Math II · S-CP.3

Understanding Conditional Probability and Its Connection to Independence

Conditional probability teaches students how evidence changes probability, which is the foundation of medical testing, risk analysis, inference, and rational decision-making.

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 understand conditional probability. Conditional probability is the probability of one event given that another event has occurred. It is written as \(P(A | B)\), read as “the probability of A given B.”

The vertical bar does not mean division by itself. It means the sample space has been restricted. Instead of considering all possible outcomes, we consider only the outcomes in event B. Then we ask what fraction of those outcomes also belong to event A.

The formula is

\[P(A | B) = P(A \cap B) / P(B)\]

as long as \(P(B) \ne 0\).

For example, suppose a card is drawn from a standard deck. Let A be “the card is a king.” Let B be “the card is a face card.” The probability of a king given that the card is a face card is not \(4/52\). Once we know the card is a face card, the relevant sample space is the 12 face cards. Among those, 4 are kings. So

\[P(A | B) = 4/12 = 1/3\].

Conditional probability is about updating the denominator. The old sample space was all 52 cards. The new sample space is only the event we are given.

The objective also asks students to connect conditional probability to independence. If A and B are independent, then knowing B occurred does not change the probability of A. In symbols,

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

If the conditional probability is different from the original probability, the events are dependent. In the card example, \(P(king) = 4/52 = 1/13\), but \(P(king | face card) = 1/3\). Knowing the card is a face card increases the probability that it is a king, so the events are not independent.

This objective is one of the most important probability ideas students will learn. It formalizes the idea that information can change probability.

Why students should learn this math

Students should learn conditional probability because real decisions almost always involve information. The question is rarely “What is the probability in general?” More often it is “What is the probability given what we now know?”

What is the probability a patient has a disease given a positive test? What is the probability a student passes given that they completed the review? What is the probability a customer cancels given that they have not used the app in 30 days? What is the probability of traffic given that it is raining? What is the probability a machine fails given that it overheated yesterday? These are conditional probability questions.

This matters because humans are notoriously bad at reasoning with evidence. People often confuse \(P(A | B)\) with \(P(B | A)\). For example, the probability of a positive test given disease may be high, but the probability of disease given a positive test may be much lower if the disease is rare. This distinction is central in medicine, law, security, and public policy.

Conditional probability is also the foundation of rational updating. New evidence should change beliefs when the evidence is related to the event. If the evidence is independent, it should not change the probability. This is the mathematical version of asking, “Does this information actually matter?”

In data and technology, conditional probability is everywhere. Recommendation systems use behavior given past behavior. Spam filters estimate probability of spam given certain words or patterns. Risk models estimate probability of default given income, credit history, and debt. Weather forecasts estimate probability of rain given atmospheric conditions. Conditional probability is one of the engines of modern prediction.

Students should learn this because it helps them interpret claims responsibly. A headline might say people with a certain condition are more likely to have an outcome. A study might report a conditional percentage. A medical test might have a high accuracy rate. Without conditional probability, students cannot parse what is being conditioned on.

The “why” is that conditional probability is the math of evidence. It teaches students how probabilities change when information changes.

The historical machinery: probability as updating information

Probability began with games of chance but expanded into reasoning under uncertainty. Conditional probability became essential because real-world uncertainty is rarely static. People receive information and update their judgments. Insurance companies update risk based on age, location, and history. Doctors update diagnoses based on symptoms and tests. Scientists update hypotheses based on data.

Bayes' theorem, a major result built from conditional probability, formalizes how to reverse conditional probabilities and update beliefs. Students do not need full Bayesian inference in this objective, but they are learning its foundation. The expression \(P(A | B)\) is one of the core building blocks.

Conditional probability also became central in statistics. Sampling, experiments, observational studies, and inference all involve conditional statements. What is the probability of data given a model? What is the probability of a treatment outcome given assignment? What is the probability of a response given demographic category?

The historical development shows why this objective matters. Probability is not just counting equally likely outcomes. It is a language for revising expectations when the known information changes.

Where this fits in the big map of mathematics

This objective follows independence and leads into two-way tables, everyday-language probability explanations, and the multiplication rule. It is the central concept in the probability sequence.

It connects to sample spaces because conditional probability restricts the sample space. Event B becomes the new universe.

