Math II · S-CP.6

Computing Conditional Probability as a Fraction of Outcomes

Conditional probability teaches students to choose the right denominator: not all outcomes, but the outcomes inside the condition.

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 compute conditional probability from outcomes. Conditional probability is not just a formula; it is a fraction formed from a restricted sample space. The event after “given” becomes the new denominator.

The standard says to compute conditional probability as the fraction of one event's outcomes that also belong to another event. In notation,

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

but the counting version is often clearer:

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

The denominator is the condition. In \(P(A | B)\), the condition is B. You are looking only at outcomes in B. Among those B outcomes, you count how many also fall in A.

For example, suppose a die is rolled. Let A be “roll an even number,” so \(A = {2, 4, 6}\). Let B be “roll a number greater than 3,” so \(B = {4, 5, 6}\). To find \(P(A | B)\), restrict the sample space to B: \({4, 5, 6}\). Among those outcomes, the even outcomes are \({4, 6}\). So

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

The original sample space had six outcomes, but the conditional probability denominator is 3 because B has three outcomes.

This objective is asking students to make that denominator shift automatic. Most conditional probability errors come from using the wrong denominator or reversing the condition. Students should learn to ask: “Given what? Which outcomes are still possible? Among those, how many satisfy the event I care about?”

Why students should learn this math

Students should learn this because conditional probability is one of the most useful tools for reasoning with evidence. In real life, we rarely make decisions with no information. We make decisions after learning something: a test result, a weather forecast, a customer's behavior, a student's attendance record, a machine warning, a symptom, a survey response, or a previous event. Conditional probability tells us how to update the probability once that information is known.

The critical skill is choosing the right denominator. Many people misuse percentages because they divide by the wrong group. If a school wants to know the pass rate among students who attended tutoring, the denominator is tutoring students, not all students. If a doctor wants to know the disease rate among people who tested positive, the denominator is positive tests, not the whole population. If a company wants to know renewal rate among active users, the denominator is active users, not all users ever acquired.

This denominator discipline is deeply practical. It affects medical decisions, legal reasoning, business analytics, education evaluation, risk assessment, and public-policy claims. A percentage without a clear denominator can mislead. Conditional probability forces the denominator into the open.

Students should also learn this because it gives meaning to two-way tables and Venn diagrams. In a two-way table, conditional probability means selecting a row or column total as the denominator. In a Venn diagram, it means zooming into one event region and asking what fraction overlaps another event. These visual models help students see the fraction rather than memorize a formula.

The “why” is that conditional probability is how data becomes evidence. It says: given this information, what fraction of the remaining possible cases have the property we care about?

The historical machinery: restriction of the sample space

Conditional probability formalizes a simple but powerful idea: information reduces the set of possible cases. If you draw a card and know it is red, the sample space is no longer all 52 cards. It is the 26 red cards. If you know a person tested positive, the relevant group is no longer everyone. It is the positive-test group.

This restriction idea became central as probability moved beyond games of chance into statistics, medicine, science, and inference. In games, sample spaces are often fixed and equally likely. In real-world reasoning, sample spaces shift as information arrives. Conditional probability became the mathematical language for that shift.

The formula \(P(A | B) = P(A \cap B) / P(B)\) can be understood as rescaling probability inside B. The event B becomes the new whole. The overlap between A and B becomes the successful part of that new whole. This idea is foundational for Bayes' theorem, Markov chains, decision trees, and statistical inference.

Students do not need the advanced machinery yet, but they are learning the central move: condition means restrict the universe.

Where this fits in the big map of mathematics

This objective follows the interpretation objective and gives the computation method. Objective 126 emphasized everyday language. Objective 127 emphasizes the actual fraction.

It connects to sample spaces and event subsets. Conditional probability is impossible without knowing what events contain and how they overlap.

It connects to two-way tables. Conditional probabilities are row or column percentages. Students can compute them from observed data.

It connects to independence. If \(P(A | B) = P(A)\), then B does not change the probability of A. Computing conditional probability gives a way to test that.

It connects to the multiplication rule. Rearranging the formula gives \(P(A \cap B) = P(B)P(A | B)\). This appears in Objective 129.

It connects to real-world inference. Many statistical claims are conditional percentages.

The big-map role is computation with meaning. Students learn that conditional probability is a fraction inside a condition.

