and \(P(A \cap B) = 0.12\). These data are consistent with independence.

Another worked example: independence in a two-way table

Suppose 200 app users are classified by whether they turned on notifications and whether they completed their daily goal.

| | Completed goal | Did not complete | Total | |---|---:|---:|---:| | Notifications on | 72 | 48 | 120 | | Notifications off | 48 | 32 | 80 | | Total | 120 | 80 | 200 |

Let A be “notifications on.” Let B be “completed goal.”

\[P(A) = 120/200 = 0.60\].
\[P(B) = 120/200 = 0.60\].
\[P(A \cap B) = 72/200 = 0.36\].

Now multiply:

\[P(A)P(B) = 0.60 \cdot 0.60 = 0.36\].

Since \(P(A \cap B) = P(A)P(B)\), the table is consistent with independence. That means completion rate among notification users is the same as the overall completion rate. Check:

\[P(B | A) = 72/120 = 0.60\].

This equals \(P(B) = 0.60\).

This does not mean notifications are useless in every possible world. It means in this data table, notification status does not change the probability of completion. A strong student learns to make exactly that claim and no stronger claim.

Why independence is not causation

Independence and causation are different ideas. If two events are independent, one does not change the probability of the other in the model or data being considered. If two events are dependent, that still does not automatically prove causation. Dependence means there is an association or probability relationship. Causation requires stronger evidence, usually from a well-designed experiment or a convincing causal structure.

For example, ice cream sales and drowning incidents may both rise in summer. They are statistically associated, but ice cream sales do not cause drowning. A hidden variable, hot weather and summer activity, affects both. Students should learn this early because probability and statistics are often misused in public claims.

Common misconceptions and how to avoid them

One common misconception is thinking independent means mutually exclusive. It does not. Mutually exclusive events cannot happen together. Independent events can happen together, but one does not change the probability of the other. In fact, if two non-impossible events are mutually exclusive, they are not independent.

Another mistake is trusting intuition instead of checking probabilities. Events that seem related may not be in a given model, and events that seem unrelated may be dependent in data.

A third mistake is comparing \(P(A \cap B)\) to \(P(A) + P(B)\) instead of \(P(A)P(B)\). Addition belongs to union reasoning; independence uses multiplication.

A fourth mistake is using counts from a table without converting consistently. You can compare counts proportionally, but probabilities must be based on the same total.

A fifth mistake is assuming independence just because events happen in different categories. Real-world processes can connect categories in hidden ways.

The big takeaway

This objective teaches students how to test whether two events are independent. The equation \(P(A \cap B) = P(A)P(B)\) means the probability of both events equals the product of their separate probabilities. Independence is not a vibe; it is a mathematical relationship. This idea prepares students for conditional probability, statistical inference, and responsible risk reasoning.

Problem Library

Problems in the App From This Objective

162 problems across 12 archetypes in the app.

compare `P(A and B)` to `P(A)P(B)`.
15 problems Warmup Practice Mixed Review Assessment
Problem 1

Test independence from probabilities P(A)=1/2, P(B)=1/3, P(A and B)=1/6.

Problem 2

Test independence from probabilities P(A)=0.4, P(B)=0.5, P(A and B)=0.3.

Problem 3

Test independence from probabilities P(A)=25%, P(B)=40%, P(A and B)=10%.

Problem 4

Test independence from probabilities P(A)=3/5, P(B)=1/2, P(A and B)=1/5.

Problem 5

Test independence from probabilities P(A)=1/4, P(B)=2/3, P(A and B)=1/6.

Problem 6

Test independence from probabilities P(A)=1/2, P(B)=1/4, P(A and B)=1/2.

Problem 7

Test independence from probabilities P(A)=0.2, P(B)=0.3, P(A and B)=0.06.

Problem 8

Test independence from probabilities P(A)=0.6, P(B)=0.2, P(A and B)=0.1.

Problem 9

Test independence from probabilities P(A)=50%, P(B)=60%, P(A and B)=30%.

Problem 10

Test independence from probabilities P(A)=10%, P(B)=20%, P(A and B)=5%.

Problem 11

Test independence from probabilities P(A)=1/5, P(B)=0.5, P(A and B)=0.1.