It connects to independence because independence means conditioning does not change probability. If \(P(A | B) = P(A)\), B gives no probability information about A. If not, B matters.

It connects to two-way tables because conditional probabilities can be computed by using row or column totals as denominators. This is often the clearest way for students to see the restricted sample space.

It connects to multiplication rules because \(P(A | B) = P(A \cap B)/P(B)\) can be rearranged as \(P(A \cap B) = P(B)P(A | B)\). That becomes the general multiplication rule.

It connects to statistics, machine learning, medical testing, and decision science. Conditional probability is one of the central ideas in any field that uses evidence to update predictions.

The big-map role is evidence. Students learn how information changes probability and when it does not.

How to execute the skill technically

To compute \(P(A | B)\), use the event after “given” as the denominator. This is the most important habit. In \(P(A | B)\), B is the new sample space.

Formula method:

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

Count method:

\[P(A | B) = number of outcomes in both A and B / number of outcomes in B\].

Example: a class has 30 students. 18 play a sport. 12 play an instrument. 6 do both. Let A be “plays a sport” and B be “plays an instrument.” Find \(P(A | B)\).

The given condition is B, so focus only on the 12 students who play an instrument. Of those, 6 also play a sport. So

\[P(A | B) = 6/12 = 1/2\].

Now find \(P(B | A)\). Focus on the 18 students who play a sport. Of those, 6 play an instrument. So

\[P(B | A) = 6/18 = 1/3\].

These are different. That is a crucial lesson: \(P(A | B)\) is not generally equal to \(P(B | A)\).

To test independence, compare conditional and original probabilities. In the example, \(P(A) = 18/30 = 3/5\). But \(P(A | B) = 1/2\). Since \(1/2 \ne 3/5\), A and B are not independent.

A worked card example: find \(P(red | queen)\). Given queen means restrict to the four queens. Two are red. So \(P(red | queen) = 2/4 = 1/2\). Since \(P(red) = 1/2\), being a queen does not change the probability of being red in a standard deck; these events are independent.

Medical-test example: why conditional probability is hard but necessary

Suppose a disease affects 1% of a population. A test is positive for 95% of people who have the disease, and it is falsely positive for 5% of people who do not have the disease. Imagine 10,000 people.

About 100 have the disease. Of those, 95 test positive.

About 9,900 do not have the disease. Of those, 5% test positive, which is 495 false positives.

So there are \(95 + 495 = 590\) positive tests total. Among those positive tests, only 95 are true disease cases. Therefore

\[P(disease | positive) = 95/590 ≈ 0.161\].

That is about 16.1%, even though the test catches 95% of disease cases. This surprises many students because they confuse \(P(positive | disease)\) with \(P(disease | positive)\). They are not the same.

This is one of the most important real-world lessons in probability. Conditional probability is not optional if people want to understand medical testing, screening, false positives, legal evidence, fraud detection, security alerts, or rare-event prediction. The base rate matters.

Conditional probability as changing the denominator

The easiest way to keep conditional probability straight is to ask: “What group am I looking inside?” In \(P(A | B)\), you are looking inside B. B becomes the denominator. Then you count how much of B is also A.

In a table, this means choose the row or column named after the given condition. In a Venn diagram, it means zoom into the region for B and ignore everything outside it. In a sentence, it means “among those where B is true, what fraction also have A?”

That denominator shift is the entire concept. Students who master it can solve most conditional probability problems. Students who do not master it may memorize formulas but still reverse the condition.

Common misconceptions and how to avoid them

The biggest misconception is reversing the condition. \(P(A | B)\) is not the same as \(P(B | A)\). Always ask which event comes after “given.”

Another mistake is using the full sample space denominator after a condition is given. Conditional probability changes the denominator.

A third mistake is thinking any two related-sounding events must be dependent. Check the probabilities.

A fourth mistake is thinking independent means impossible to occur together. That is mutually exclusive, not independent.

A fifth mistake is ignoring base rates. In medical or risk contexts, the overall frequency of the event matters greatly when interpreting conditional probability.

The big takeaway

Conditional probability is probability after information. \(P(A | B)\) asks for the probability of A when B is known to have happened. The event B becomes the new sample space. Independence is the special case where the new information does not change the probability. This idea is central to evidence, data, medical testing, inference, and rational decision-making.

