Sample Spaces and Events

Probability is the mathematics of uncertainty. Before we can calculate the probability of an event, we need to know all the outcomes that could possibly occur.

The complete collection of possible outcomes is called the sample space.

For example, when rolling a standard six-sided die:

S = {1, 2, 3, 4, 5, 6}

If we are interested in rolling an even number, the event is:

E = {2, 4, 6}

The sample space describes everything that could happen, while an event describes the outcomes we are interested in.

Outcomes, Sample Spaces, and Events

These three ideas form the foundation of probability.

Outcome

An outcome is one possible result of an experiment.

Examples:

• rolling a 4
• flipping heads
• drawing the queen of hearts
• selecting a red marble

Sample Space

The sample space is the set of all possible outcomes.

It is commonly represented by:

S

Event

An event is a collection of one or more outcomes from the sample space.

For a die:

S = {1, 2, 3, 4, 5, 6}

An event A representing "roll an odd number" is:

A = {1, 3, 5}

An event B representing "roll a number greater than 4" is:

B = {5, 6}

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Constructing a Sample Space

A sample space should include:

• every possible outcome
• each outcome only once
• clearly distinguishable outcomes

For a coin toss:

S = {H, T}

where:

H = heads
T = tails

For choosing a day of the weekend:

S = {Saturday, Sunday}

For selecting a letter from the word MATH:

S = {M, A, T, H}

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Sample Spaces and Set Notation

Because a sample space is a set, we normally write its outcomes inside braces:

{ }

For example:

S = {1, 2, 3, 4, 5, 6}

Individual outcomes are separated by commas.

If:

A = {2, 4, 6}

then A represents an event containing three outcomes.

The notation:

2 ∈ A

means:

"2 is an element of A."

The notation:

3 ∉ A

means:

"3 is not an element of A."

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Simple Events

A simple event contains only one outcome.

Suppose:

S = {1, 2, 3, 4, 5, 6}

The event:

A = {4}

is a simple event because it contains exactly one outcome.

Examples include:

• rolling exactly 5
• drawing the ace of spades
• flipping heads once
• selecting one specific person

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Compound Events

A compound event contains more than one outcome.

For a die:

A = {2, 4, 6}

represents the event:

"roll an even number."

This is a compound event because three different outcomes satisfy the condition.

Another event:

B = {4, 5, 6}

could represent:

"roll a number greater than 3."

Compound events are extremely common in probability.

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Counting Outcomes

The notation:

n(S)

means the number of outcomes in the sample space.

For:

S = {1, 2, 3, 4, 5, 6}

we have:

n(S) = 6

If:

A = {2, 4, 6}

then:

n(A) = 3

When all outcomes are equally likely:

P(A) = n(A)/n(S)

Therefore:

P(A) = 3/6

= 1/2

This connects sample spaces directly to theoretical probability.

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Two Coin Tosses

Suppose we toss a coin twice.

Simply writing:

{H, T}

is no longer enough because we need to record the result of both tosses.

The sample space is:

S = {HH, HT, TH, TT}

There are four possible outcomes.

Notice that:

HT ≠ TH

because the order is different.

Events with Two Coins

Using:

S = {HH, HT, TH, TT}

consider the event:

A = exactly one head

Then:

A = {HT, TH}

So:

n(A) = 2

and:

n(S) = 4

Therefore:

P(A) = 2/4

= 1/2

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At Least and At Most

Probability questions often use phrases such as at least and at most.

At Least

"At least one head" means:

one or more heads

For two tosses:

A = {HH, HT, TH}

The only excluded outcome is TT.

Therefore:

P(A) = 3/4

At Most

"At most one head" means:

zero or one head

So:

B = {HT, TH, TT}

Again:

P(B) = 3/4

Understanding this language is essential when constructing events correctly.

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Using a Table to Construct a Sample Space

Suppose we roll two six-sided dice.

There are too many outcomes to list casually, so a table is useful.

Each outcome can be represented as an ordered pair:

(first die, second die)

For example:

(2, 5)

means:

first die = 2
second die = 5

A systematic sample space begins:

(1,1), (1,2), (1,3), ..., (1,6)

(2,1), (2,2), (2,3), ..., (2,6)

and continues through:

(6,1), (6,2), ..., (6,6)

There are:

6 × 6 = 36 outcomes

Why Ordered Pairs Matter

When rolling two dice:

(2, 5)

and:

(5, 2)

are different outcomes.

They produce the same total, but the individual dice show different results.

This distinction is important.

There are 36 equally likely ordered outcomes, but there are only 11 possible sums:

2, 3, 4, ..., 12

These sums are not equally likely.

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Example: Sum of 7

Let event A be:

"the sum of two dice is 7."

Then:

A = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

Therefore:

n(A) = 6

Since:

n(S) = 36

we have:

P(A) = 6/36

= 1/6

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Why Some Dice Sums Are More Likely

Consider a sum of 2.

Only one outcome produces it:

(1,1)

So:

P(sum of 2) = 1/36

Now consider a sum of 7.

Six outcomes produce it.

Therefore:

P(sum of 7) = 6/36

= 1/6

So although the numbers 2 and 7 are both possible sums, they are not equally likely.

This illustrates why constructing the complete sample space can be so useful.

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Tree Diagrams

A tree diagram is another systematic method for constructing sample spaces.

Suppose we toss a coin and then roll a three-sided spinner labelled:

1, 2, 3

The first stage has:

H or T

Each of these branches then has:

1, 2, or 3

The outcomes are:

S = {H1, H2, H3, T1, T2, T3}

Therefore:

n(S) = 6

Tree diagrams are especially useful for experiments that occur in stages.

The Multiplication Principle

If an experiment has several stages, the number of possible outcomes can often be found by multiplication.

Suppose:

• Stage 1 has 3 possible outcomes.
• Stage 2 has 4 possible outcomes.

Then:

total outcomes = 3 × 4 = 12

If a meal includes:

3 starters

4 main courses

2 desserts

then the number of possible meals is:

3 × 4 × 2 = 24

This is sometimes called the fundamental counting principle.

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Events as Sets

Events are sets, so we can use set notation to describe relationships between them.

Suppose:

S = {1, 2, 3, 4, 5, 6}

Let:

A = {2, 4, 6}

represent even numbers.

Let:

B = {4, 5, 6}

represent numbers greater than 3.

These events overlap.

The outcome 4 belongs to both.

The outcome 6 also belongs to both.

This leads to important set operations.

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Intersection: A AND B

The intersection of two events contains outcomes that belong to both events.

Notation:

A ∩ B

Using:

A = {2, 4, 6}

B = {4, 5, 6}

we get:

A ∩ B = {4, 6}

This represents:

"even AND greater than 3."

Union: A OR B

The union contains outcomes belonging to A, B, or both.

Notation:

A ∪ B

Using:

A = {2, 4, 6}

B = {4, 5, 6}

we obtain:

A ∪ B = {2, 4, 5, 6}

Notice that 4 and 6 are written only once.

Sets do not list duplicate elements.

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Complement: NOT A

The complement of an event contains all outcomes in the sample space that are not in the event.

It can be written as:

Aᶜ

Suppose:

S = {1, 2, 3, 4, 5, 6}

and:

A = {2, 4, 6}

Then:

Aᶜ = {1, 3, 5}

If A means:

"roll an even number"

then Aᶜ means:

"do not roll an even number."

In this case:

"roll an odd number."

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Complement Probability

Because an event either occurs or does not occur:

P(A) + P(Aᶜ) = 1

Therefore:

P(Aᶜ) = 1 − P(A)

This is called the complement rule.

For example, if:

P(A) = 0.7

then:

P(Aᶜ) = 1 − 0.7

= 0.3

This becomes extremely useful for problems involving phrases such as:

"at least one"

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Mutually Exclusive Events

Two events are mutually exclusive if they cannot occur at the same time.

For a single die roll:

A = "roll a 2"

B = "roll a 5"

A and B cannot both occur on the same roll.

Therefore:

A ∩ B = ∅

The symbol:

∅

means the empty set.

It contains no outcomes.