How to execute the skill technically

Use the following routine:

  1. Identify the event after the bar. That is the condition and denominator.
  2. Count how many outcomes are in the condition.
  3. Count how many of those outcomes also satisfy the event before the bar.
  4. Form the fraction and interpret it.

Example: A class has 25 students. 15 are in band, 10 play a sport, and 6 do both. Let A be “plays a sport” and B be “in band.” Find \(P(A | B)\).

The condition is B, in band. There are 15 band students. Of those, 6 also play a sport. So

\[P(A | B) = 6/15 = 2/5\].

Interpretation: among students in band, 40% play a sport.

Find \(P(B | A)\). Now the condition is A, plays a sport. There are 10 sport students. Of those, 6 are in band. So

\[P(B | A) = 6/10 = 3/5\].

These are different because the denominator changed.

Another example: draw one card from a standard deck. Find \(P(heart | red)\). Given red means the denominator is 26 red cards. Of those, 13 are hearts. So \(P(heart | red) = 13/26 = 1/2\).

Find \(P(red | heart)\). Given heart means the denominator is 13 hearts. All hearts are red, so \(P(red | heart) = 13/13 = 1\).

These two examples show why condition direction matters.

Using two-way tables to compute conditional probability

Suppose a table summarizes 200 customers.

| | Renewed | Did not renew | Total | |---|---:|---:|---:| | Used feature | 84 | 36 | 120 | | Did not use feature | 32 | 48 | 80 | | Total | 116 | 84 | 200 |

Find \(P(renewed | used feature)\). The condition is used feature, so use the row total 120 as denominator. Among those, 84 renewed. So

\[P(renewed | used feature) = 84/120 = 0.70\].

Find \(P(used feature | renewed)\). The condition is renewed, so use the column total 116. Among those, 84 used the feature. So

\[P(used feature | renewed) = 84/116 ≈ 0.724\].

The two conditional probabilities are close here but not identical. They answer different questions. One is a renewal rate among feature users. The other is a feature-usage rate among renewed customers.

A website/app should make this denominator shift visible by highlighting the relevant row or column when the user selects the condition. Students should see the denominator change before they calculate.

Another worked example: conditional probability from a Venn diagram

Suppose a survey of 80 students shows that 35 students play a sport, 28 students are in a club, and 14 students do both. Let A be “plays a sport” and B be “is in a club.”

Find \(P(A | B)\). The condition is B, so the denominator is the number of students in a club: 28. Of those 28 club students, 14 also play a sport. Therefore

\[P(A | B) = 14/28 = 1/2\].

Find \(P(B | A)\). Now the condition is A, so the denominator is the number of sport students: 35. Of those 35 sport students, 14 are also in a club. Therefore

\[P(B | A) = 14/35 = 2/5\].

The overlap is the same in both calculations, but the denominator changes. This is the whole reason the two conditional probabilities differ.

A Venn diagram helps. The overlap region has 14. The whole B circle has 28. The whole A circle has 35. When computing \(P(A | B)\), cover everything outside B. When computing \(P(B | A)\), cover everything outside A. The visible universe changes.

Interpreting conditional probability carefully

A conditional probability should always be written as a sentence. \(P(A | B) = 1/2\) is not complete understanding. The sentence is: “Among students in a club, one-half also play a sport.” That sentence names the denominator group first.

This is important because conditional probabilities often appear in arguments. “Among users who completed onboarding, 65% returned the next day” is a conditional probability. It does not say that 65% of all users returned. It says that within the onboarding-completion group, the return rate was 65%.

Good probability writing should make the condition visible. Use phrases like “among,” “given,” “out of those who,” or “within the group.” If the sentence does not reveal the denominator, it is probably too vague.

Common misconceptions and how to avoid them

The biggest mistake is using the full sample space denominator. In conditional probability, the condition is the denominator.

Another mistake is reversing \(P(A | B)\) and \(P(B | A)\). The event after the bar controls the denominator.

A third mistake is assuming conditional probability is always smaller than ordinary probability. It can be smaller, larger, or equal.

A fourth mistake is forgetting the overlap. The numerator must satisfy both events, not just the event before the bar.

A fifth mistake is treating the formula as abstract when a counting interpretation is available. For many student problems, “out of the condition group, how many also...” is the clearest method.