Problem Library

Problems in the App From This Objective

177 problems across 12 archetypes in the app.

divide favorable outcomes within given condition.
15 problems Warmup Practice Mixed Review Assessment
Problem 1

Compute conditional probability from counts: 30 students play sports, 12 of those play soccer.

Problem 2

Compute conditional probability from counts: table has 50 total, 20 in B, 8 in A and B.

Open in simulator
Problem 3

Compute conditional probability from counts: 10 red cards among 26 black-or-red cards? condition is red or face with 16 outcomes and 6 favorable.

Problem 4

Compute conditional probability from counts: in group B there are b outcomes, and j are also in A.

Problem 5

Compute conditional probability from counts: Out of 40 students who took an exam, 25 passed. 15 of those who passed scored above 90.

Problem 6

Compute conditional probability from counts: A bag contains 100 marbles. 60 are red, and 20 of the red marbles are striped.

Problem 7

Compute conditional probability from counts: In a survey of 200 people, 120 prefer coffee. Of those who prefer coffee, 80 also prefer tea.

Problem 8

Compute conditional probability from counts: From a standard deck of 52 cards, if a card drawn is a face card (Jack, Queen, King), what is the probability it is a King? There are 12 face cards and 4 Kings.

Problem 9

Compute conditional probability from counts: A dartboard has 3 concentric circles. The inner circle has area 10, the middle ring has area 20, and the outer ring has area 30. If a dart lands in the middle or outer ring, what is the probability it landed in the middle ring? Total area for condition is 20+30=50, favorable is 20.

Problem 10

Compute conditional probability from counts: Given P(B)=0.4 and P(A and B)=0.1, with a total sample space of 100 outcomes. This means 40 outcomes are in B, and 10 outcomes are in A and B.

Problem 11

Compute conditional probability from counts: Among 't' total items, 'c' items satisfy condition C. Out of those 'c' items, 'f' items also satisfy event F.

Problem 12

Compute conditional probability from counts: In a class of 25 students, 15 are girls. 10 of the girls wear glasses.

Problem 13

Compute conditional probability from counts: There are 75 items in a set. 30 items are in event B. 18 items are in both A and B.

Problem 14

Compute conditional probability from counts: Out of 50 animals in a zoo enclosure, 35 are mammals. 14 of the mammals are carnivores.

Problem 15

Compute conditional probability from counts: A survey found that among 150 people who own pets, 90 own dogs. Of those who own dogs, 60 also own cats.

use `P(A|B)=P(A and B)/P(B)`.
15 problems Warmup Practice Mixed Review Assessment
Problem 16

Compute conditional probability from probabilities: P(A and B)=0.18, P(B)=0.30.

Problem 17

Compute conditional probability from probabilities: P(A and B)=1/8, P(B)=1/2.

Problem 18

Compute conditional probability from probabilities: P(A and B)=12%, P(B)=40%.

Problem 19

Compute conditional probability from probabilities: P(A and B)=j, P(B)=p.

Problem 20

Compute conditional probability from probabilities: P(E and F)=0.25, P(F)=0.50.

Problem 21

Compute conditional probability from probabilities: P(G and H)=0.12, P(H)=0.60.

Problem 22

Compute conditional probability from probabilities: P(I and J)=2/9, P(J)=1/3.

Problem 23

Compute conditional probability from probabilities: P(K and L)=3/10, P(L)=3/5.

Problem 24

Compute conditional probability from probabilities: P(M and N)=20%, P(N)=80%.

Problem 25

Compute conditional probability from probabilities: P(O and P)=10%, P(P)=30%.

Problem 26

Compute conditional probability from probabilities: P(Q and R)=x, P(R)=y.

Problem 27

Compute conditional probability from probabilities: P(S and T)=0.05, P(T)=0.25.

Problem 28

Compute conditional probability from probabilities: P(U and V)=1/5, P(V)=1/2.

Problem 29

Compute conditional probability from probabilities: P(W and X)=15%, P(X)=50%.

Problem 30

Compute conditional probability from probabilities: P(Y and Z)=a/4, P(Z)=a/2.

Open in simulator
identify condition and target event.
15 problems Warmup Practice Mixed Review Assessment
Problem 31

Interpret the conditional probability notation P(A|B) in words.