Problem 12

Test independence from probabilities P(A)=0.7, P(B)=10%, P(A and B)=0.05.

Problem 13

Test independence from probabilities P(A)=2/5, P(B)=1/4, P(A and B)=1/10.

Problem 14

Test independence from probabilities P(A)=0.8, P(B)=0.3, P(A and B)=0.2.

Open in simulator
Problem 15

Test independence from probabilities P(A)=20%, P(B)=30%, P(A and B)=6%.

multiply probabilities.
15 problems Warmup Practice Mixed Review Assessment
Problem 16

Find the joint probability if events are independent: P(A)=1/4 and P(B)=2/3.

Problem 17

Find the joint probability if events are independent: P(A)=0.3 and P(B)=0.5.

Problem 18

Find the joint probability if events are independent: P(A)=20% and P(B)=40%.

Problem 19

Find the joint probability if events are independent: P(A)=p and P(B)=q.

Problem 20

Find the joint probability if events are independent: P(A)=1/2 and P(B)=1/3.

Problem 21

Find the joint probability if events are independent: P(A)=3/5 and P(B)=1/4.

Problem 22

Find the joint probability if events are independent: P(A)=0.2 and P(B)=0.4.

Problem 23

Find the joint probability if events are independent: P(A)=0.6 and P(B)=0.7.

Problem 24

Find the joint probability if events are independent: P(A)=10% and P(B)=30%.

Problem 25

Find the joint probability if events are independent: P(A)=50% and P(B)=60%.

Problem 26

Find the joint probability if events are independent: P(A)=1/5 and P(B)=0.5.

Problem 27

Find the joint probability if events are independent: P(A)=1/10 and P(B)=20%.

Problem 28

Find the joint probability if events are independent: P(A)=0.25 and P(B)=50%.

Problem 29

Find the joint probability if events are independent: P(A)=2x and P(B)=1/2.

Open in simulator
Problem 30

Find the joint probability if events are independent: P(A)=x and P(B)=y.

compute marginal and joint probabilities.
12 problems Warmup Practice Mixed Review Assessment
Problem 31

Determine independence from the two-way table data total 100; A=40, B=50, A and B=20.

Problem 32

Determine independence from the two-way table data total 200; A=80, B=60, A and B=30.

Open in simulator
Problem 33

Determine independence from the two-way table data total 50; A=10, B=25, A and B=5.

Problem 34

Determine independence from the two-way table data total n; A count a, B count b, overlap count j.

Problem 35

Determine independence from the two-way table data total 100; A=30, B=60, A and B=18.

Problem 36

Determine independence from the two-way table data total 100; A=30, B=60, A and B=15.

Problem 37

Determine independence from the two-way table data total 120; A=40, B=60, A and B=20.

Problem 38

Determine independence from the two-way table data total 120; A=40, B=60, A and B=25.

Problem 39

Determine independence from the two-way table data total 500; A=100, B=250, A and B=50.

Problem 40

Determine independence from the two-way table data total 500; A=100, B=250, A and B=60.

Problem 41

Determine independence from the two-way table data total 70; A=14, B=35, A and B=7.

Problem 42

Determine independence from the two-way table data total 70; A=14, B=35, A and B=8.

compute region probabilities and compare.
12 problems Warmup Practice Mixed Review Assessment
Problem 43

Determine independence from Venn diagram data total 100; A only 20, overlap 10, B only 30, outside 40.

Problem 44

Determine independence from Venn diagram data total 50; A only 5, overlap 5, B only 20, outside 20.

Problem 45

Determine independence from Venn diagram data A and B are disjoint with positive probabilities.

Problem 46

Determine independence from Venn diagram data total n with A only x, overlap y, B only z.

Problem 47

Determine independence from Venn diagram data total 100; A only 20, overlap 20, B only 30, outside 30.

Problem 48

Determine independence from Venn diagram data total 200; A only 50, overlap 10, B only 80, outside 60.

Problem 49

Determine independence from Venn diagram data total 20; A only 6, overlap 4, B only 4, outside 6.

Open in simulator
Problem 50

Determine independence from Venn diagram data total 60; A only 15, overlap 5, B only 20, outside 20.

Problem 51

Determine independence from Venn diagram data total 70; A only 14, overlap 14, B only 21, outside 21.

