Math II · S-CP.8

Applying and Interpreting the General Multiplication Rule

The general Multiplication Rule shows how to find the probability of combined events when one event changes the chance of the next.

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 apply the general Multiplication Rule:

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

Equivalently,

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

It can also be written as

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

The rule says that the probability of both A and B equals the probability of one event times the probability of the other event after the first event is known. The conditional probability is what makes it general. If the events are independent, then \(P(B | A) = P(B)\), and the rule simplifies to \(P(A and B) = P(A)P(B)\). But when events are dependent, the conditional probability must be used.

For example, suppose a bag contains 3 red marbles and 2 blue marbles. Two marbles are drawn without replacement. What is the probability both are red? Let A be “first marble is red.” Let B be “second marble is red.” \(P(A) = 3/5\). After a red marble is removed, 2 red marbles remain out of 4 total marbles. So \(P(B | A) = 2/4\). Therefore

\[P(A and B) = (3/5)(2/4) = 6/20 = 3/10\].

The second probability changed because the first event changed the bag. That is why conditional probability is needed.

This objective is about combined events, especially sequential or dependent ones. Students need to know when simple multiplication is appropriate and when the general rule is required.

Why students should learn this math

Students should learn the general Multiplication Rule because many real events happen in sequences where earlier events affect later probabilities. Drawing cards without replacement changes the deck. A machine that overheats may be more likely to fail next. A customer who ignores several reminders may be less likely to renew. A patient with one symptom may be more likely to have another. A student who misses the first prerequisite skill may be more likely to miss the next one. The probability of a chain of events often requires conditional thinking.

Simple multiplication works only when events are independent. But real life is full of dependence. If we multiply probabilities as if every event is independent, we can get very wrong answers. Risk analysis, medical diagnosis, quality control, fraud detection, sports prediction, and app analytics all require careful conditional multiplication.

The rule also helps students understand tree diagrams. Each branch has a probability, and probabilities along a path are multiplied. But the branch probabilities after the first split may be conditional. The probability of a full path is the product of the probabilities along that path. This is one of the most useful visual tools in probability.

For example, a company might model the probability that a user opens an email and then buys a product. The probability of buying is likely different among users who opened the email than among those who did not. The probability of “opens and buys” is \(P(opens)P(buys | opens)\). That is directly the general Multiplication Rule.

The “why” is that combined events often depend on order and information. The general Multiplication Rule is the math of probability chains.

The historical machinery: probability trees and dependent events

The multiplication of probabilities has roots in early probability problems involving repeated trials. When trials are independent, multiplying probabilities is intuitive: the chance of heads then heads is \(1/2 \cdot 1/2\). But many problems involve changing conditions. Drawing without replacement is a classic example. After the first draw, the sample space changes.

As probability developed, conditional probability became the natural way to handle this. The general Multiplication Rule is essentially the conditional probability formula rearranged. Since

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

multiplying both sides by \(P(A)\) gives

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

So the rule is not arbitrary. It comes directly from the definition of conditional probability.

Tree diagrams became a practical way to represent sequential probability. Each branch records a conditional probability based on the path so far. Multiplying along branches gives path probabilities. Adding path probabilities can then find broader events. This combination of multiplication and addition supports many probability models.

Where this fits in the big map of mathematics

This objective follows conditional probability and the Addition Rule. Students already know how to compute \(P(B | A)\) and how to handle unions. Now they use conditional probability to compute intersections.

It connects to independence. The independent multiplication rule is a special case. Students should understand that \(P(A)P(B)\) is valid only when \(P(B | A) = P(B)\).

It connects to two-way tables. From a table, students can compute \(P(A)\) and \(P(B | A)\) and multiply them to recover \(P(A \cap B)\).

It connects to combinatorics and counting. Sequential probability often depends on whether choices are made with or without replacement.

It connects to decision trees, risk models, and Bayesian reasoning. The general Multiplication Rule is one of the foundations for modeling chains of uncertain events.