Problem 32

Interpret the conditional probability notation P(B|A) in words.

Problem 33

Interpret the conditional probability notation P(likes soccer | plays sports) in words.

Problem 34

Interpret the conditional probability notation P(red | face card) in words.

Problem 35

Interpret the conditional probability notation P(X|Y) in words.

Problem 36

Interpret the conditional probability notation P(E1|E2) in words.

Problem 37

Interpret the conditional probability notation P(rain | cloudy) in words.

Problem 38

Interpret the conditional probability notation P(sick | positive test) in words.

Problem 39

Interpret the conditional probability notation P(pass | studied) in words.

Problem 40

Interpret the conditional probability notation P(defective | factory A) in words.

Open in simulator
Problem 41

Interpret the conditional probability notation P(second head | first head) in words.

Problem 42

Interpret the conditional probability notation P(age > 50 | female) in words.

Problem 43

Interpret the conditional probability notation P(dog owner | cat owner) in words.

Problem 44

Interpret the conditional probability notation P(successful | practiced) in words.

Problem 45

Interpret the conditional probability notation P(sunny | no clouds) in words.

identify given condition.
15 problems Warmup Practice Mixed Review Assessment
Problem 46

Translate the verbal conditional probability into notation: probability of A given B.

Open in simulator
Problem 47

Translate the verbal conditional probability into notation: probability a student plays soccer given the student plays sports.

Problem 48

Translate the verbal conditional probability into notation: probability a card is a king given it is a face card.

Problem 49

Translate the verbal conditional probability into notation: probability of passing given studying.

Problem 50

Translate the verbal conditional probability into notation: probability of E given F.

Problem 51

Translate the verbal conditional probability into notation: probability a person has a disease given they test positive.

Problem 52

Translate the verbal conditional probability into notation: probability of rain if it's cloudy.

Problem 53

Translate the verbal conditional probability into notation: probability of a student being male among those who study math.

Problem 54

Translate the verbal conditional probability into notation: probability of rolling an even number given it's greater than 3.

Problem 55

Translate the verbal conditional probability into notation: probability a card is red given it's a heart.

Problem 56

Translate the verbal conditional probability into notation: probability a customer buys product A given they viewed product B.

Problem 57

Translate the verbal conditional probability into notation: probability of sunshine given it's summer.

Problem 58

Translate the verbal conditional probability into notation: probability of getting an A given consistent effort.

Problem 59

Translate the verbal conditional probability into notation: probability of event X given event Y.

Problem 60

Translate the verbal conditional probability into notation: probability a student passes the exam given they attended all lectures.

recognize reversed conditions generally differ.
12 problems Warmup Practice Mixed Review Assessment
Problem 61

Compare P(A|B) and P(B|A) from A and B count 12, A total 30, B total 20.

Problem 62

Compare P(A|B) and P(B|A) from A and B count 8, A total 16, B total 16.

Problem 63

Compare P(A|B) and P(B|A) from 10 students are athletes and musicians, 25 are athletes, 40 are musicians.

Problem 64

Compare P(A|B) and P(B|A) from joint j, A total a, B total b.

Problem 65

Compare P(A|B) and P(B|A) from A and B count 5, A total 10, B total 20.

Problem 66

Compare P(A|B) and P(B|A) from 10 people like both coffee and tea, 50 people like coffee, 20 people like tea.

Problem 67

Compare P(A|B) and P(B|A) from The joint probability of X and Y is 0.1, P(X)=0.2, P(Y)=0.5.

Problem 68

Compare P(A|B) and P(B|A) from A and B count 7, A total 14, B total 14.

Problem 69

Compare P(A|B) and P(B|A) from 3 students passed both math and science, 30 students passed math, 10 students passed science.

Open in simulator
Problem 70

Compare P(A|B) and P(B|A) from joint probability of E and F is 0.05, P(E)=0.1, P(F)=0.1.

Problem 71

Compare P(A|B) and P(B|A) from A and B count 20, A total 40, B total 80.

Problem 72

Compare P(A|B) and P(B|A) from 6 people own both a car and a bike, 18 people own a car, 12 people own a bike.

compare `P(A|B)` to `P(A)`.
15 problems Warmup Practice Mixed Review Assessment
Problem 73

Use conditional probability to test independence from P(A)=0.40 and P(A|B)=0.40.