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Events That Are Not Mutually Exclusive

Suppose:

A = "roll an even number"

B = "roll a number greater than 3"

Then:

A = {2, 4, 6}

B = {4, 5, 6}

Since:

A ∩ B = {4, 6}

the events can occur together.

Therefore, they are not mutually exclusive.

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Certain and Impossible Events

The sample space itself represents a certain event.

Therefore:

P(S) = 1

The empty set represents an impossible event.

Therefore:

P(∅) = 0

All probabilities must satisfy:

0 ≤ P(A) ≤ 1

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Venn Diagrams

A Venn diagram provides a visual representation of events.

Usually:

• the rectangle represents the sample space
• circles represent events
• overlapping regions represent intersections
• everything inside either circle represents a union
• everything outside an event represents its complement

Venn diagrams become especially useful when several events overlap.

Example Using a Venn Diagram

Suppose a group of 30 students includes:

18 who play football

12 who play basketball

5 who play both

Let:

F = football

B = basketball

Since 5 students belong to both groups:

n(F ∩ B) = 5

Students who play only football:

18 − 5 = 13

Students who play only basketball:

12 − 5 = 7

Students who play at least one sport:

13 + 5 + 7

= 25

Therefore, students who play neither:

30 − 25

= 5

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The Addition Rule

When events overlap:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

Why subtract the intersection?

Because outcomes belonging to both A and B would otherwise be counted twice.

If events are mutually exclusive:

P(A ∩ B) = 0

so the rule simplifies to:

P(A ∪ B) = P(A) + P(B)

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Example: Playing Sports

Using the previous example:

30 students total

18 play football

12 play basketball

5 play both

Probability that a randomly selected student plays football or basketball:

P(F ∪ B) = 18/30 + 12/30 − 5/30

= 25/30

= 5/6

The subtraction prevents the five students who play both sports from being counted twice.

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Systematic Listing

When constructing a sample space, it is important to work systematically.

Suppose a café offers:

Bread:

white, wholegrain

Filling:

cheese, chicken, tuna

The possible sandwiches are:

white-cheese

white-chicken

white-tuna

wholegrain-cheese

wholegrain-chicken

wholegrain-tuna

There are:

2 × 3 = 6 possibilities

A systematic approach helps prevent missing or repeating outcomes.

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Three Coin Tosses

Suppose a fair coin is tossed three times.

The sample space is:

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Therefore:

n(S) = 8

This also follows from:

2 × 2 × 2 = 8

or:

2³ = 8

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Example: Exactly Two Heads

Let A represent:

"exactly two heads."

From the sample space:

A = {HHT, HTH, THH}

Therefore:

n(A) = 3

and:

P(A) = 3/8

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Example: At Least One Head

Using three coin tosses:

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

The event "at least one head" includes seven outcomes.

We could count all seven.

But there is an easier method.

The complement of "at least one head" is:

no heads

which is:

TTT

Therefore:

P(at least one H)

= 1 − P(TTT)

= 1 − 1/8

= 7/8

This is a powerful use of the complement rule.

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Simple and Compound Experiments

A simple experiment involves one stage.

Examples:

• roll one die
• flip one coin
• draw one card

A compound experiment involves two or more stages.

Examples:

• flip two coins
• roll two dice
• select a card and then spin a spinner
• choose a shirt and trousers
• toss a coin three times

Compound experiments usually have larger sample spaces and benefit from tables or tree diagrams.

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Ordered and Unordered Outcomes

Sometimes order matters.

Suppose two students are selected for:

captain and vice-captain

Choosing:

Alice as captain and Ben as vice-captain

is different from:

Ben as captain and Alice as vice-captain.

Order matters because the roles are different.

But if two students are simply selected to form a team, then:

Alice and Ben

is the same pair as:

Ben and Alice.

Before constructing a sample space, always ask:

Does order matter?

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Equally Likely Outcomes

The simple probability formula:

P(A) = n(A)/n(S)

works directly when the outcomes in S are equally likely.

For a fair die:

{1, 2, 3, 4, 5, 6}

are equally likely.

But consider a spinner where half the circle is red and the remaining half is divided between blue and green.

Writing:

S = {red, blue, green}

does not mean each colour has probability 1/3.

The outcomes listed by name are not necessarily equally likely.

This distinction is important when solving probability problems.

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Sample Spaces and Real-World Decisions

Sample spaces are not limited to coins and dice.

They can be used to analyse:

• medical testing outcomes
• weather possibilities
• manufacturing defects
• sports results
• genetics
• insurance risks
• survey responses
• transportation choices
• quality control
• computer algorithms

The central idea remains the same:

identify all possible outcomes before calculating probabilities.

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A Systematic Problem-Solving Strategy

When solving a sample-space problem:

Identify the Experiment

What process is taking place?

Identify Each Stage

Is there one action or several?

Decide Whether Order Matters

Would changing the order create a different outcome?

Construct the Sample Space

Use:

• a list
• a table
• a tree diagram
• ordered pairs
• set notation

Define the Event

Identify exactly which outcomes satisfy the condition.

Count the Outcomes

Determine:

n(S) and n(A)

Calculate Probability

If outcomes are equally likely:

P(A) = n(A)/n(S)

Check

Make sure:

0 ≤ P(A) ≤ 1

and that no possible outcomes were missed.

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Common Mistakes

Confusing an outcome with an event

An outcome is one possible result.

An event may contain one or many outcomes.

Leaving outcomes out of the sample space

A sample space must include every possible outcome.

Counting an outcome twice

Each distinct outcome should appear only once.

Assuming order never matters

HT and TH are different outcomes when recording two sequential coin tosses.

Assuming order always matters

If two people are simply selected as a pair, AB and BA may represent the same outcome.

Confusing AND with OR

A ∩ B means A AND B.

A ∪ B means A OR B.

Forgetting the overlap

When finding A ∪ B, outcomes in A ∩ B should not be counted twice.

Misinterpreting "at least"

"At least 2" means:

2 or more

not exactly 2.

Misinterpreting "at most"

"At most 2" means:

2 or fewer

Assuming listed outcomes are equally likely

The way outcomes are named does not guarantee equal probability.

Confusing mutually exclusive with independent

Mutually exclusive events cannot occur together.

Independent events are events where one does not change the probability of the other.

These are different ideas.

Using n(A)/n(S) when outcomes are not equally likely

This formula requires equally likely outcomes.

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Key Terms

Probability experiment — A process that produces an uncertain outcome.

Outcome — One possible result of a probability experiment.

Sample space — The set containing all possible outcomes of an experiment.

Event — A set of outcomes satisfying a particular condition.

Simple event — An event containing exactly one outcome.

Compound event — An event containing more than one outcome.

Set — A collection of distinct objects or outcomes.

Element — An individual member of a set.

Set notation — Mathematical notation used to describe sets and their relationships.

n(S) — The number of outcomes in sample space S.

n(A) — The number of outcomes belonging to event A.

Intersection — Outcomes belonging to both events.

A ∩ B — The intersection of events A and B; A AND B.

Union — Outcomes belonging to at least one of two events.

A ∪ B — The union of events A and B; A OR B.

Complement — All outcomes in the sample space that do not belong to an event.

Aᶜ — The complement of event A.

Empty set — A set containing no elements, written ∅.

Mutually exclusive events — Events that cannot occur simultaneously.

Certain event — An event with probability 1.

Impossible event — An event with probability 0.

Equally likely outcomes — Outcomes that have the same probability of occurring.

Ordered pair — Two values written in a particular order, such as (2,5).

Tree diagram — A branching diagram showing possible outcomes at successive stages.

Venn diagram — A diagram using regions to represent sets and their relationships.

Fundamental counting principle — A method of counting outcomes by multiplying the number of possibilities at successive stages.

Complement rule — The relationship P(Aᶜ) = 1 − P(A).

Addition rule — A rule used to find the probability of A or B while accounting for overlap.

At least — A stated amount or more.

At most — A stated amount or less.

Compound experiment — A probability experiment involving multiple stages.