The big-map role is intersection through sequence. Students learn how to calculate the probability that multiple events occur together when probabilities may change along the way.

How to execute the skill technically

Use the rule:

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

Step 1: Identify the first event and its probability. Step 2: Identify the probability of the second event given the first. Step 3: Multiply. Step 4: Interpret the path.

Example: A deck has 52 cards. Two cards are drawn without replacement. What is the probability both are aces?

Let A be “first card is an ace.” \(P(A) = 4/52\).

Let B be “second card is an ace.” Given A, there are 3 aces left among 51 cards, so \(P(B | A) = 3/51\).

Therefore

\[P(A \cap B) = (4/52)(3/51) = 12/2652 = 1/221\].

Another example: A quality-control system flags 8% of products for review. Among flagged products, 30% are actually defective. The probability a product is flagged and defective is

\[P(flagged and defective) = P(flagged)P(defective | flagged) = 0.08 \cdot 0.30 = 0.024\].

So 2.4% of all products are expected to be both flagged and defective.

Notice the interpretation: the 30% is not out of all products; it is among flagged products. Multiplication combines the restricted probability with the probability of entering that restricted group.

Tree diagram example

Suppose 60% of users open a lesson reminder. Among users who open it, 50% complete the lesson. Among users who do not open it, 20% complete the lesson.

The probability a user opens and completes is

\[0.60 \cdot 0.50 = 0.30\].

The probability a user does not open and completes is

\[0.40 \cdot 0.20 = 0.08\].

The total probability of completion is the sum of the two paths:

\[0.30 + 0.08 = 0.38\].

This example shows how multiplication and addition work together. Multiply along a path. Add across different paths that lead to the desired outcome.

For an app, this should be interactive. Students should drag branch probabilities and watch path probabilities update. That visual makes the general Multiplication Rule much less abstract.

Another worked example: dependent medical events

Suppose 4% of people in a screened group have a condition. Among people with the condition, 90% test positive. What is the probability that a randomly selected person both has the condition and tests positive?

Let A be “has the condition.” Let B be “tests positive.”

\[P(A) = 0.04\].
\[P(B | A) = 0.90\].

So

\[P(A \cap B) = P(A)P(B | A) = 0.04 \cdot 0.90 = 0.036\].

So 3.6% of the screened group both has the condition and tests positive.

This example is powerful because it shows what the multiplication rule actually does. It starts with the whole population, restricts to the 4% with the condition, then takes 90% of that restricted group. The product is a path probability.

General Multiplication Rule in reverse

The rule can be rearranged. If

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

then when \(P(A) \ne 0\),

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

So the conditional probability formula and the multiplication rule are the same relationship viewed in different directions. Conditional probability divides a joint probability by the condition. The multiplication rule builds a joint probability from a condition and a conditional probability.

This is helpful for students who feel like probability has too many formulas. There are fewer ideas than there appear to be. The formula changes form depending on which quantity is unknown.

Multi-stage extension

The same logic extends to more than two events. For three events,

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

This says multiply along the path, updating the condition each time. For example, drawing three aces in a row without replacement from a standard deck has probability

\[(4/52)(3/51)(2/50)\].

Each probability depends on the previous successful draws. Students do not need advanced notation to understand the idea: after each event, the sample space changes, so the next probability must be updated.

This multi-stage view is important for tree diagrams, simulations, and real-world risk chains. Many events unfold step by step, not all at once.

Connection to independence

The general Multiplication Rule also explains the independence rule. If A and B are independent, then \(P(B | A) = P(B)\). Substituting into the general rule gives

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

So students do not need to memorize two unrelated multiplication rules. There is one general rule. The independent-events version is what happens when the condition does not change the second probability.

A key caution is that many students learn “multiply for and” too broadly. The safer rule is: for “and,” multiply, but make sure the later probability is conditional if the events are dependent.