Problem 52

Determine independence from Venn diagram data total 100; A only 5, overlap 80, B only 10, outside 5.

Problem 53

Determine independence from Venn diagram data total 120; A only 18, overlap 12, B only 36, outside 54.

Problem 54

Determine independence from Venn diagram data total 150; A only 40, overlap 5, B only 60, outside 45.

state that one event does not change likelihood of another.
15 problems Warmup Practice Mixed Review Assessment
Problem 55

Explain independence in probability language for coin flip result and die roll result.

Problem 56

Explain independence in probability language for drawing a card without replacement twice.

Problem 57

Explain independence in probability language for student grade level and preferred lunch from a survey.

Problem 58

Explain independence in probability language for machine A working and machine B working with separate power sources.

Problem 59

Explain independence in probability language for rolling a 6 on a die and then rolling a 3 on the same die.

Problem 60

Explain independence in probability language for a person's height and their favorite color.

Problem 61

Explain independence in probability language for the weather in London and the stock market in New York.

Problem 62

Explain independence in probability language for getting heads on two separate coin flips.

Problem 63

Explain independence in probability language for drawing two red marbles from a bag without replacement.

Problem 64

Explain independence in probability language for a student passing a test and having studied for it.

Problem 65

Explain independence in probability language for a car's speed and its fuel efficiency.

Problem 66

Explain independence in probability language for getting a flat tire and driving over a nail.

Problem 67

Explain independence in probability language for a person's age and their preference for a certain music genre from a survey.

Problem 68

Explain independence in probability language for two friends living in different cities getting sick on the same day.

Open in simulator
Problem 69

Explain independence in probability language for the color of a car and the driver's gender from a large dataset.

understand nonempty independent intersections except zero-probability cases.
12 problems Warmup Practice Mixed Review Assessment
Problem 70

Distinguish independence from mutually exclusive events in A and B are mutually exclusive and both have positive probability.

Problem 71

Distinguish independence from mutually exclusive events in coin heads and die shows 6.

Problem 72

Distinguish independence from mutually exclusive events in rolling even and rolling odd on one die.

Problem 73

Distinguish independence from mutually exclusive events in P(A)=0.4, P(B)=0.5, P(A and B)=0.2.

Problem 74

Distinguish independence from mutually exclusive events in Drawing two red cards consecutively from a standard deck without replacement.

Problem 75

Distinguish independence from mutually exclusive events in Rolling a 1 and rolling a 6 on a single fair die.

Open in simulator
Problem 76

Distinguish independence from mutually exclusive events in Flipping a coin twice, getting heads on the first flip and tails on the second flip.

Problem 77

Distinguish independence from mutually exclusive events in P(X)=0.6, P(Y)=0.7, P(X or Y)=0.9.

Problem 78

Distinguish independence from mutually exclusive events in A student passing a math test and failing the same math test.

Problem 79

Distinguish independence from mutually exclusive events in A person owning a dog and a person owning a cat, assuming no correlation between pet ownership.

Problem 80

Distinguish independence from mutually exclusive events in In a city, 40% of people own a car, 30% own a bicycle, and 20% own both.

Problem 81

Distinguish independence from mutually exclusive events in Rolling a 7 on a standard six-sided die and flipping a coin to get heads.

multiply probabilities across independent trials.
15 problems Warmup Practice Mixed Review Assessment
Problem 82

Use independence to calculate probability of repeated events: flip heads three times.

Problem 83

Use independence to calculate probability of repeated events: roll a 6 twice in a row.

Problem 84

Use independence to calculate probability of repeated events: component works with probability 0.9 for two independent components.

Problem 85

Use independence to calculate probability of repeated events: make a free throw with probability 0.75 on three independent attempts.

Problem 86

Use independence to calculate probability of repeated events: flip tails four times in a row.

Open in simulator
Problem 87

Use independence to calculate probability of repeated events: roll an even number on a standard die three times in a row.

Problem 88

Use independence to calculate probability of repeated events: draw a red card from a standard deck, replace it, and draw a red card again.

Problem 89

Use independence to calculate probability of repeated events: a machine produces a defective item with probability 0.05 for two independent items.

Problem 90

Use independence to calculate probability of repeated events: a basketball player makes a shot with probability 0.6 on two independent attempts.