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Key Takeaways

• An outcome is one possible result.
• The sample space contains every possible outcome.
• An event contains the outcomes that satisfy a particular condition.
• Sample spaces are commonly written using set notation.
• A simple event contains one outcome.
• A compound event contains multiple outcomes.
• The number of outcomes in S is written n(S).
• When outcomes are equally likely:

P(A) = n(A)/n(S)

• Sample spaces should be constructed systematically so outcomes are not missed or duplicated.
• Tables are useful for experiments involving two variables, such as rolling two dice.
• Tree diagrams are useful for experiments involving several stages.
• The fundamental counting principle can determine the number of possible outcomes without listing all of them.
• Order sometimes matters and sometimes does not.
• For two coin tosses, HT and TH are distinct outcomes.
• A ∩ B means A AND B.
• A ∪ B means A OR B.
• Aᶜ means NOT A.
• Mutually exclusive events have no outcomes in common.
• The empty set ∅ represents an impossible event.
• The sample space itself represents a certain event.
• The complement rule is:

P(Aᶜ) = 1 − P(A)

• Complement reasoning is especially useful for "at least one" problems.
• The addition rule is:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

• Listed outcomes are not automatically equally likely.
• A complete sample space often makes a difficult probability problem much easier to understand.
• Sample spaces and events provide the foundation for more advanced probability.

A useful strategy is:

Experiment → Sample Space → Event → Count → Probability

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Check Your Understanding

1. Define an outcome.

2. Define a sample space.

3. Define an event.

4. Write the sample space for tossing one coin.

5. Write the sample space for rolling one standard die.

6. For a die roll, write the event A representing an even number.

7. Is A in Question 6 a simple or compound event?

8. Write the sample space for tossing a coin twice.

9. Write the event representing exactly one head.

10. Calculate the probability of exactly one head.

11. Write the event representing at least one head.

12. Calculate its probability.

13. A coin is tossed three times. How many outcomes are in the sample space?

14. List the complete sample space for three coin tosses.

15. Write the event representing exactly two heads.

16. Calculate its probability.

17. Two standard dice are rolled. How many ordered outcomes are possible?

18. List all outcomes that produce a sum of 8.

19. Calculate the probability of rolling a sum of 8.

20. Explain why a sum of 7 is more likely than a sum of 2 when rolling two fair dice.

21. Let:

A = {1, 2, 3, 4}

B = {3, 4, 5, 6}

Find:

A ∩ B

22. Find:

A ∪ B

23. If:

S = {1, 2, 3, 4, 5, 6}

and:

A = {2, 4, 6}

find:

Aᶜ

24. Explain what it means for two events to be mutually exclusive.

25. If:

P(A) = 0.35

find:

P(Aᶜ)

26. A restaurant offers 4 starters, 6 main courses, and 3 desserts. How many different three-course meals are possible?

27. Explain why a tree diagram can be useful for a compound experiment.

28. Explain the difference between "at least 2" and "at most 2."

29. Explain why the formula P(A) = n(A)/n(S) cannot automatically be used when outcomes are not equally likely.

30. Challenge: A student rolls a standard six-sided die and then tosses a fair coin.

a. Construct the complete sample space using outcomes such as (1,H).
b. Determine n(S).
c. Let A be the event "the die shows an even number." List A.
d. Let B be the event "the coin shows heads." List B.
e. Find A ∩ B.
f. Find A ∪ B.
g. Find Aᶜ.
h. Calculate P(A).
i. Calculate P(B).
j. Calculate P(A ∩ B).
k. Calculate P(A ∪ B).
l. Determine whether A and B are mutually exclusive and explain your answer.
m. Find the probability that the die shows an odd number OR the coin shows tails.
n. Explain how a tree diagram could represent the same sample space.
o. Explain why constructing the complete sample space helps reduce errors in this problem.

Probability Rules

Probability rules allow us to calculate the likelihood of more complicated events without always listing every possible outcome. They are especially useful when a problem contains words such as or, and, not, at least, or both.

The most important rules in this topic are:

• the addition rule for events involving OR
• the multiplication rule for events involving AND
• the complement rule for events involving NOT
• rules for recognizing mutually exclusive and independent events

Choosing the correct rule depends on understanding what the problem is asking.

Probability Is Between 0 and 1

Every probability must satisfy:

0 ≤ P(A) ≤ 1

A probability of:

0 means the event is impossible.

1 means the event is certain.

Values between 0 and 1 represent different degrees of likelihood.

For example:

P(A) = 0.75

means that event A has a 75% probability of occurring.

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The Addition Rule: OR

The addition rule is used when we want the probability that:

A OR B occurs

In probability notation, "A or B" is written:

A ∪ B

The symbol ∪ means union.

For two events that may overlap:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

The intersection:

A ∩ B

represents outcomes that belong to both A and B.

Why Do We Subtract the Intersection?

Suppose:

P(A) = 0.55

P(B) = 0.45

P(A ∩ B) = 0.20

If we simply add:

0.55 + 0.45 = 1.00

we have counted the intersection twice.

Therefore:

P(A ∪ B)

= 0.55 + 0.45 − 0.20

= 0.80

So there is an 80% probability that A or B or both occur.

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Understanding OR

In probability, OR usually includes both.

If we ask for:

A or B

we usually mean:

• A occurs
• B occurs
• both A and B occur

This is called an inclusive OR.

For example, if A means "plays football" and B means "plays basketball," then A ∪ B includes students who play both sports.

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Worked Example: Addition Rule

In a class:

P(student studies Biology) = 0.60

P(student studies Chemistry) = 0.45

P(student studies both) = 0.25

Find the probability that a randomly selected student studies Biology or Chemistry.

Use:

P(B ∪ C)

= P(B) + P(C) − P(B ∩ C)

= 0.60 + 0.45 − 0.25

= 0.80

Therefore, the probability is:

80%

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Mutually Exclusive Events

Two events are mutually exclusive if they cannot occur at the same time.

For example, when rolling one die:

A = roll a 2

B = roll a 5

A single roll cannot be both 2 and 5.

Therefore:

P(A ∩ B) = 0

For mutually exclusive events, the addition rule simplifies to:

P(A ∪ B) = P(A) + P(B)

Example: Mutually Exclusive Events

Roll a fair six-sided die.

Find the probability of rolling:

2 or 5

P(2) = 1/6

P(5) = 1/6

Since the events cannot occur together:

P(2 or 5)

= 1/6 + 1/6

= 2/6

= 1/3

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Events That Are Not Mutually Exclusive

Suppose:

A = roll an even number

B = roll a number greater than 3

For a standard die:

A = {2, 4, 6}

B = {4, 5, 6}

The events overlap:

A ∩ B = {4, 6}

Therefore, they are not mutually exclusive.

We cannot simply add their probabilities because outcomes 4 and 6 would be counted twice.

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Worked Example: Overlapping Events

Using the previous example:

P(A) = 3/6

P(B) = 3/6

P(A ∩ B) = 2/6

Therefore:

P(A ∪ B)

= 3/6 + 3/6 − 2/6

= 4/6

= 2/3

We can check this directly:

A ∪ B = {2, 4, 5, 6}

Four of the six outcomes satisfy the condition.

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The Multiplication Rule: AND

The multiplication rule is used when we want the probability that:

A AND B occur

This is written:

P(A ∩ B)

For independent events:

P(A ∩ B) = P(A) × P(B)

The word AND is often a clue that multiplication may be required.

However, we must first determine whether the events are independent.

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Independent Events

Two events are independent when the occurrence of one event does not change the probability of the other.

For example:

• toss a coin
• roll a die

The result of the coin does not affect the die.

Suppose we want:

heads AND a 6

P(H) = 1/2

P(6) = 1/6

Therefore:

P(H and 6)

= 1/2 × 1/6

= 1/12

Example: Two Coin Tosses

A fair coin is tossed twice.

Find the probability of:

heads AND heads

Each toss is independent.

P(H on first toss) = 1/2

P(H on second toss) = 1/2

Therefore:

P(HH)

= 1/2 × 1/2

= 1/4

The complete sample space confirms this:

S = {HH, HT, TH, TT}

Only one of four outcomes is HH.