Common misconceptions and how to avoid them

One common mistake is using \(P(A)P(B)\) when events are dependent. Use \(P(A)P(B | A)\) unless independence is known.

Another mistake is reversing the condition. \(P(B | A)\) means probability of B after A, not probability of A after B.

A third mistake is failing to update counts in without-replacement problems.

A fourth mistake is adding probabilities along a sequence instead of multiplying. Along one path, multiply.

A fifth mistake is multiplying path probabilities and then forgetting to interpret what combined event the product represents.

The big takeaway

The general Multiplication Rule finds the probability that two events both occur. It uses conditional probability because the first event may change the chance of the second. The rule \(P(A \cap B) = P(A)P(B | A)\) is the foundation for sequential probability, tree diagrams, risk chains, and many real-world models of dependent events.

Problem Library

Problems in the App From This Objective

177 problems across 12 archetypes in the app.

multiply marginal and conditional probability.
15 problems Warmup Practice Mixed Review Assessment
Problem 1

Compute joint probability using P(A and B)=P(A)P(B|A) from P(A)=0.40 and P(B|A)=0.25.

Problem 2

Compute joint probability using P(A and B)=P(A)P(B|A) from P(A)=1/3 and P(B|A)=3/5.

Problem 3

Compute joint probability using P(A and B)=P(A)P(B|A) from P(first defective)=0.10 and P(second defective|first defective)=0.20.

Problem 4

Compute joint probability using P(A and B)=P(A)P(B|A) from P(A)=p and P(B|A)=q.

Open in simulator
Problem 5

Compute joint probability using P(A and B)=P(A)P(B|A) from P(A)=0.6 and P(B|A)=0.5.

Problem 6

Compute joint probability using P(A and B)=P(A)P(B|A) from P(A)=0.8 and P(B|A)=0.75.

Problem 7

Compute joint probability using P(A and B)=P(A)P(B|A) from P(A)=1/2 and P(B|A)=1/4.

Problem 8

Compute joint probability using P(A and B)=P(A)P(B|A) from P(A)=2/3 and P(B|A)=1/2.

Problem 9

Compute joint probability using P(A and B)=P(A)P(B|A) from P(A)=20% and P(B|A)=50%.

Problem 10

Compute joint probability using P(A and B)=P(A)P(B|A) from P(A)=75% and P(B|A)=40%.

Problem 11

Compute joint probability using P(A and B)=P(A)P(B|A) from P(X)=0.3 and P(Y|X)=r.

Problem 12

Compute joint probability using P(A and B)=P(A)P(B|A) from P(E)=s and P(F|E)=0.9.

Problem 13

Compute joint probability using P(A and B)=P(A)P(B|A) from P(A)=0.25 and P(B|A)=1/5.

Problem 14

Compute joint probability using P(A and B)=P(A)P(B|A) from P(A)=3/4 and P(B|A)=0.8.

Problem 15

Compute joint probability using P(A and B)=P(A)P(B|A) from P(rain)=0.3 and P(traffic|rain)=0.8.

choose condition direction.
15 problems Warmup Practice Mixed Review Assessment
Problem 16

Compute joint probability using P(B)P(A|B) from P(B)=0.30 and P(A|B)=0.60.

Problem 17

Compute joint probability using P(B)P(A|B) from P(B)=1/2 and P(A|B)=2/5.

Problem 18

Compute joint probability using P(B)P(A|B) from P(has coupon)=25% and P(buys item|coupon)=80%.

Problem 19

Compute joint probability using P(B)P(A|B) from P(B)=b and P(A|B)=c.

Problem 20

Compute joint probability using P(B)P(A|B) from P(B)=0.5 and P(A|B)=0.4.

Problem 21

Compute joint probability using P(B)P(A|B) from P(B)=0.75 and P(A|B)=0.2.

Problem 22

Compute joint probability using P(B)P(A|B) from P(B)=1/3 and P(A|B)=3/4.

Problem 23

Compute joint probability using P(B)P(A|B) from P(B)=2/7 and P(A|B)=1/2.