Problem 91

Use independence to calculate probability of repeated events: a light bulb has a 0.99 probability of working for three independent bulbs.

Problem 92

Use independence to calculate probability of repeated events: guess a correct answer on a multiple-choice question with 5 options twice in a row.

Problem 93

Use independence to calculate probability of repeated events: a traffic light is green 40% of the time for two independent observations.

Problem 94

Use independence to calculate probability of repeated events: roll a number greater than 4 on a standard die three times in a row.

Problem 95

Use independence to calculate probability of repeated events: a computer program crashes with a probability of 0.1 on two independent runs.

Problem 96

Use independence to calculate probability of repeated events: draw an ace from a standard deck, replace it, and draw an ace again.

compare observed joint to product.
15 problems Warmup Practice Mixed Review Assessment
Problem 97

Identify dependent events from changed probability in draw a card, keep it, then draw another card.

Problem 98

Identify dependent events from changed probability in survey data shows P(likes math)=0.60 but P(likes math | in club)=0.80.

Problem 99

Identify dependent events from changed probability in select one marble without replacement, then select another.

Problem 100

Identify dependent events from changed probability in P(A)=0.5 and P(A|B)=0.5.

Problem 101

Identify dependent events from changed probability in picking a red crayon from a box of 10 assorted crayons, not putting it back, then picking another crayon.

Problem 102

Identify dependent events from changed probability in P(student passes test)=0.70 but P(student passes test | attended tutoring)=0.85.

Problem 103

Identify dependent events from changed probability in removing a defective component from a batch of 20, then selecting another component from the remaining 19.

Problem 104

Identify dependent events from changed probability in a person eats one cookie from a jar containing 12 cookies, then eats another cookie from the same jar.

Problem 105

Identify dependent events from changed probability in selecting a team captain from a group of 12 people, then selecting a vice-captain from the remaining 11 people.

Problem 106

Identify dependent events from changed probability in P(player scores next point | previous point scored)=0.60, but P(player scores next point | previous point missed)=0.40.

Problem 107

Identify dependent events from changed probability in drawing a face card from a deck of 52 cards, not replacing it, then drawing another card.

Problem 108

Identify dependent events from changed probability in a student chooses a number from 1 to 20 without replacement, then chooses another number from the remaining set.

Problem 109

Identify dependent events from changed probability in rolling a standard six-sided die, then flipping a fair coin.

Problem 110

Identify dependent events from changed probability in drawing a card from a deck, replacing it, then drawing another card.

Problem 111

Identify dependent events from changed probability in P(getting a cold)=0.20 but P(getting a cold | exposed to virus)=0.50.

Open in simulator
evaluate replacement, repeated trials, or linked categories.
12 problems Warmup Practice Mixed Review Assessment
Problem 112

Decide whether the context suggests independence before calculating: rolling a die and flipping a coin.

Problem 113

Decide whether the context suggests independence before calculating: drawing two cards without replacement.

Problem 114

Decide whether the context suggests independence before calculating: randomly selecting two students with replacement from the same list.

Problem 115

Decide whether the context suggests independence before calculating: height category and shoe size in a survey.

Problem 116

Decide whether the context suggests independence before calculating: drawing a red ball from a bag, replacing it, then drawing another ball.

Problem 117

Decide whether the context suggests independence before calculating: selecting a president and a vice-president from a club of 20 members.

Problem 118

Decide whether the context suggests independence before calculating: the outcome of a coin flip and the outcome of a spinner.

Open in simulator
Problem 119

Decide whether the context suggests independence before calculating: a person's income level and their preferred mode of transportation.

Problem 120

Decide whether the context suggests independence before calculating: rolling a die twice.

Problem 121

Decide whether the context suggests independence before calculating: taking two socks from a drawer without looking and without replacing the first.

Problem 122

Decide whether the context suggests independence before calculating: whether a student passes a math test and whether they pass an English test.

Problem 123

Decide whether the context suggests independence before calculating: the color of a car passing by and the number rolled on a die inside a house.

solve for marginal or joint probability.
15 problems Warmup Practice Mixed Review Assessment
Problem 124

Find the missing probability using independence equation P(A and B)=0.12 and P(A)=0.3.