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Example: Three Coin Tosses

Find the probability of obtaining three heads in a row.

P(HHH)

= 1/2 × 1/2 × 1/2

= 1/8

The multiplication rule can therefore be extended across several independent stages.

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Dependent Events

Events are dependent when one event changes the probability of another.

A common example is drawing objects without replacement.

Suppose a bag contains:

3 red marbles

2 blue marbles

There are 5 marbles altogether.

A marble is selected and not replaced.

The probability of selecting red first is:

3/5

If a red marble is selected, only:

2 red marbles

remain among:

4 total marbles.

Therefore:

P(second red | first red) = 2/4

The second probability has changed.

The events are dependent.

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Conditional Probability

The notation:

P(B | A)

means:

the probability of B given that A has already occurred

For dependent events, the multiplication rule becomes:

P(A ∩ B) = P(A) × P(B | A)

This is the general multiplication rule.

The vertical bar | can be read as:

"given"

---

Worked Example: Without Replacement

A bag contains:

5 red marbles

3 blue marbles

Two marbles are selected without replacement.

Find the probability that both are red.

First red:

P(R₁) = 5/8

After one red marble has been removed:

4 red marbles remain

7 total marbles remain

Therefore:

P(R₂ | R₁) = 4/7

Multiply:

P(RR)

= 5/8 × 4/7

= 20/56

= 5/14

---

With Replacement

Now suppose the first marble is returned to the bag before the second marble is selected.

The probabilities do not change.

P(first red) = 5/8

P(second red) = 5/8

Therefore:

P(RR)

= 5/8 × 5/8

= 25/64

With replacement often makes repeated selections independent.

Without replacement usually makes them dependent.

Independent Does Not Mean Mutually Exclusive

These two ideas are often confused.

Mutually Exclusive

The events cannot happen together.

Independent

One event does not affect the probability of the other.

These are very different ideas.

For example, on a single die roll:

A = roll an even number

B = roll an odd number

These events are mutually exclusive.

If A happens, B definitely cannot happen.

Therefore, knowing A occurred dramatically changes the probability of B.

So the events are not independent.

---

Testing for Independence

Two events A and B are independent if:

P(A ∩ B) = P(A) × P(B)

For example:

P(A) = 0.5

P(B) = 0.4

If:

P(A ∩ B) = 0.2

then:

P(A) × P(B)

= 0.5 × 0.4

= 0.2

Therefore, A and B are independent.

---

The Complement Rule

The complement of event A means that A does not occur.

It can be written:

Aᶜ

Since either A occurs or it does not:

P(A) + P(Aᶜ) = 1

Therefore:

P(Aᶜ) = 1 − P(A)

For example:

P(rain) = 0.30

Then:

P(no rain)

= 1 − 0.30

= 0.70

Why Complements Are So Useful

Complements are particularly useful when a question asks for:

at least one

Sometimes calculating every successful possibility is difficult.

Instead:

P(at least one) = 1 − P(none)

This can turn a complicated calculation into a simple one.

---

Worked Example: At Least One Head

A fair coin is tossed three times.

Find the probability of obtaining at least one head.

We could count:

• exactly one head
• exactly two heads
• exactly three heads

But it is easier to find the complement.

The complement of:

"at least one head"

is:

"no heads"

which means:

TTT

Probability:

P(TTT)

= 1/2 × 1/2 × 1/2

= 1/8

Therefore:

P(at least one H)

= 1 − 1/8

= 7/8

---

Another At-Least-One Example

Suppose the probability that a machine produces a defective item is:

0.04

Assume successive items are independent.

What is the probability that at least one of three items is defective?

It is easier to calculate:

P(no defective items)

Probability one item is not defective:

1 − 0.04 = 0.96

For three items:

P(none defective)

= 0.96³

= 0.884736

Therefore:

P(at least one defective)

= 1 − 0.884736

= 0.115264

≈ 0.115

or approximately:

11.5%

---

Probability Trees

Tree diagrams are extremely useful for multi-step probability problems.

Each branch represents a possible outcome.

To use a probability tree:

Multiply along branches.

Add between appropriate final branches.

This connects the multiplication and addition rules.

Worked Tree Example

Suppose:

P(rain) = 0.30

P(no rain) = 0.70

If it rains:

P(bus late | rain) = 0.40

If it does not rain:

P(bus late | no rain) = 0.10

We want the probability that the bus is late.

There are two ways this can happen:

rain AND late

or:

no rain AND late

First path:

P(rain and late)

= 0.30 × 0.40

= 0.12

Second path:

P(no rain and late)

= 0.70 × 0.10

= 0.07

Now add:

P(late)

= 0.12 + 0.07

= 0.19

So the probability that the bus is late is:

19%

---

Multiply Along, Add Across

A useful probability-tree rule is:

AND → multiply

OR → add

For a path:

A AND B

multiply the branch probabilities.

For several separate paths that satisfy the event:

Path 1 OR Path 2

add their probabilities.

This is not a substitute for understanding the rules, but it is a useful problem-solving reminder.

---

Example: Two Dice

Two fair six-sided dice are rolled.

Find the probability of rolling:

a 6 on the first die AND an even number on the second die

P(first die = 6) = 1/6

P(second die even) = 3/6 = 1/2

The rolls are independent.

Therefore:

P(6 and even)

= 1/6 × 1/2

= 1/12

---

Addition and Multiplication Together

Many probability problems require more than one rule.

Suppose two fair coins are tossed.

Find the probability of getting:

exactly one head

This can occur in two ways:

HT

or:

TH

Calculate each path:

P(HT)

= 1/2 × 1/2

= 1/4

P(TH)

= 1/2 × 1/2

= 1/4

Now add:

P(exactly one H)

= 1/4 + 1/4

= 1/2

We multiplied within each path and added the successful paths.

---

A More Complex Example

A bag contains:

4 red marbles

3 blue marbles

Two marbles are drawn without replacement.

Find the probability that the marbles are different colours.

There are two possible paths:

red then blue

or:

blue then red

Red then Blue

P(RB)

= 4/7 × 3/6

= 12/42

= 2/7

Blue then Red

P(BR)

= 3/7 × 4/6

= 12/42

= 2/7

Now add:

P(different colours)

= 2/7 + 2/7

= 4/7

---

Another Method Using the Complement

For the same bag, we could calculate:

P(different colours)

= 1 − P(same colour)

Same colour means:

RR or BB

P(RR)

= 4/7 × 3/6

= 2/7

P(BB)

= 3/7 × 2/6

= 1/7

Therefore:

P(same)

= 2/7 + 1/7

= 3/7

So:

P(different)

= 1 − 3/7

= 4/7

Different approaches can lead to the same result.

---

Cards and Probability

A standard deck contains:

52 cards

with:

• 4 suits
• 13 cards per suit
• 26 red cards
• 26 black cards
• 4 aces

Suppose one card is selected.

Probability of an ace:

4/52 = 1/13

Probability of a red card:

26/52 = 1/2

Some card events overlap.

For example, there are two red aces.

So:

P(ace or red)

= P(ace) + P(red) − P(red ace)

= 4/52 + 26/52 − 2/52

= 28/52

= 7/13

---

Cards Without Replacement

Suppose two cards are drawn without replacement.

Find the probability that both are aces.

First ace:

P(A₁) = 4/52

Second ace given first ace:

P(A₂ | A₁) = 3/51

Therefore:

P(two aces)

= 4/52 × 3/51

= 12/2652

= 1/221

The probability changes after the first card because it is not replaced.

---

At Least One Using Complements

Suppose two cards are drawn with replacement.

Find the probability of drawing at least one ace.

Probability of not drawing an ace:

P(not ace)

= 48/52

= 12/13

Probability of no aces in two independent draws:

P(no aces)

= 12/13 × 12/13

= 144/169

Therefore:

P(at least one ace)

= 1 − 144/169

= 25/169

The complement method avoids separately calculating:

• exactly one ace
• exactly two aces

---

Conditional Probability

Conditional probability asks how probability changes when we already know something.