Problem 24

Compute joint probability using P(B)P(A|B) from P(rain)=40% and P(traffic|rain)=75%.

Open in simulator
Problem 25

Compute joint probability using P(B)P(A|B) from P(student passes)=90% and P(gets scholarship|passes)=10%.

Problem 26

Compute joint probability using P(B)P(A|B) from P(X)=x and P(Y|X)=y.

Problem 27

Compute joint probability using P(B)P(A|B) from P(event E)=0.6 and P(event F|event E)=0.5.

Problem 28

Compute joint probability using P(B)P(A|B) from P(red card)=1/2 and P(face card|red card)=3/13.

Problem 29

Compute joint probability using P(B)P(A|B) from P(B)=0.25 and P(A|B)=1/2.

Problem 30

Compute joint probability using P(B)P(A|B) from P(B)=60% and P(A|B)=0.5.

update probability after first event.
15 problems Warmup Practice Mixed Review Assessment
Problem 31

Use the multiplication rule for dependent events without replacement: draw two aces from a 52-card deck without replacement.

Problem 32

Use the multiplication rule for dependent events without replacement: draw two red marbles from 5 red and 7 blue without replacement.

Problem 33

Use the multiplication rule for dependent events without replacement: select two seniors from 8 seniors and 12 juniors without replacement.

Open in simulator
Problem 34

Use the multiplication rule for dependent events without replacement: draw a heart then a spade without replacement.

Problem 35

Use the multiplication rule for dependent events without replacement: draw two blue balls from a bag containing 6 blue and 4 red balls without replacement.

Problem 36

Use the multiplication rule for dependent events without replacement: draw two non-face cards from a standard 52-card deck without replacement.

Problem 37

Use the multiplication rule for dependent events without replacement: select two defective items from a batch of 3 defective and 17 non-defective items without replacement.

Problem 38

Use the multiplication rule for dependent events without replacement: draw two black socks from a drawer with 7 black and 5 white socks without replacement.

Problem 39

Use the multiplication rule for dependent events without replacement: draw a king then a queen from a 52-card deck without replacement.

Problem 40

Use the multiplication rule for dependent events without replacement: choose two girls from a class of 10 girls and 15 boys without replacement.

Problem 41

Use the multiplication rule for dependent events without replacement: pick two green candies from a jar containing 8 green, 5 yellow, and 7 red candies without replacement.

Problem 42

Use the multiplication rule for dependent events without replacement: draw two even numbers from a set of numbers {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} without replacement.

Problem 43

Use the multiplication rule for dependent events without replacement: select two red apples from a basket of 4 red and 3 green apples without replacement.

Problem 44

Use the multiplication rule for dependent events without replacement: draw a club then a diamond from a 52-card deck without replacement.

Problem 45

Use the multiplication rule for dependent events without replacement: choose two freshmen from a group of 6 freshmen, 8 sophomores, and 10 juniors without replacement.

recognize conditional probability equals marginal.
15 problems Warmup Practice Mixed Review Assessment
Problem 46

Use the multiplication rule for independent events: flip heads and roll a 6.

Problem 47

Use the multiplication rule for independent events: roll two 3s in two die rolls.

Problem 48

Use the multiplication rule for independent events: machine A works with probability 0.90 and machine B works with probability 0.80 independently.

Problem 49

Use the multiplication rule for independent events: event A probability p and independent event B probability q.

Problem 50

Use the multiplication rule for independent events: draw a King from a deck, replace it, then draw a Queen.

Problem 51

Use the multiplication rule for independent events: flip heads twice in a row.

Problem 52

Use the multiplication rule for independent events: roll an even number on a die and then roll a number greater than 4 on a second die.

Problem 53

Use the multiplication rule for independent events: a student passes math with 0.8 probability and science with 0.7 probability, independently.

Problem 54

Use the multiplication rule for independent events: event X probability x and independent event Y probability y.