The big takeaway

Conditional probability is a fraction inside a restricted sample space. \(P(A | B)\) means: among outcomes in B, what fraction also belong to A? The denominator is not the whole sample space; it is the given condition. Mastering this denominator shift is the key to probability with evidence.

Problem Library

Problems in the App From This Objective

177 problems across 12 archetypes in the app.

restrict to given event and count overlap.
15 problems Warmup Practice Mixed Review Assessment
Problem 1

Compute conditional probability from finite sample space S={1,2,3,4,5,6}; find P(even | greater than 3).

Problem 2

Compute conditional probability from finite sample space two coins; find P(two heads | first is heads).

Problem 3

Compute conditional probability from finite sample space one card from ace, king, queen, jack; find P(king | face card).

Problem 4

Compute conditional probability from finite sample space S has condition set B with b outcomes and overlap A and B with j outcomes.

Problem 5

Compute conditional probability from finite sample space a single die roll; find P(prime | odd).

Problem 6

Compute conditional probability from finite sample space two fair dice are rolled; find P(first die is 3 | sum is 7).

Problem 7

Compute conditional probability from finite sample space a bag contains 3 red, 2 blue, and 5 green marbles; find P(blue | not red).

Problem 8

Compute conditional probability from finite sample space a card is drawn from a standard 52-card deck; find P(face card | heart).

Problem 9

Compute conditional probability from finite sample space a class has 10 boys and 15 girls. 3 boys wear glasses and 5 girls wear glasses. A student is chosen at random; find P(boy | wears glasses).

Problem 10

Compute conditional probability from finite sample space a letter is chosen randomly from the word MISSISSIPPI; find P('I' | is a vowel).

Problem 11

Compute conditional probability from finite sample space S={1,2,3,4,5,6,7,8,9,10}; find P(even | multiple of 3).

Problem 12

Compute conditional probability from finite sample space two coins are flipped; find P(two heads | at least one head).

Open in simulator
Problem 13

Compute conditional probability from finite sample space a day of the week is chosen randomly; find P(weekday | starts with 'T').

Problem 14

Compute conditional probability from finite sample space an urn contains 4 red, 6 blue, and 2 yellow balls; find P(red | not blue).

Problem 15

Compute conditional probability from finite sample space two fair dice are rolled; find P(sum is 6 | product is even).

divide intersection by condition region.
15 problems Warmup Practice Mixed Review Assessment
Problem 16

Compute conditional probability from Venn diagram data A only 4, overlap 6, B only 10; find P(A|B).

Problem 17

Compute conditional probability from Venn diagram data A only 9, overlap 3, B only 6; find P(B|A).

Problem 18

Compute conditional probability from Venn diagram data overlap j and B only b; find P(A|B).

Problem 19

Compute conditional probability from Venn diagram data A only 5 and overlap 15; find P(B|A).

Problem 20

Compute conditional probability from Venn diagram data A only 7, overlap 3, B only 5; find P(A|B).

Problem 21

Compute conditional probability from Venn diagram data A only 12, overlap 4, B only 8; find P(B|A).

Problem 22

Compute conditional probability from Venn diagram data A only 10, overlap 5, B only 15, outside 2; find P(A|B).

Problem 23

Compute conditional probability from Venn diagram data A only 6, overlap 2, B only 4, outside 1; find P(B|A).

Problem 24

Compute conditional probability from Venn diagram data A only 20, overlap 10, B only 30; find P(A|B).

Problem 25

Compute conditional probability from Venn diagram data A only 25, overlap 15, B only 35; find P(B|A).

Problem 26

Compute conditional probability from Venn diagram data A only x, overlap y, B only z; find P(A|B).

Problem 27

Compute conditional probability from Venn diagram data A only p, overlap q, B only r; find P(B|A).

Problem 28

Compute conditional probability from Venn diagram data A only 10, overlap 8, total B 18; find P(A|B).

Open in simulator
Problem 29

Compute conditional probability from Venn diagram data overlap 7, B only 12, total A 15; find P(B|A).

Problem 30

Compute conditional probability from Venn diagram data A only 1, B only 2, overlap 3, outside 4; find P(A|B).

use row or column denominator.
15 problems Warmup Practice Mixed Review Assessment
Problem 31

Compute conditional probability from a two-way table: freshman row total 20, freshman bus cell 8; find P(bus|freshman).