Suppose:

P(A ∩ B) = 0.18

and:

P(A) = 0.30

Then:

P(B | A) = P(A ∩ B)/P(A)

Therefore:

P(B | A)

= 0.18/0.30

= 0.60

Knowing that A occurred changes the sample space we are considering.

---

Example: Students

Suppose a school survey finds:

60 students study Chemistry.

24 of those Chemistry students also study Physics.

If we know that a randomly selected student studies Chemistry, what is the probability that the student also studies Physics?

P(Physics | Chemistry)

= 24/60

= 0.40

or:

40%

The condition "studies Chemistry" restricts the group we are considering.

---

Reading Probability Language

Probability problems often contain clues.

These clues are useful, but always think about the situation before selecting a rule.

---

Multi-Step Probability Strategy

For a complicated problem, use the following approach.

Identify the Events

Clearly define A, B, and any other events.

Translate the Language

Ask whether the question involves:

• AND
• OR
• NOT
• GIVEN
• AT LEAST

Check Relationships

Determine whether events are:

• mutually exclusive
• overlapping
• independent
• dependent

Choose a Representation

Consider using:

• a sample space
• a Venn diagram
• a probability tree
• a table

Apply the Rules

Use addition, multiplication, conditional probability, or complements as needed.

Check the Answer

The final probability must satisfy:

0 ≤ P ≤ 1

---

Real-World Probability Rules

Probability rules are used far beyond dice and cards.

They are important in:

Medicine

Estimating the probability of diseases, test outcomes, or treatment responses.

Engineering

Calculating the probability that components or systems fail.

Weather

Combining probabilities of different weather conditions.

Insurance

Estimating the probability of claims and financial risk.

Manufacturing

Predicting defective products and quality-control outcomes.

Genetics

Calculating probabilities of inherited traits.

Computing

Analysing algorithms, networks, reliability, and artificial intelligence.

Finance

Evaluating uncertainty and investment risk.

---

Example: System Reliability

Suppose two independent components must both work for a machine to operate.

Component A works with probability:

0.98

Component B works with probability:

0.95

For the system to work:

A AND B must work.

Therefore:

P(system works)

= 0.98 × 0.95

= 0.931

So the system has a:

93.1%

probability of operating successfully.

Its failure probability is:

1 − 0.931

= 0.069

or:

6.9%

---

Example: Redundant Systems

Suppose two independent backup systems each have a failure probability of:

0.05

The overall system fails only if:

both systems fail

Therefore:

P(both fail)

= 0.05 × 0.05

= 0.0025

So:

P(at least one works)

= 1 − 0.0025

= 0.9975

or:

99.75%

This shows why backup systems can greatly improve reliability.

Common Mistakes

Adding probabilities when the problem says AND

AND often requires multiplication.

Multiplying probabilities when the problem says OR

OR often requires addition.

Forgetting the intersection in the addition rule

For overlapping events, simply adding P(A) and P(B) double-counts the overlap.

Assuming events are mutually exclusive

Check whether they can happen together.

Confusing mutually exclusive and independent

Mutually exclusive means:

cannot occur together

Independent means:

one does not affect the other

Using the simple multiplication rule for dependent events

When events are dependent, probabilities change after earlier outcomes.

Forgetting whether an object is replaced

"With replacement" and "without replacement" can produce different probabilities.

Forgetting the complement

"At least one" problems are often much easier using:

1 − P(none)

Adding probabilities along a tree branch

Probabilities along a path are multiplied.

Multiplying different successful paths

Alternative successful paths are generally added.

Forgetting to adjust totals without replacement

If an object is removed, the number of remaining objects decreases.

Giving a probability greater than 1

This always indicates an error.

Rounding too early

Keep accurate values until the final step.

---

Key Terms

Addition rule — A rule for finding the probability that one event or another occurs.

Multiplication rule — A rule for finding the probability that events occur together.

Union — The event A or B or both, written A ∪ B.

Intersection — The event A and B, written A ∩ B.

Complement — The event that A does not occur, written Aᶜ.

Complement rule — The relationship P(Aᶜ) = 1 − P(A).

Mutually exclusive events — Events that cannot occur simultaneously.

Overlapping events — Events that share one or more possible outcomes.

Independent events — Events for which the occurrence of one does not change the probability of the other.

Dependent events — Events for which one event changes the probability of another.

Conditional probability — The probability of an event given that another event has occurred.

P(B | A) — The probability of B given A.

With replacement — Returning a selected object before making another selection.

Without replacement — Not returning a selected object before another selection.

Probability tree — A branching diagram showing possible sequences and their probabilities.

Branch probability — The probability associated with one branch of a probability tree.

Sample space — The complete set of possible outcomes.

Event — A set of outcomes satisfying a specified condition.

Intersection probability — The probability that two events both occur.

At least one — One or more occurrences of an event.

At most — No more than a stated number.

Certain event — An event with probability 1.

Impossible event — An event with probability 0.

Reliability — The probability that a system or component performs successfully.

Inclusive OR — An OR statement that includes the possibility that both events occur.

---

Key Takeaways

• Probability rules allow complicated events to be analysed efficiently.
• Every probability must lie between 0 and 1.
• OR usually indicates the addition rule.
• AND usually indicates the multiplication rule.
• For overlapping events, the addition rule subtracts the intersection so it is not counted twice.
• Mutually exclusive events cannot occur together.
• For mutually exclusive events, the intersection probability is zero.
• Independent events do not affect one another's probabilities.
• Dependent events do affect one another's probabilities.
• Independent and mutually exclusive do not mean the same thing.
• For independent events, probabilities can be multiplied directly.
• For dependent events, use conditional probabilities.
• P(B | A) means the probability of B given that A has occurred.
• Drawing without replacement usually creates dependent events.
• Drawing with replacement often produces independent events.
• The complement rule is:

P(Aᶜ) = 1 − P(A)

• "At least one" problems are often easiest using:

P(at least one) = 1 − P(none)

• Probability trees are useful for multi-stage problems.
• On a probability tree:

multiply along a path

and:

add appropriate alternative paths

• Multi-step problems may require several probability rules.
• A sample space, tree diagram, table, or Venn diagram can help organize complicated problems.
• Probability rules are widely used in medicine, engineering, insurance, genetics, manufacturing, finance, computing, and risk analysis.

A useful decision process is:

OR → Think Addition

AND → Think Multiplication

NOT → Think Complement

GIVEN → Think Conditional Probability

AT LEAST ONE → Consider the Complement

For example:

P(heads) = 1/2

P(rolling 6) = 1/6

Therefore:

P(heads and 6)

= 1/2 × 1/6

= 1/12

---

Testing Whether Events Are Independent

Suppose:

P(A) = 0.40

P(B) = 0.50

P(A ∩ B) = 0.20

If A and B are independent, we should have:

P(A) × P(B)

= 0.40 × 0.50

= 0.20

This equals the given intersection probability.

Therefore, A and B are independent.

---

Testing Independence Using Conditional Probability

We can also test independence by asking whether:

P(A | B) = P(A)

Suppose:

P(A) = 0.40

and:

P(A | B) = 0.40

Knowing B occurred has not changed the probability of A.

Therefore, this supports the conclusion that A and B are independent.

But if:

P(A | B) = 0.65

then knowing B occurred changed the probability of A.

The events are dependent.

---

Independent vs Mutually Exclusive

These ideas must not be confused.

Independent Events

One event does not affect the probability of the other.

They can occur together.

Mutually Exclusive Events

The events cannot occur together.

For mutually exclusive events:

P(A ∩ B) = 0

If A occurs, B cannot occur.

Therefore, if both events have nonzero probability, mutually exclusive events are not independent.

Example: Mutually Exclusive but Not Independent

Roll one standard die.

Let:

A = roll a number less than 3

A = {1, 2}

Let:

B = roll a number greater than 4

B = {5, 6}

These events cannot happen simultaneously.

Therefore:

P(A ∩ B) = 0

But:

P(A) = 2/6

and:

P(B) = 2/6

If they were independent:

P(A ∩ B)

would equal:

2/6 × 2/6

= 1/9

But the actual intersection probability is 0.