Problem 55

Use the multiplication rule for independent events: draw a red ball from a bag with 3 red, 2 blue, 5 green, replace it, then draw a blue ball.

Problem 56

Use the multiplication rule for independent events: it rains tomorrow with 0.3 probability and is windy with 0.4 probability, independently.

Problem 57

Use the multiplication rule for independent events: guess correctly on two multiple choice questions, each with 4 options.

Problem 58

Use the multiplication rule for independent events: Team A wins with 0.6 probability and Team B wins with 0.7 probability in their respective independent games.

Open in simulator
Problem 59

Use the multiplication rule for independent events: Component A has a 0.05 defect rate and Component B has a 0.02 defect rate, independently.

Problem 60

Use the multiplication rule for independent events: draw a 7 from a deck, replace it, then draw an Ace.

rearrange multiplication rule.
15 problems Warmup Practice Mixed Review Assessment
Problem 61

Find the missing conditional probability from joint and marginal probabilities P(A and B)=0.18 and P(A)=0.30.

Problem 62

Find the missing conditional probability from joint and marginal probabilities P(A and B)=1/8 and P(B)=1/2.

Problem 63

Find the missing conditional probability from joint and marginal probabilities joint 12%, condition event 40%.

Problem 64

Find the missing conditional probability from joint and marginal probabilities P(A and B)=j and P(A)=p.

Problem 65

Find the missing conditional probability from joint and marginal probabilities P(A and B)=0.24 and P(B)=0.80.

Problem 66

Find the missing conditional probability from joint and marginal probabilities P(A and B)=1/10 and P(A)=1/2.

Problem 67

Find the missing conditional probability from joint and marginal probabilities Joint probability is 15%, probability of B is 50%.

Problem 68

Find the missing conditional probability from joint and marginal probabilities P(X and Y)=0.06 and P(Y)=0.20.

Open in simulator
Problem 69

Find the missing conditional probability from joint and marginal probabilities P(E1 intersect E2)=3/20 and P(E2)=3/5.

Problem 70

Find the missing conditional probability from joint and marginal probabilities The probability of A and B occurring is 21%, the probability of A is 70%.

Problem 71

Find the missing conditional probability from joint and marginal probabilities P(A and B)=0.075 and P(B)=0.25.

Problem 72

Find the missing conditional probability from joint and marginal probabilities Given P(A intersect B)=2/9 and P(A)=4/9.

Problem 73

Find the missing conditional probability from joint and marginal probabilities Joint event probability is 6%, condition event probability is 20%.

Problem 74

Find the missing conditional probability from joint and marginal probabilities P(A and B)=k and P(B)=m.

Problem 75

Find the missing conditional probability from joint and marginal probabilities The joint probability of X and Y is 0.14, and P(X) is 0.35.

multiply along branches.
15 problems Warmup Practice Mixed Review Assessment
Problem 76

Interpret the multiplication rule in a tree diagram from branch A has 0.40, then B after A has 0.25.

Problem 77

Interpret the multiplication rule in a tree diagram from first red 5/12, second red after first red 4/11.

Problem 78

Interpret the multiplication rule in a tree diagram from pass first test 0.80, pass second given first 0.90.

Problem 79

Interpret the multiplication rule in a tree diagram from branch probability p then conditional branch q.

Problem 80

Interpret the multiplication rule in a tree diagram from event X occurs with 0.6, then event Y occurs given X with 0.3.

Problem 81

Interpret the multiplication rule in a tree diagram from draw a blue marble 1/3, then draw another blue marble 1/3 (with replacement).

Problem 82

Interpret the multiplication rule in a tree diagram from probability of rain is 0.7, probability of traffic jam given rain is 0.85.

Problem 83

Interpret the multiplication rule in a tree diagram from first card is a heart 13/52, second card is a heart given first was heart 12/51.

Open in simulator
Problem 84

Interpret the multiplication rule in a tree diagram from P(A) = 0.2, P(B|A) = 0.5.