Problem 32

Compute conditional probability from a two-way table: bus column total 30, freshman bus cell 12; find P(freshman|bus).

Problem 33

Compute conditional probability from a two-way table: row total r, cell c.

Open in simulator
Problem 34

Compute conditional probability from a two-way table: column total k, cell c.

Problem 35

Compute conditional probability from a two-way table: sports column total 50, male sports cell 30; find P(male|sports).

Problem 36

Compute conditional probability from a two-way table: male row total 70, male sports cell 30; find P(sports|male).

Problem 37

Compute conditional probability from a two-way table: music column total 40, female music cell 25; find P(female|music).

Problem 38

Compute conditional probability from a two-way table: female row total 60, female music cell 25; find P(music|female).

Problem 39

Compute conditional probability from a two-way table: math column total 80, 10th grade math cell 20; find P(10th grade|math).

Problem 40

Compute conditional probability from a two-way table: 10th grade row total 75, 10th grade math cell 20; find P(math|10th grade).

Problem 41

Compute conditional probability from a two-way table: apartment column total 60, dog apartment cell 15; find P(dog|apartment).

Problem 42

Compute conditional probability from a two-way table: dog row total 45, dog apartment cell 15; find P(apartment|dog).

Problem 43

Compute conditional probability from a two-way table: work column total 100, car work cell 70; find P(car|work).

Problem 44

Compute conditional probability from a two-way table: car row total 85, car work cell 70; find P(work|car).

Problem 45

Compute conditional probability from a two-way table: agree column total 90, under 30 agree cell 40; find P(under 30|agree).

use joint percentage over condition percentage.
15 problems Warmup Practice Mixed Review Assessment
Problem 46

Compute conditional probability from percentages P(A and B)=12%, P(B)=30%.

Problem 47

Compute conditional probability from percentages joint is 0.18 and condition is 0.60.

Problem 48

Compute conditional probability from percentages 8% are seniors in band; 20% are in band.

Problem 49

Compute conditional probability from percentages joint j% and condition b%.

Problem 50

Compute conditional probability from percentages P(X and Y)=15%, P(Y)=50%.

Problem 51

Compute conditional probability from percentages P(C and D)=0.05, P(D)=0.25.

Problem 52

Compute conditional probability from percentages 10% of students are both on the honor roll and play sports; 40% play sports.

Problem 53

Compute conditional probability from percentages joint probability is 0.06 and the condition probability is 0.15.

Problem 54

Compute conditional probability from percentages P(E and F)=21%, P(F)=70%.

Problem 55

Compute conditional probability from percentages P(M and N)=36%, P(N)=45%.

Problem 56

Compute conditional probability from percentages 18% of people like coffee and tea; 60% like tea.

Open in simulator
Problem 57

Compute conditional probability from percentages joint is 0.04 and condition is 0.08.

Problem 58

Compute conditional probability from percentages P(G and H)=10%, P(H)=25%.

Problem 59

Compute conditional probability from percentages 5% of employees are managers who work remotely; 20% of employees work remotely.

Problem 60

Compute conditional probability from percentages P(S and T)=0.12, P(T)=0.48.

multiply conditional rate by condition total.
15 problems Warmup Practice Mixed Review Assessment
Problem 61

Find the joint count from conditional probability and condition count: P(A|B)=0.40 and B count is 50.

Problem 62

Find the joint count from conditional probability and condition count: P(success|practice)=75% and 80 practiced.

Problem 63

Find the joint count from conditional probability and condition count: P(A|B)=2/5 and B count is 30.

Problem 64

Find the joint count from conditional probability and condition count: P(A|B)=c and B count is b.

Open in simulator
Problem 65

Find the joint count from conditional probability and condition count: P(X|Y)=0.25 and Y count is 100.

Problem 66

Find the joint count from conditional probability and condition count: P(rain|cloudy)=60% and 30 days were cloudy.

Problem 67

Find the joint count from conditional probability and condition count: P(pass|studied)=1/3 and 45 students studied.

Problem 68

Find the joint count from conditional probability and condition count: The probability of finding a bug given a new feature is 0.15, and 20 new features were released.