Therefore, they are not independent.

---

With Replacement

Suppose a bag contains:

4 red marbles

6 blue marbles.

A marble is selected, recorded, and then returned to the bag.

The probability of red on the first draw is:

4/10

After replacement, the bag again contains:

4 red

6 blue.

Therefore:

P(red second | red first)

= 4/10

The first draw did not change the second probability.

The draws are independent.

---

Without Replacement

Now suppose the first marble is not returned.

If the first marble was red:

3 red marbles remain

9 total marbles remain.

Therefore:

P(red second | red first)

= 3/9

= 1/3

This differs from the original probability:

4/10 = 2/5

The events are dependent.

---

Probability Trees and Conditional Probability

Probability trees are excellent for representing conditional probabilities.

Suppose a bag contains:

3 red marbles

2 blue marbles.

Two marbles are drawn without replacement.

On the first draw:

P(R) = 3/5

P(B) = 2/5

If the first is red:

P(R second | R first) = 2/4

P(B second | R first) = 2/4

If the first is blue:

P(R second | B first) = 3/4

P(B second | B first) = 1/4

Notice that the second-stage probabilities depend on what happened first.

Worked Tree Example

Using the bag containing:

3 red

2 blue

Find the probability of drawing:

red then blue

P(R then B)

= P(R) × P(B | R)

= 3/5 × 2/4

= 6/20

= 3/10

Now find:

blue then red

P(B then R)

= 2/5 × 3/4

= 6/20

= 3/10

Therefore, the probability of getting one marble of each colour is:

3/10 + 3/10

= 3/5

---

Conditional Probability in Medical Testing

Conditional probability is extremely important in medical testing.

Consider two different questions:

What is the probability of testing positive given that a person has a disease?

and:

What is the probability of having the disease given that a person tested positive?

These are not the same probability.

Symbolically:

P(positive | disease)

is not generally equal to:

P(disease | positive)

This distinction is one of the most important applications of conditional probability.

---

Example: Medical Screening

Suppose 10,000 people are screened.

Assume:

100 have a particular condition.

Of those 100:

90 test positive.

Of the 9,900 people without the condition:

495 also test positive.

Then:

total positive tests

= 90 + 495

= 585

If a person tests positive, the probability that they actually have the condition is:

P(condition | positive)

= 90/585

≈ 0.154

or approximately:

15.4%

Even though the test correctly identified 90% of people with the condition, a positive result does not mean there is a 90% probability that the person has the condition.

The base rate matters.

---

Why Base Rates Matter

A base rate is the overall frequency of an event in a population.

If a condition is very rare, even a fairly accurate test can produce many false positives compared with true positives.

Conditional probability therefore helps us interpret:

• medical tests
• security screening
• fraud detection
• quality control
• diagnostic systems

It reminds us to ask:

"What population are we conditioning on?"

---

Conditional Probability in Manufacturing

Suppose a factory has two machines.

Machine A produces:

60% of the products.

Machine B produces:

40%.

Suppose:

2% of Machine A's products are defective.

5% of Machine B's products are defective.

The probability that a randomly selected product is both:

from Machine A AND defective

is:

0.60 × 0.02

= 0.012

The probability that it is:

from Machine B AND defective

is:

0.40 × 0.05

= 0.020

Therefore:

P(defective)

= 0.012 + 0.020

= 0.032

or:

3.2%

---

Given That the Product Is Defective

Now suppose we discover that a product is defective.

What is the probability that it came from Machine B?

Of the probability associated with defective products:

0.020 came from Machine B

while:

0.032 is the total defective probability.

Therefore:

P(B | defective)

= 0.020/0.032

= 0.625

or:

62.5%

Even though Machine B produces only 40% of the products, it accounts for 62.5% of the defective products because its defect rate is higher.

This demonstrates the power of conditional probability.

---

Conditional Probability in Weather

Suppose historical data show:

P(traffic delay) = 0.20

but:

P(traffic delay | heavy rain) = 0.55

The probability increases substantially when heavy rain occurs.

This tells us that traffic delays and heavy rain are dependent events.

The condition provides useful information about the probability of the outcome.

---

Conditional Probability in Sports

Suppose a basketball player makes:

75% of all free throws.

But historical data show:

P(make | final minute) = 0.68

The player's overall success rate and conditional success rate are different.

The condition:

"final minute"

may be associated with a different performance level.

However, probability alone does not automatically prove why the difference occurs. Other factors may be involved.

---

Conditional Probability and Surveys

Suppose a survey of 500 people gives:

Find:

P(public transport | under 30)

= 140/200

= 0.70

Now find:

P(under 30 | public transport)

= 140/260

≈ 0.538

Again:

70% ≠ 53.8%

The order of the condition matters.

---

Conditional Probability and "Given That"

Whenever you see:

• given
• given that
• knowing that
• among those who
• of the people who
• assuming that

you should consider conditional probability.

For example:

"Among students who study physics, what proportion also study chemistry?"

translates naturally into:

P(Chemistry | Physics)

---

Conditional Probability and "Of Those"

Suppose:

"40% of students play a sport. Of those students, 30% play football."

The phrase:

"of those students"

indicates a condition.

We can write:

P(sport) = 0.40

P(football | sport) = 0.30

Therefore:

P(sport and football)

= 0.40 × 0.30

= 0.12

So 12% of all students belong to both categories.

---

Using Complements with Conditional Probability

Complements can also be used within a condition.

Suppose:

P(A | B) = 0.72

Then:

P(Aᶜ | B)

= 1 − 0.72

= 0.28

Within the restricted sample space B, either A occurs or A does not occur.

This is useful in many multi-step problems.

---

Multi-Step Example

A school reports that:

65% of students participate in extracurricular activities.

Of those students:

40% participate in sports.

Of the sports participants:

30% participate in basketball.

Find the probability that a randomly selected student participates in extracurricular activities, sports, and basketball.

Multiply along the sequence:

0.65 × 0.40 × 0.30

= 0.078

Therefore:

7.8%

of all students are predicted to belong to all three groups under these conditions.

---

A Practical Problem-Solving Strategy

When solving conditional probability problems:

Identify the Condition

Look for the words:

given that

Ask:

What information am I being told is already true?

Restrict the Sample Space

Focus only on outcomes satisfying the condition.

Identify the Desired Event

Within the restricted group, what outcome are you trying to find?

Use Counts or Probabilities

From a table:

desired outcomes within condition / total outcomes within condition

Or use the conditional probability formula.

Check for Dependence

Ask whether knowing one event occurred changes the probability of another.

Interpret the Answer

State what the probability means in the context of the problem.

---

Common Mistakes

Reversing the Condition

P(A | B) is not generally equal to P(B | A).

Always identify which event comes after the vertical bar.

Using the Original Sample Space

Once B is given, the relevant sample space becomes B.

Dividing by the Wrong Event

For P(A | B), the denominator relates to B.

Assuming Events Are Independent

Do not assume independence simply because two events have different names.

Assuming Events Are Dependent

Some events really are independent. Check whether one changes the probability of the other.

Confusing Independence and Mutual Exclusivity

Independent events can occur together.

Mutually exclusive events cannot.

Forgetting to Change Probabilities Without Replacement

Removing an item changes both the number of favourable outcomes and the total number of outcomes.

Changing Probabilities With Replacement

If an object is replaced and the system is restored to its original state, probabilities generally remain unchanged.

Confusing Test Accuracy with Probability of Having a Condition

P(positive | condition) and P(condition | positive) are different.

Ignoring Base Rates

A rare event can remain unlikely even after receiving evidence associated with it.

Reading a Two-Way Table in the Wrong Direction

Identify the condition first, then use the corresponding row or column total.

Forgetting Context

A probability such as 0.65 should be interpreted as:

65% of the restricted group

not necessarily 65% of the entire population.

---

Key Terms

Conditional probability — The probability of an event occurring given that another event has occurred.

Condition — Information assumed or known to be true when calculating a probability.

P(A | B) — The probability of A given B.