Open in simulator
Problem 74

Use conditional probability to test independence from P(A)=0.30 and P(A|B)=0.50.

Problem 75

Use conditional probability to test independence from P(B)=2/5 and P(B|A)=2/5.

Problem 76

Use conditional probability to test independence from P(A)=a and P(A|B)=c.

Problem 77

Use conditional probability to test independence from P(B)=0.65 and P(B|A)=0.65.

Problem 78

Use conditional probability to test independence from P(B)=0.25 and P(B|A)=0.30.

Problem 79

Use conditional probability to test independence from P(A)=1/4 and P(A|B)=1/4.

Problem 80

Use conditional probability to test independence from P(A)=1/2 and P(A|B)=1/3.

Problem 81

Use conditional probability to test independence from P(C)=0.1 and P(C|D)=0.1.

Problem 82

Use conditional probability to test independence from P(E)=3/4 and P(E|F)=1/2.

Problem 83

Use conditional probability to test independence from P(X)=p and P(X|Y)=p.

Problem 84

Use conditional probability to test independence from P(Y)=q and P(Y|X)=r.

Problem 85

Use conditional probability to test independence from P(A)=0.75 and P(A|B)=3/4.

Problem 86

Use conditional probability to test independence from P(B)=0.6 and P(B|A)=0.5.

Problem 87

Use conditional probability to test independence from P(E)=1/3 and P(E|F)=1/2.

state condition does not alter likelihood.
15 problems Warmup Practice Mixed Review Assessment
Problem 88

Explain independence as unchanged conditional probability for coin result and die result.

Problem 89

Explain independence as unchanged conditional probability for grade level and club membership.

Problem 90

Explain independence as unchanged conditional probability for drawing with replacement.

Problem 91

Explain independence as unchanged conditional probability for machine failures on separate circuits.

Open in simulator
Problem 92

Explain independence as unchanged conditional probability for weather in London and weather in New York.

Problem 93

Explain independence as unchanged conditional probability for rolling two separate dice.

Problem 94

Explain independence as unchanged conditional probability for drawing a card and then drawing another after replacement.

Problem 95

Explain independence as unchanged conditional probability for flipping two coins.

Problem 96

Explain independence as unchanged conditional probability for gender of a first child and gender of a second child.

Problem 97

Explain independence as unchanged conditional probability for birthdays of two randomly chosen people.

Problem 98

Explain independence as unchanged conditional probability for traffic light status at two intersections miles apart.

Problem 99

Explain independence as unchanged conditional probability for choosing a number from one hat and a number from another hat.

Problem 100

Explain independence as unchanged conditional probability for a student's shoe size and their favorite color.

Problem 101

Explain independence as unchanged conditional probability for battery life of two devices operating independently.

Problem 102

Explain independence as unchanged conditional probability for outcome of two separate lottery tickets.

rearrange conditional formula.
15 problems Warmup Practice Mixed Review Assessment
Problem 103

Find the missing joint probability from conditional probability data P(A|B)=0.60 and P(B)=0.25.

Problem 104

Find the missing joint probability from conditional probability data P(A|B)=2/5 and P(B)=1/2.

Problem 105

Find the missing joint probability from conditional probability data P(success|practice)=75% and P(practice)=40%.

Problem 106

Find the missing joint probability from conditional probability data P(A|B)=c and P(B)=p.

Problem 107

Find the missing joint probability from conditional probability data P(X|Y)=0.75 and P(Y)=0.40.

Problem 108

Find the missing joint probability from conditional probability data P(E|F)=0.8 and P(F)=0.15.

Problem 109

Find the missing joint probability from conditional probability data P(C|D)=3/4 and P(D)=1/3.

Problem 110

Find the missing joint probability from conditional probability data P(G|H)=1/2 and P(H)=2/3.

Problem 111

Find the missing joint probability from conditional probability data P(rain|cloudy)=60% and P(cloudy)=70%.

Problem 112

Find the missing joint probability from conditional probability data P(pass|study)=90% and P(study)=80%.

Problem 113

Find the missing joint probability from conditional probability data P(M|N)=0.5 and P(N)=1/4.