Problem 85

Interpret the multiplication rule in a tree diagram from flip heads 1/2, then flip heads again 1/2.

Problem 86

Interpret the multiplication rule in a tree diagram from 60% chance of success on first trial, 75% chance of success on second given first.

Problem 87

Interpret the multiplication rule in a tree diagram from student passes math 0.75, student passes science given math 0.8.

Problem 88

Interpret the multiplication rule in a tree diagram from pick a green ball 2/5, then pick a red ball given green 1/4.

Problem 89

Interpret the multiplication rule in a tree diagram from probability of component A failing is 0.05, probability of component B failing given A failed is 0.1.

Problem 90

Interpret the multiplication rule in a tree diagram from probability of drawing a face card 12/52, probability of drawing another face card given first was face card 11/51.

organize sequential events and branch probabilities.
15 problems Warmup Practice Mixed Review Assessment
Problem 91

Build a tree diagram from conditional probabilities P(A)=0.30, P(not A)=0.70, P(B|A)=0.50, P(B|not A)=0.20.

Problem 92

Build a tree diagram from conditional probabilities bag has 3 red and 2 blue, draw twice without replacement.

Problem 93

Build a tree diagram from conditional probabilities P(success)=0.60 and independent second success 0.60.

Problem 94

Build a tree diagram from conditional probabilities first event probabilities p and 1-p, conditional second q and r.

Problem 95

Build a tree diagram from conditional probabilities P(sunny)=0.7, P(cloudy)=0.3, P(play golf|sunny)=0.9, P(play golf|cloudy)=0.4.

Problem 96

Build a tree diagram from conditional probabilities standard deck of 52 cards, draw two cards without replacement, first is red, second is red.

Open in simulator
Problem 97

Build a tree diagram from conditional probabilities urn has 4 green and 6 yellow balls, draw twice with replacement.

Problem 98

Build a tree diagram from conditional probabilities P(disease)=0.01, P(no disease)=0.99, P(positive test|disease)=0.95, P(positive test|no disease)=0.10.

Problem 99

Build a tree diagram from conditional probabilities flip a fair coin twice.

Problem 100

Build a tree diagram from conditional probabilities P(defect A)=0.05, P(no defect A)=0.95, P(defect B|defect A)=0.80, P(defect B|no defect A)=0.15.

Problem 101

Build a tree diagram from conditional probabilities P(team wins first game)=0.6, P(team loses first game)=0.4, P(wins second|wins first)=0.75, P(wins second|loses first)=0.5.

Problem 102

Build a tree diagram from conditional probabilities P(event X)=1/3, P(event Y|event X)=1/2, P(event Y|not event X)=2/5.

Problem 103

Build a tree diagram from conditional probabilities jar has 5 blue, 3 red, 2 green marbles, draw two without replacement.

Problem 104

Build a tree diagram from conditional probabilities P(rain tomorrow)=0.3, P(no rain tomorrow)=0.7, P(heavy traffic|rain)=0.8, P(light traffic|no rain)=0.7.

Problem 105

Build a tree diagram from conditional probabilities P(pass math)=0.8, P(fail math)=0.2, P(pass science|pass math)=0.9, P(pass science|fail math)=0.5.

identify when probabilities change between stages.
15 problems Warmup Practice Mixed Review Assessment
Problem 106

Compare dependent and independent joint probabilities in draw two red cards with replacement versus without replacement.

Problem 107

Compare dependent and independent joint probabilities in two coin flips versus two marble draws without replacement.

Problem 108

Compare dependent and independent joint probabilities in component failures independent versus shared power supply.

Problem 109

Compare dependent and independent joint probabilities in event A probability p, independent B probability q, dependent B after A probability r.

Problem 110

Compare dependent and independent joint probabilities in Drawing two blue marbles from a bag of 5 blue and 5 red, with replacement versus without replacement.