Given that — Language indicating that the sample space should be restricted.

Restricted sample space — The smaller collection of outcomes remaining after a condition is applied.

Intersection — Outcomes belonging to both events A and B.

A ∩ B — The event that A and B both occur.

Independent events — Events where one event does not change the probability of another.

Dependent events — Events where one event changes the probability of another.

Mutually exclusive events — Events that cannot occur simultaneously.

With replacement — Returning a selected item before making another selection.

Without replacement — Not returning a selected item before another selection.

Conditional frequency — The frequency of an outcome within a specified subgroup.

Two-way table — A table organizing data according to two categorical variables.

Contingency table — Another name for a two-way table used to examine relationships between categorical variables.

Probability tree — A branching diagram representing sequences of events and their probabilities.

Branch — One possible outcome or stage represented on a probability tree.

Base rate — The overall frequency or probability of an event within a population.

False positive — A positive test result when the condition being tested for is actually absent.

True positive — A positive test result when the condition is actually present.

Complement — All outcomes in the relevant sample space where an event does not occur.

Joint probability — The probability that two or more events occur together.

Marginal probability — The probability of a single event without conditioning on another event.

Sample space — The complete collection of possible outcomes before conditions are applied.

---

Key Takeaways

• Conditional probability describes probability when additional information is known.
• P(A | B) is read "the probability of A given B."
• The event after the vertical bar is the condition.
• A condition effectively creates a new, restricted sample space.
• Conditional probability compares the intersection with the given event.
• When using counts:

P(A | B) = n(A ∩ B)/n(B)

• P(A | B) and P(B | A) are usually different.
• Reversing the condition changes the denominator.
• Conditional probability is closely connected to the multiplication rule.
• For dependent events, later probabilities depend on earlier outcomes.
• Drawing without replacement usually produces dependent events.
• Drawing with replacement often produces independent events.
• Independent events satisfy:

P(A | B) = P(A)

• If knowing B changes the probability of A, the events are dependent.
• Independent events and mutually exclusive events are different concepts.
• Probability trees are useful for representing conditional probabilities in multi-stage experiments.
• Two-way tables make conditional probabilities easier to identify from data.
• "Given that," "among those who," and "of those" often indicate conditional probability.
• Base rates are important when interpreting conditional information.
• Test accuracy is not the same as the probability of having a condition after receiving a positive result.
• Conditional probability is widely used in medicine, manufacturing, insurance, weather forecasting, surveys, sports, engineering, and risk analysis.
• Every conditional probability should be interpreted in terms of the restricted group, not automatically the entire population.

A useful way to think about conditional probability is:

Condition → Restrict the sample space → Find the desired proportion

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Check Your Understanding

1. Define conditional probability in your own words.

2. What does the notation P(A | B) mean?

3. What does the vertical bar | mean?

4. Explain why conditional probability changes the sample space.

5. Suppose:

P(B) = 0.50

P(A ∩ B) = 0.20

Calculate:

P(A | B)

6. A class has 40 students. Of these, 25 study Biology and 15 of the Biology students also study Chemistry. Find:

P(Chemistry | Biology)

7. Explain why P(A | B) is not generally equal to P(B | A).

8. A school has:

80 football players

50 basketball players

30 students who play both.

Find:

P(basketball | football)

9. Find:

P(football | basketball)

10. Explain why your answers to Questions 8 and 9 are different.

11. A bag contains 7 red and 3 blue marbles. Two are selected without replacement. Find:

P(second red | first red)

12. Are the two selections in Question 11 independent or dependent? Explain.

13. Repeat Question 11 if the first marble is replaced before the second draw.

14. Explain how replacement changes the relationship between the events.

15. Suppose:

P(A) = 0.60

P(B) = 0.50

P(A ∩ B) = 0.30

Determine whether A and B are independent.

16. Suppose:

P(A) = 0.40

and:

P(A | B) = 0.65

Are A and B independent? Explain.

17. Explain the difference between independent and mutually exclusive events.

18. Why are two mutually exclusive events with nonzero probabilities not independent?

19. A survey gives:

Calculate:

P(bicycle | under 18)

20. Calculate:

P(under 18 | bicycle)

21. Explain why Questions 19 and 20 have different answers.

22. A machine produces 70% of a factory's products. Of its products, 3% are defective. Find the probability that a randomly selected product was made by this machine and is defective.

23. Explain why conditional probability is useful when interpreting medical tests.

24. Explain why:

P(positive | disease)

and:

P(disease | positive)

are different questions.

25. What is a base rate?

26. Suppose P(A | B) = 0.82. Find P(Aᶜ | B).

27. Explain how a probability tree can display conditional probabilities.

28. Give one real-world example of dependent events.

29. Give one real-world example of independent events.

30. Challenge: A company uses two factories to manufacture a component.

Factory A produces 70% of the components and has a 2% defect rate.

Factory B produces 30% of the components and has a 6% defect rate.

a. Find the probability that a randomly selected component comes from Factory A and is defective.
b. Find the probability that it comes from Factory B and is defective.
c. Find the overall probability that a component is defective.
d. A component is found to be defective. Find the probability that it came from Factory A.
e. Find the probability that it came from Factory B.
f. Explain why the answers to parts d and e are not simply 70% and 30%.
g. Which factory contributes more defective components overall?
h. Explain why Factory B can contribute a large proportion of defects even though it produces fewer components.
i. Draw or describe a probability tree representing the situation.
j. Explain where conditional probability appears in the tree.
k. Suppose Factory B reduces its defect rate to 2%. Recalculate the overall defect probability.
l. Explain how the change affects P(Factory B | defective).
m. Explain how this example demonstrates the importance of base rates.

Worked Example: Two Red Marbles

Using the bag:

3 red

2 blue

Find the probability of drawing two red marbles without replacement.

Follow the RR path:

First red:

3/5

Second red given first red:

2/4

Therefore:

P(RR)

= 3/5 × 2/4

= 6/20

= 3/10

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Worked Example: One of Each Colour

Using the same bag, find the probability of drawing one red and one blue.

There are two possible sequences:

RB

or:

BR

Red then Blue

P(RB)

= 3/5 × 2/4

= 6/20

= 3/10

Blue then Red

P(BR)

= 2/5 × 3/4

= 6/20

= 3/10

Now add the two paths:

P(one of each)

= 3/10 + 3/10

= 3/5

---

With Replacement

Now suppose the first marble is returned to the bag before the second draw.

The bag always contains:

3 red

2 blue

Therefore, on both draws:

P(R) = 3/5

P(B) = 2/5

The probabilities do not change.

For example:

P(RR)

= 3/5 × 3/5

= 9/25

Replacement often makes repeated selections independent.

---

Without Replacement vs With Replacement

This distinction is extremely important.

With Replacement

The selected object is returned.

The total number of objects returns to its original value.

The probabilities generally remain the same.

Without Replacement

The selected object is not returned.

The total decreases.

The composition may change.

Later probabilities therefore depend on earlier results.

A Three-Stage Tree

Tree diagrams can contain more than two stages.

Suppose a fair coin is tossed three times.

There are:

2 × 2 × 2 = 8

possible final outcomes:

HHH
HHT
HTH
HTT
THH
THT
TTH
TTT

Each individual path has probability:

1/2 × 1/2 × 1/2

= 1/8

A larger tree helps ensure that no possible sequence is forgotten.

---

Worked Example: Exactly Two Heads

For three fair coin tosses, exactly two heads can occur as:

HHT

HTH

THH

Each has probability:

1/8

Therefore:

P(exactly two heads)

= 1/8 + 1/8 + 1/8

= 3/8

Again:

multiply along each path

then:

add the successful paths

---

At Least One Event

Tree diagrams can also help with at least one problems.

Suppose a coin is tossed three times.

We want:

P(at least one head)

We could add all paths containing a head.

But there are seven such paths.

It is easier to use the complement.

The only outcome with no heads is:

TTT

P(TTT)

= 1/2 × 1/2 × 1/2

= 1/8

Therefore:

P(at least one H)

= 1 − 1/8

= 7/8

Sometimes a tree helps us recognize when the complement is the easier method.