Problem 69

Find the joint count from conditional probability and condition count: P(defective|batch)=2% and 500 items in the batch.

Problem 70

Find the joint count from conditional probability and condition count: P(correct|review)=3/4 and 120 assignments were reviewed.

Problem 71

Find the joint count from conditional probability and condition count: P(win|home)=0.7 and 40 home games played.

Problem 72

Find the joint count from conditional probability and condition count: P(like|seen)=50% and 60 people saw the ad.

Problem 73

Find the joint count from conditional probability and condition count: P(E|F)=p and F count is n.

Problem 74

Find the joint count from conditional probability and condition count: P(success|try)=1/k and k_count tried.

Problem 75

Find the joint count from conditional probability and condition count: P(event_A|event_B)=d and event_B count is N.

rearrange conditional fraction.
15 problems Warmup Practice Mixed Review Assessment
Problem 76

Find the condition total from joint count and conditional probability: joint count 18 and P(A|B)=0.60.

Problem 77

Find the condition total from joint count and conditional probability: 12 people are in A and B, and P(A|B)=2/5.

Problem 78

Find the condition total from joint count and conditional probability: 45 successes are 75% of those who practiced.

Problem 79

Find the condition total from joint count and conditional probability: joint count j and P(A|B)=c.

Problem 80

Find the condition total from joint count and conditional probability: joint count 25 and P(X|Y)=0.5.

Problem 81

Find the condition total from joint count and conditional probability: 20 students passed both tests, and P(passed test A | passed test B)=1/4.

Problem 82

Find the condition total from joint count and conditional probability: 36 items are defective and from batch B, and 30% of batch B items are defective.

Problem 83

Find the condition total from joint count and conditional probability: If 54 people have both characteristic C and D, and the probability of having C given D is 0.9.

Open in simulator
Problem 84

Find the condition total from joint count and conditional probability: 15 events occurred in both categories, and the conditional probability of one given the other is 3/8.

Problem 85

Find the condition total from joint count and conditional probability: joint occurrence of 120 and P(E|F)=0.25.

Problem 86

Find the condition total from joint count and conditional probability: 72 employees are both full-time and have benefits, and 80% of full-time employees have benefits.

Problem 87

Find the condition total from joint count and conditional probability: 6 items are in set A and B, and P(A|B)=1/3.

Problem 88

Find the condition total from joint count and conditional probability: A survey found 108 respondents who like both coffee and tea, and 0.45 is the conditional probability of liking coffee given liking tea.

Problem 89

Find the condition total from joint count and conditional probability: 21 successful trials were observed in group X, and 35% of group X trials were successful.

Problem 90

Find the condition total from joint count and conditional probability: The intersection of events M and N has a count of 48. P(M|N) is 4/7.

compute and compare rates for groups.
15 problems Warmup Practice Mixed Review Assessment
Problem 91

Compare two conditional probabilities from P(pass|studied)=0.85 and P(pass|did not study)=0.45.

Problem 92

Compare two conditional probabilities from P(bus|freshman)=8/20 and P(bus|sophomore)=15/30.

Problem 93

Compare two conditional probabilities from P(A|B)=0.30 and P(A|not B)=0.30.

Problem 94

Compare two conditional probabilities from Group 1 has 12/40 and group 2 has 18/45.

Problem 95

Compare two conditional probabilities from P(rain|cloudy)=0.60 and P(rain|clear)=0.10.

Problem 96

Compare two conditional probabilities from P(success|treatment)=0.75 and P(success|placebo)=0.80.

Problem 97

Compare two conditional probabilities from P(defect|machine A)=0.05 and P(defect|machine B)=0.05.

Problem 98

Compare two conditional probabilities from P(like|male)=25/50 and P(like|female)=35/60.

Problem 99

Compare two conditional probabilities from P(flu|vaccinated)=5/100 and P(flu|unvaccinated)=15/100.

Problem 100

Compare two conditional probabilities from P(pass|online)=15/25 and P(pass|in-person)=18/30.

Problem 101

Compare two conditional probabilities from P(satisfied|product A)=70% and P(satisfied|product B)=65%.

Problem 102

Compare two conditional probabilities from P(error|method X)=12% and P(error|method Y)=8%.

Open in simulator
Problem 103

Compare two conditional probabilities from P(return|customer type 1)=20% and P(return|customer type 2)=20%.