Problem 111

Compare dependent and independent joint probabilities in Drawing two kings from a standard deck of 52 cards, with replacement versus without replacement.

Problem 112

Compare dependent and independent joint probabilities in Flipping a coin twice versus picking two specific students from a class of 20 without replacement.

Problem 113

Compare dependent and independent joint probabilities in Two independent events A and B versus event A influencing event B.

Open in simulator
Problem 114

Compare dependent and independent joint probabilities in Selecting two defective light bulbs from a batch of 100 (5 defective) with replacement versus without replacement.

Problem 115

Compare dependent and independent joint probabilities in The probability of rain in two distant cities on the same day versus the probability of rain on two consecutive days in the same city.

Problem 116

Compare dependent and independent joint probabilities in Two unrelated software modules failing versus one module's failure increasing the chance of another's failure.

Problem 117

Compare dependent and independent joint probabilities in Drawing two even numbers from the set {1, 2,., 10} with replacement versus without replacement.

Problem 118

Compare dependent and independent joint probabilities in Randomly selecting two people for an opinion poll with replacement versus without replacement from a small group of 10.

Problem 119

Compare dependent and independent joint probabilities in Two independent security alarms triggering versus one alarm triggering and causing a cascade of others.

Problem 120

Compare dependent and independent joint probabilities in Probability of a student passing two different, unrelated exams versus passing a prerequisite exam and then a subsequent advanced exam.

multiply sequence probabilities.
15 problems Warmup Practice Mixed Review Assessment
Problem 121

Compute probability of the ordered compound event: red then blue from 4 red and 6 blue without replacement.

Problem 122

Compute probability of the ordered compound event: heads then tails then heads.

Problem 123

Compute probability of the ordered compound event: defective then good from 3 defective and 17 good without replacement.

Problem 124

Compute probability of the ordered compound event: A then B then C with conditional probabilities p, q, r.

Problem 125

Compute probability of the ordered compound event: drawing a King, then a Queen, then an Ace from a standard deck without replacement.

Problem 126

Compute probability of the ordered compound event: drawing two red marbles then one blue marble from a bag with 5 red and 3 blue marbles without replacement.

Problem 127

Compute probability of the ordered compound event: rolling a 6 then a 1 then a 3 on a standard six-sided die.

Problem 128

Compute probability of the ordered compound event: tails then heads then tails then heads.

Problem 129

Compute probability of the ordered compound event: picking a green, then a yellow, then a green from 3 green, 2 yellow, and 5 blue candies without replacement.

Problem 130

Compute probability of the ordered compound event: X then Y with conditional probabilities a, b.

Problem 131

Compute probability of the ordered compound event: E1 then E2 then E3 then E4 with conditional probabilities p1, p2, p3, p4.

Open in simulator
Problem 132

Compute probability of the ordered compound event: selecting a girl then a boy from a group of 7 girls and 5 boys without replacement.

Problem 133

Compute probability of the ordered compound event: drawing a heart then a spade from a standard deck with replacement.

Problem 134

Compute probability of the ordered compound event: drawing two black cards then one red card from a standard deck without replacement.

Problem 135

Compute probability of the ordered compound event: event A (prob 0.3) then event B (prob 0.7).

combine multiplication and addition across paths.
15 problems Warmup Practice Mixed Review Assessment
Problem 136

Compute probability of unordered compound event using multiple paths: one head and one tail in two coin flips.

Problem 137

Compute probability of unordered compound event using multiple paths: one red and one blue from 4 red and 6 blue without replacement.

Problem 138

Compute probability of unordered compound event using multiple paths: at least one success in two independent trials with success probability p.

Problem 139

Compute probability of unordered compound event using multiple paths: one defective and one good from 3 defective and 17 good without replacement.

Problem 140

Compute probability of unordered compound event using multiple paths: exactly two heads in three coin flips.

Problem 141

Compute probability of unordered compound event using multiple paths: exactly one six in two rolls of a fair die.