Problem 114

Find the missing joint probability from conditional probability data P(Q|R)=2/5 and P(R)=0.5.

Problem 115

Find the missing joint probability from conditional probability data P(X|Y)=x and P(Y)=y.

Problem 116

Find the missing joint probability from conditional probability data P(A|B)=0.2x and P(B)=0.5.

Open in simulator
Problem 117

Find the missing joint probability from conditional probability data P(event1|event2)=5/6 and P(event2)=3/5.

update sample space after first event.
15 problems Warmup Practice Mixed Review Assessment
Problem 118

Use conditional probability in a without-replacement context: draw two red cards from a 52-card deck without replacement.

Problem 119

Use conditional probability in a without-replacement context: draw a blue marble then another blue from 5 blue and 3 red without replacement.

Problem 120

Use conditional probability in a without-replacement context: draw a king after first drawing an ace and not replacing it.

Problem 121

Use conditional probability in a without-replacement context: select two defective parts from 6 defective and 14 good without replacement.

Open in simulator
Problem 122

Use conditional probability in a without-replacement context: draw two Queens from a standard 52-card deck without replacement.

Problem 123

Use conditional probability in a without-replacement context: select two green apples from a basket containing 7 green and 5 red apples without replacement.

Problem 124

Use conditional probability in a without-replacement context: choose two students who are seniors from a class of 10 seniors and 15 juniors without replacement.

Problem 125

Use conditional probability in a without-replacement context: draw a red marble then a green marble from a bag with 4 red, 3 green, and 2 blue marbles without replacement.

Problem 126

Use conditional probability in a without-replacement context: draw two face cards from a 52-card deck without replacement.

Problem 127

Use conditional probability in a without-replacement context: select two mystery novels from a shelf containing 8 mystery, 5 fantasy, and 7 sci-fi novels without replacement.

Problem 128

Use conditional probability in a without-replacement context: pick two yellow candies from a jar containing 6 yellow, 4 orange, and 5 purple candies without replacement.

Problem 129

Use conditional probability in a without-replacement context: draw a Spade then a Heart from a 52-card deck without replacement.

Problem 130

Use conditional probability in a without-replacement context: select two even-numbered balls from a bag containing balls numbered 1 through 10 without replacement.

Problem 131

Use conditional probability in a without-replacement context: choose a girl then a boy from a group of 8 girls and 7 boys without replacement.

Problem 132

Use conditional probability in a without-replacement context: pick two black socks from a drawer containing 3 black, 2 white, and 1 striped sock without replacement.

restrict denominator to condition region.
15 problems Warmup Practice Mixed Review Assessment
Problem 133

Interpret conditional probability from Venn diagram data A only 8, overlap 12, B only 10.

Problem 134

Interpret conditional probability from Venn diagram data A only 5, overlap 15, B only 20.

Problem 135

Interpret conditional probability from Venn diagram data overlap j, B only b.

Problem 136

Interpret conditional probability from Venn diagram data overlap 6, A only 9, outside 15.

Problem 137

Interpret conditional probability from Venn diagram data A only 10, overlap 5, B only 15.

Problem 138

Interpret conditional probability from Venn diagram data A only 7, overlap 3, B only 12.

Problem 139

Interpret conditional probability from Venn diagram data A only 10, overlap 8, total B 20.

Problem 140

Interpret conditional probability from Venn diagram data B only 15, overlap 7, total A 25.

Problem 141

Interpret conditional probability from Venn diagram data X only 12, overlap 6, Y only 18.

Problem 142

Interpret conditional probability from Venn diagram data X only 9, overlap 4, Y only 11.

Problem 143

Interpret conditional probability from Venn diagram data A only 15, overlap 5, B only 20, outside 10.

Problem 144

Interpret conditional probability from Venn diagram data A only 11, overlap 9, B only 14, outside 7.

Problem 145

Interpret conditional probability from Venn diagram data A only x, overlap y, B only z.

Problem 146

Interpret conditional probability from Venn diagram data overlap m, total B n.

Problem 147

Interpret conditional probability from Venn diagram data overlap 10, B only 8, total A 30.

Open in simulator
compare conditional rates across groups.
15 problems Warmup Practice Mixed Review Assessment
Problem 148

Decide whether conditional probability supports the claim: Claim: club members are more likely to prefer math; P(math|club)=0.70 and P(math|not club)=0.45.