Problem 104

Compare two conditional probabilities from P(approve|group A)=0.75 and P(approve|group B)=4/5.

Problem 105

Compare two conditional probabilities from P(attend|event 1)=0.40 and P(attend|event 2)=50%.

state denominator explicitly.
15 problems Warmup Practice Mixed Review Assessment
Problem 106

Interpret conditional probability as a fraction of a subgroup: P(likes art | senior)=18/30.

Problem 107

Interpret conditional probability as a fraction of a subgroup: P(A|B)=j/b.

Problem 108

Interpret conditional probability as a fraction of a subgroup: P(pass|studied)=0.90.

Problem 109

Interpret conditional probability as a fraction of a subgroup: P(red|face card)=6/12.

Problem 110

Interpret conditional probability as a fraction of a subgroup: P(has flu | tested positive)=0.95.

Problem 111

Interpret conditional probability as a fraction of a subgroup: P(defective | from batch A)=7/100.

Problem 112

Interpret conditional probability as a fraction of a subgroup: P(eats meat | vegetarian)=0/50.

Problem 113

Interpret conditional probability as a fraction of a subgroup: P(sunny | cloudy)=0.

Problem 114

Interpret conditional probability as a fraction of a subgroup: P(voted | registered)=0.75.

Problem 115

Interpret conditional probability as a fraction of a subgroup: P(female | student)=150/300.

Open in simulator
Problem 116

Interpret conditional probability as a fraction of a subgroup: P(blue eyes | blonde hair)=12/20.

Problem 117

Interpret conditional probability as a fraction of a subgroup: P(wins | plays well)=0.8.

Problem 118

Interpret conditional probability as a fraction of a subgroup: P(late | heavy traffic)=15/25.

Problem 119

Interpret conditional probability as a fraction of a subgroup: P(likes coffee | likes tea)=0.6.

Problem 120

Interpret conditional probability as a fraction of a subgroup: P(prime | number < 10)=4/9.

parse "given," "among," and "of those."
12 problems Warmup Practice Mixed Review Assessment
Problem 121

Choose the correct conditional denominator from wording: among students who play sports, what fraction play soccer?.

Problem 122

Choose the correct conditional denominator from wording: of those who passed, what fraction studied?.

Problem 123

Choose the correct conditional denominator from wording: given that the card is red, what is the probability it is a king?.

Problem 124

Choose the correct conditional denominator from wording: what fraction of club members are seniors?.

Problem 125

Choose the correct conditional denominator from wording: among the animals at the shelter, what fraction are cats?.

Problem 126

Choose the correct conditional denominator from wording: of those who attended the concert, what fraction bought merchandise?.

Problem 127

Choose the correct conditional denominator from wording: given that a student passed the test, what is the probability they studied for it?.

Problem 128

Choose the correct conditional denominator from wording: what percentage of the total budget was spent on marketing?.

Problem 129

Choose the correct conditional denominator from wording: among the books on the shelf, what is the probability one is a novel?.

Problem 130

Choose the correct conditional denominator from wording: of those who prefer coffee, what fraction also like espresso?.

Open in simulator
Problem 131

Choose the correct conditional denominator from wording: given that a fruit is an apple, what is the probability it is red?.

Problem 132

Choose the correct conditional denominator from wording: what proportion of the surveyed individuals are under 30?.

compare conditional rates and interpret.
15 problems Warmup Practice Mixed Review Assessment
Problem 133

Use conditional probability to make a decision: P(recover|treatment)=0.70, P(recover|control)=0.55.

Problem 134

Use conditional probability to make a decision: P(defect|supplier A)=0.03, P(defect|supplier B)=0.08.

Problem 135

Use conditional probability to make a decision: P(pass|course 1)=0.82, P(pass|course 2)=0.81.

Problem 136

Use conditional probability to make a decision: P(risk|choice A)=0.12, P(risk|choice B)=0.20.

Problem 137

Use conditional probability to make a decision: P(success|method X)=0.95, P(success|method Y)=0.88.

Problem 138

Use conditional probability to make a decision: P(satisfaction|product A)=0.65, P(satisfaction|product B)=0.75.

Open in simulator
Problem 139

Use conditional probability to make a decision: P(positive test|drug A)=0.99, P(positive test|drug B)=0.99.