Problem 125

Find the missing probability using independence equation P(A and B)=1/10 and P(B)=1/2.

Problem 126

Find the missing probability using independence equation P(A)=0.25 and P(B)=0.8.

Problem 127

Find the missing probability using independence equation P(A and B)=j and P(A)=p.

Problem 128

Find the missing probability using independence equation P(A and B)=0.35 and P(B)=0.7.

Problem 129

Find the missing probability using independence equation P(A and B)=0.06 and P(A)=0.2.

Problem 130

Find the missing probability using independence equation P(A)=0.4 and P(B)=0.6.

Problem 131

Find the missing probability using independence equation P(A and B)=1/6 and P(B)=1/3.

Problem 132

Find the missing probability using independence equation P(A and B)=3/20 and P(A)=3/4.

Problem 133

Find the missing probability using independence equation P(A)=2/5 and P(B)=1/4.

Problem 134

Find the missing probability using independence equation P(A and B)=x and P(B)=y.

Problem 135

Find the missing probability using independence equation P(A)=m and P(B)=n.

Open in simulator
Problem 136

Find the missing probability using independence equation P(A and B)=0.15 and P(A)=3/10.

Problem 137

Find the missing probability using independence equation P(A and B)=1/8 and P(B)=0.5.

Problem 138

Find the missing probability using independence equation P(A)=0.125 and P(B)=0.4.

explain independent/dependent conclusion with evidence.
12 problems Warmup Practice Mixed Review Assessment
Problem 139

Interpret the independence test result in context: survey result shows P(A and B)=P(A)P(B) for grade and club membership.

Problem 140

Interpret the independence test result in context: P(prefers bus and grade 10) is greater than product of marginals.

Open in simulator
Problem 141

Interpret the independence test result in context: coin and die product test matches.

Problem 142

Interpret the independence test result in context: card draw without replacement product test fails.

Problem 143

Interpret the independence test result in context: P(rain | cloudy) equals P(rain).

Problem 144

Interpret the independence test result in context: P(passing exam | studied) is much higher than P(passing exam).

Problem 145

Interpret the independence test result in context: observed frequency of (male and prefers coffee) equals expected frequency if independent.

Problem 146

Interpret the independence test result in context: P(owns dog and owns cat) is not equal to P(owns dog) * P(owns cat).

Problem 147

Interpret the independence test result in context: a chi-squared test for independence between hair color and eye color yielded a p-value of 0.35.

Problem 148

Interpret the independence test result in context: the probability of a car having an accident is higher on icy roads than on dry roads.

Problem 149

Interpret the independence test result in context: rolling a 6 on a die and flipping heads on a coin satisfy the independence product rule.

Problem 150

Interpret the independence test result in context: drawing two red marbles from a bag without replacement shows P(R1 and R2)!= P(R1)P(R2).

catch wrong denominator, confusing union/joint, and mutual-exclusive mistakes.
12 problems Warmup Practice Mixed Review Assessment
Problem 151

Correct the independence-test error: A student compares P(A union B) to P(A)P(B).

Problem 152

Correct the independence-test error: A student uses the overlap count as P(A) without dividing by total.

Open in simulator
Problem 153

Correct the independence-test error: A student says mutually exclusive positive-probability events are independent.

Problem 154

Correct the independence-test error: A student uses a row total as denominator for P(A) and an overall total for P(B).

Problem 155

Correct the independence-test error: A student uses P(A) as the joint probability P(A and B) in the independence test.

Problem 156

Correct the independence-test error: A student checks independence by seeing if P(A or B) = P(A)P(B).

Problem 157

Correct the independence-test error: A student claims P(A and B) = P(A)P(B) for mutually exclusive events.

Problem 158

Correct the independence-test error: A student calculates P(A and B) by dividing the count of (A and B) by the total count of A, instead of the overall total.

Problem 159

Correct the independence-test error: A student uses P(A|B) in place of P(A) when checking if P(A and B) = P(A)P(B).

Problem 160

Correct the independence-test error: A student attempts to prove independence by checking if P(A or B) = P(A) + P(B).

Problem 161

Correct the independence-test error: A student assumes that if events A and B are independent, then P(A and B) must be 0.

Problem 162

Correct the independence-test error: A student uses P(B|A) as the marginal probability P(B) in the independence test.