Problem 149

Decide whether conditional probability supports the claim: Claim: treatment lowers risk; P(risk|treatment)=0.12 and P(risk|no treatment)=0.08.

Problem 150

Decide whether conditional probability supports the claim: Claim: events are independent; P(A)=0.30 and P(A|B)=0.30.

Problem 151

Decide whether conditional probability supports the claim: Claim: B causes A because P(A|B)>P(A).

Problem 152

Decide whether conditional probability supports the claim: Claim: students who study daily are more likely to pass; P(pass|study daily)=0.90 and P(pass|not study daily)=0.60.

Problem 153

Decide whether conditional probability supports the claim: Claim: eating breakfast makes you more alert; P(alert|breakfast)=0.75 and P(alert|no breakfast)=0.80.

Problem 154

Decide whether conditional probability supports the claim: Claim: older cars are less likely to pass inspection; P(pass|older car)=0.40 and P(pass|newer car)=0.70.

Problem 155

Decide whether conditional probability supports the claim: Claim: being left-handed makes you less likely to be an engineer; P(engineer|left-handed)=0.15 and P(engineer|right-handed)=0.12.

Open in simulator
Problem 156

Decide whether conditional probability supports the claim: Claim: gender and preference for sci-fi movies are independent; P(sci-fi|male)=0.55 and P(sci-fi|female)=0.55.

Problem 157

Decide whether conditional probability supports the claim: Claim: owning a pet is independent of living in an apartment; P(pet|apartment)=0.30 and P(pet|house)=0.60.

Problem 158

Decide whether conditional probability supports the claim: Claim: regular exercise causes better health because P(good health|exercise)>P(good health|no exercise).

Problem 159

Decide whether conditional probability supports the claim: Claim: people with a college degree are more likely to earn over $70k; P(>$70k|degree)=0.65 and P(>$70k|no degree)=0.35.

Problem 160

Decide whether conditional probability supports the claim: Claim: smokers are less likely to live past 70; P(live past 70|smoker)=0.30 and P(live past 70|non-smoker)=0.60.

Problem 161

Decide whether conditional probability supports the claim: Claim: eye color and hair color are independent; P(blue eyes|blonde hair)=0.40 and P(blue eyes|brown hair)=0.20.

Problem 162

Decide whether conditional probability supports the claim: Claim: eating organic food prevents illness because P(illness|organic food) is lower than P(illness|non-organic food).

catch reversed condition, wrong denominator, and independence confusion.
15 problems Warmup Practice Mixed Review Assessment
Problem 163

Correct the conditional-probability error: A student computes P(A|B) by dividing by the total sample space.

Problem 164

Correct the conditional-probability error: A student treats P(A|B) and P(B|A) as always equal.

Problem 165

Correct the conditional-probability error: A student says P(A|B)=P(A) proves dependence.

Problem 166

Correct the conditional-probability error: A student uses P(A or B) in the numerator for P(A|B).

Problem 167

Correct the conditional-probability error: A student calculates P(B|A) when the problem asks for P(A|B).

Problem 168

Correct the conditional-probability error: To find P(X|Y), a student divides P(X and Y) by P(X).

Problem 169

Correct the conditional-probability error: A student computes P(A|B) as P(A) assuming independence, but it's not given.

Problem 170

Correct the conditional-probability error: A student incorrectly identifies the 'given that' event as the numerator event.

Problem 171

Correct the conditional-probability error: A student states that if two events are mutually exclusive, they must be independent.

Problem 172

Correct the conditional-probability error: When calculating P(E|F), a student uses P(E) in the numerator instead of P(E and F).

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Problem 173

Correct the conditional-probability error: A student counts all occurrences of event A for P(A|B), not just those within B.

Problem 174

Correct the conditional-probability error: A student finds P(A and B) by multiplying P(A|B) by P(A).

Problem 175

Correct the conditional-probability error: A student provides P(A and B) as the answer when asked for P(A|B).

Problem 176

Correct the conditional-probability error: To find P(C|D), a student uses P(D') as the denominator.

Problem 177

Correct the conditional-probability error: A student claims that P(A|B) can never be greater than P(A).