Problem 140

Use conditional probability to make a decision: P(error|process 1)=0.01, P(error|process 2)=0.005.

Problem 141

Use conditional probability to make a decision: P(conversion|campaign A)=0.05, P(conversion|campaign B)=0.048.

Problem 142

Use conditional probability to make a decision: P(rejection|batch 1)=0.15, P(rejection|batch 2)=0.10.

Problem 143

Use conditional probability to make a decision: P(improvement|therapy A)=0.60, P(improvement|therapy B)=0.58.

Problem 144

Use conditional probability to make a decision: P(return|strategy X)=0.10, P(return|strategy Y)=0.12.

Problem 145

Use conditional probability to make a decision: P(completion|group A)=0.78, P(completion|group B)=0.78.

Problem 146

Use conditional probability to make a decision: P(adverse event|medication P)=0.02, P(adverse event|medication Q)=0.015.

Problem 147

Use conditional probability to make a decision: P(engagement|platform A)=0.45, P(engagement|platform B)=0.50.

know probabilities range from 0 to 1 and denominator cannot be zero.
15 problems Warmup Practice Mixed Review Assessment
Problem 148

Identify impossible conditional probability values from P(A|B)=1.2.

Problem 149

Identify impossible conditional probability values from P(A|B)=-0.1.

Problem 150

Identify impossible conditional probability values from P(A|B) when P(B)=0.

Problem 151

Identify impossible conditional probability values from P(A|B)=0.75 with P(B)>0.

Problem 152

Identify impossible conditional probability values from P(A|B)=0 with P(B)>0.

Problem 153

Identify impossible conditional probability values from P(X|Y)=1 with P(Y)>0.

Problem 154

Identify impossible conditional probability values from P(E|F)=1/2 with P(F)>0.

Open in simulator
Problem 155

Identify impossible conditional probability values from P(C|D)=0.3 with P(D)>0.

Problem 156

Identify impossible conditional probability values from P(G|H)=-1/4.

Problem 157

Identify impossible conditional probability values from P(I|J)=5/4.

Problem 158

Identify impossible conditional probability values from P(K|L)=110%.

Problem 159

Identify impossible conditional probability values from P(M|N)=-5%.

Problem 160

Identify impossible conditional probability values from P(R|S)=0.5 when P(S)=0.

Problem 161

Identify impossible conditional probability values from P(U|V) when P(V)=0.

Problem 162

Identify impossible conditional probability values from P(W|Z)=45% with P(Z)>0.

catch wrong denominator, wrong intersection, and percent/count mix-ups.
15 problems Warmup Practice Mixed Review Assessment
Problem 163

Correct the conditional-probability computation error: A student divides overlap by total instead of by the condition region.

Open in simulator
Problem 164

Correct the conditional-probability computation error: A student uses A only instead of A and B for P(A|B).

Problem 165

Correct the conditional-probability computation error: A student mixes 12 people with 30% as if both were counts.

Problem 166

Correct the conditional-probability computation error: A student computes P(A|B) when the wording asks P(B|A).

Problem 167

Correct the conditional-probability computation error: A student uses the probability of the target event as the denominator for conditional probability.

Problem 168

Correct the conditional-probability computation error: A student uses the probability of the intersection as the denominator for P(A|B).

Problem 169

Correct the conditional-probability computation error: A student uses the probability of the union as the denominator for P(A|B).

Problem 170

Correct the conditional-probability computation error: A student uses P(A) * P(B) as the numerator for P(A|B), assuming independence.

Problem 171

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

Problem 172

Correct the conditional-probability computation error: A student uses the complement of the condition P(B') as the denominator for P(A|B).

Problem 173

Correct the conditional-probability computation error: A student computes P(A and B) instead of P(A|B).

Problem 174

Correct the conditional-probability computation error: A student reverses the numerator and denominator in the conditional probability formula, calculating P(B) / P(A and B) for P(A|B).

Problem 175

Correct the conditional-probability computation error: A student misinterprets 'P(A given B)' as P(A) + P(B).

Problem 176

Correct the conditional-probability computation error: A student computes P(B) / P(A) for P(A|B).

Problem 177

Correct the conditional-probability computation error: A student computes P(A) instead of P(A|B) when a condition is clearly stated.