Problem 142

Compute probability of unordered compound event using multiple paths: one King and one Queen from a standard 52-card deck without replacement.

Problem 143

Compute probability of unordered compound event using multiple paths: at least one success in three independent trials with success probability p.

Problem 144

Compute probability of unordered compound event using multiple paths: one red and one green from 5 red, 3 blue, 2 green marbles without replacement.

Problem 145

Compute probability of unordered compound event using multiple paths: exactly one success in two independent trials with success probabilities p1 and p2.

Problem 146

Compute probability of unordered compound event using multiple paths: one even and one odd number in two rolls of a fair die.

Problem 147

Compute probability of unordered compound event using multiple paths: at least one four in two rolls of a fair die.

Open in simulator
Problem 148

Compute probability of unordered compound event using multiple paths: one blue and one yellow from 7 blue and 3 yellow marbles without replacement.

Problem 149

Compute probability of unordered compound event using multiple paths: exactly one head in three coin flips.

Problem 150

Compute probability of unordered compound event using multiple paths: at least one red from 2 red and 8 green balls in two draws without replacement.

explain both event sequence and result.
15 problems Warmup Practice Mixed Review Assessment
Problem 151

Interpret joint probability from context: P(pass and studied)=0.54.

Problem 152

Interpret joint probability from context: P(red then blue)=2/9.

Problem 153

Interpret joint probability from context: P(A and B)=0.12.

Problem 154

Interpret joint probability from context: P(machine works and sensor works)=0.855.

Problem 155

Interpret joint probability from context: P(rain and cold)=0.3.

Problem 156

Interpret joint probability from context: P(sunny and warm)=60%.

Problem 157

Interpret joint probability from context: P(heads then tails)=1/4.

Problem 158

Interpret joint probability from context: P(defective and returned)=0.02.

Problem 159

Interpret joint probability from context: P(first pick blue then second pick green)=0.15.

Problem 160

Interpret joint probability from context: P(healthy and exercises)=3/5.

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Problem 161

Interpret joint probability from context: P(student passes and attends class)=75%.

Problem 162

Interpret joint probability from context: P(draw King then Queen)=1/221.

Problem 163

Interpret joint probability from context: P(event X and event Y)=0.05.

Problem 164

Interpret joint probability from context: P(successful login then transaction complete)=90%.

Problem 165

Interpret joint probability from context: P(car starts and drives smoothly)=7/8.

catch failure to update denominator, reversed condition, and independent/dependent confusion.
12 problems Warmup Practice Mixed Review Assessment
Problem 166

Correct the multiplication-rule error: A student does not update the denominator after drawing without replacement.

Problem 167

Correct the multiplication-rule error: A student multiplies P(A) by P(A|B) to find P(A and B).

Problem 168

Correct the multiplication-rule error: A student adds branch probabilities along a tree path.

Problem 169

Correct the multiplication-rule error: A student treats dependent events as independent without justification.

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Problem 170

Correct the multiplication-rule error: A student updates only the denominator, not the numerator, for specific items when drawing without replacement.

Problem 171

Correct the multiplication-rule error: A student confuses P(A|B) with P(B|A) when setting up the multiplication rule for dependent events.

Problem 172

Correct the multiplication-rule error: A student multiplies probabilities of mutually exclusive events to find the probability of 'A or B'.

Problem 173

Correct the multiplication-rule error: A student omits a necessary conditional probability term when calculating the probability of a sequence of three or more dependent events.

Problem 174

Correct the multiplication-rule error: A student uses conditional probabilities for events that are independent because items are replaced after each draw.

Problem 175

Correct the multiplication-rule error: A student reverses the order of events when applying the multiplication rule P(A and B) = P(A)P(B|A).

Problem 176

Correct the multiplication-rule error: A student calculates the probability of 'at least one' event by directly multiplying or adding individual probabilities.

Problem 177

Correct the multiplication-rule error: A student fails to multiply the probabilities of all sequential events along a single desired path in a probability tree.