---

Unequal Branch Probabilities

Tree diagrams are not limited to equally likely outcomes.

Suppose:

P(rain) = 0.30

P(no rain) = 0.70

If it rains:

P(bus late | rain) = 0.50

P(bus on time | rain) = 0.50

If it does not rain:

P(bus late | no rain) = 0.10

P(bus on time | no rain) = 0.90

The probabilities at each branching point still add to 1.

---

Worked Example: Bus Is Late

The bus can be late through two paths:

rain AND late

or:

no rain AND late

Rain and late:

0.30 × 0.50

= 0.15

No rain and late:

0.70 × 0.10

= 0.07

Therefore:

P(late)

= 0.15 + 0.07

= 0.22

So the probability that the bus is late is:

22%

This problem uses both multiplication and addition.

---

Interpreting a Tree Backwards

Sometimes we know the final outcome and want to determine something about an earlier event.

Using the previous example:

P(rain and late) = 0.15

P(late) = 0.22

Suppose we know that the bus was late.

What is the probability that it rained?

P(rain | late)

= 0.15/0.22

≈ 0.682

So approximately:

68.2%

This connects probability trees with conditional probability.

---

Tree Diagrams in Real Situations

Probability trees can represent:

• medical test results
• manufacturing defects
• weather conditions
• sporting outcomes
• genetics
• insurance risks
• quality control
• repeated selections
• customer behaviour
• reliability of systems

The branches do not have to represent coins, dice, or marbles.

They can represent any sequence of uncertain events.

Probability Tables

A probability table organizes outcomes and probabilities into rows and columns.

Tables are especially useful when:

• two categorical variables are involved
• survey results are given
• events overlap
• conditional probabilities are needed
• many outcomes must be organized systematically

One common type is a two-way table.

---

Two-Way Tables

Suppose 100 students are surveyed about whether they play a sport.

The table contains:

cell values — individual categories

row totals — totals across rows

column totals — totals down columns

grand total — total number of observations

Here:

grand total = 100

Finding Probability from a Table

Using the previous table:

59 of 100 students play a sport.

Therefore:

P(plays sport)

= 59/100

= 0.59

There are:

50 Grade 9 students.

Therefore:

P(Grade 9)

= 50/100

= 0.50

There are:

32 students who are Grade 9 AND play a sport.

Therefore:

P(Grade 9 ∩ sport)

= 32/100

= 0.32

---

Conditional Probability from a Table

Suppose we want:

P(plays sport | Grade 9)

The condition Grade 9 restricts us to the Grade 9 row.

There are:

50 Grade 9 students

and:

32 play a sport.

Therefore:

P(sport | Grade 9)

= 32/50

= 0.64

or:

64%

The denominator is the total for the condition.

---

Reversing the Condition

Now find:

P(Grade 9 | plays sport)

There are:

59 students who play a sport

and:

32 are in Grade 9.

Therefore:

P(Grade 9 | sport)

= 32/59

≈ 0.542

or approximately:

54.2%

These probabilities are different because the conditions are different.

---

Completing a Probability Table

Sometimes some values are missing.

Suppose:

Start with Grade 9:

28 + ? = 45

Therefore:

walk = 17

Grade 10 total:

32 + 23 = 55

Bus total:

28 + 32 = 60

Walk total:

17 + 23 = 40

The completed table is:

Always check that the row and column totals agree with the grand total.

---

Joint Probabilities

A joint probability describes two conditions occurring together.

For example:

P(Grade 9 AND bus)

Using the table:

28 students satisfy both conditions.

Therefore:

P(Grade 9 ∩ bus)

= 28/100

= 0.28

Joint probabilities correspond to the interior cells of a two-way table when divided by the grand total.

---

Marginal Probabilities

A marginal probability describes one variable without conditioning on another.

For example:

P(bus)

= 60/100

= 0.60

The term marginal comes from the fact that these totals are often shown along the margins of a table.

Examples include:

• row totals
• column totals

---

Conditional Probabilities

A conditional probability restricts attention to one row or column.

For example:

P(bus | Grade 9)

= 28/45

≈ 0.622

But:

P(Grade 9 | bus)

= 28/60

≈ 0.467

The same cell value appears in both calculations, but the denominators are different.

---

Tables for Two Dice

Tables are also useful for constructing sample spaces.

Suppose two standard dice are rolled.

A 6 × 6 table can show all:

36

ordered outcomes.

The rows can represent the first die.

The columns can represent the second die.

The cells can show:

• ordered pairs
• sums
• products
• differences

For example, if the table shows sums, the value 7 appears six times.

Therefore:

P(sum of 7)

= 6/36

= 1/6

Worked Example: Sum Greater Than 9

Two fair dice are rolled.

Possible sums greater than 9 are:

10, 11, 12

Sum of 10:

(4,6), (5,5), (6,4)

→ 3 outcomes

Sum of 11:

(5,6), (6,5)

→ 2 outcomes

Sum of 12:

(6,6)

→ 1 outcome

Total favourable outcomes:

3 + 2 + 1 = 6

Therefore:

P(sum > 9)

= 6/36

= 1/6

A table makes these outcomes easier to identify systematically.

---

Frequency Tables and Probability Tables

A frequency table contains counts.

For example:

Total observations:

24 + 16 + 10 = 50

Experimental probabilities can be calculated by dividing each frequency by the total.

P(red)

= 24/50

= 0.48

P(blue)

= 16/50

= 0.32

P(green)

= 10/50

= 0.20

These probabilities add to:

0.48 + 0.32 + 0.20

= 1

---

Probability Distribution Tables

A probability table can also directly show probabilities.

Suppose X represents the number of goals scored by a team.

The probabilities must add to:

1

Check:

0.15 + 0.30 + 0.35 + 0.15 + 0.05

= 1.00

This is an important check for any probability distribution.

---

Finding a Missing Probability

Suppose:

Since all probabilities must add to 1:

0.20 + 0.35 + 0.30 + P(3) = 1

0.85 + P(3) = 1

Therefore:

P(3) = 0.15

---

Trees vs Tables

Tree diagrams and tables organize probability information differently.

Neither method is always better.

The best representation depends on the problem.

---

When Should You Use a Tree?

A tree diagram is particularly useful when:

• events happen one after another
• order matters
• probabilities change after earlier outcomes
• objects are drawn with or without replacement
• conditional probabilities appear at different stages
• there are several possible paths to an outcome

Examples:

• drawing two cards
• choosing several marbles
• medical test followed by diagnosis
• rain followed by transportation outcome
• multiple manufacturing stages

---

When Should You Use a Table?

A table is particularly useful when:

• data are already grouped into categories
• two variables are being compared
• survey data are given
• conditional proportions are required
• row and column totals are useful
• the sample space forms a grid

Examples:

• grade level vs transport method
• gender vs activity choice
• treatment vs outcome
• product type vs defect status
• first die vs second die

---

Sequential Probability Problems

A sequential probability problem involves events occurring in a particular order.

For example:

A box contains:

4 white balls

3 black balls

2 balls are selected without replacement.

Find:

P(white then black)

First white:

P(W) = 4/7

After selecting white:

3 black remain

6 total remain.

Therefore:

P(B | W) = 3/6

Multiply:

P(WB)

= 4/7 × 3/6

= 12/42

= 2/7

---

Order Can Matter

Using the same box:

P(white then black)

= 4/7 × 3/6

= 2/7

Now:

P(black then white)

= 3/7 × 4/6

= 2/7

In this particular case the probabilities happen to be equal.

But the sequences:

WB

and:

BW

are still different paths.

If the question asks for:

one of each colour

we must include both:

P(one of each)

= 2/7 + 2/7

= 4/7

---

Multi-Step Real-World Example

Suppose a factory uses two production lines.

Line A produces:

70% of items.

Line B produces:

30%.

The defect rates are:

P(defect | A) = 0.02

P(defect | B) = 0.06

The probability of an item being from Line A and defective is:

0.70 × 0.02

= 0.014

For Line B:

0.30 × 0.06

= 0.018

Therefore:

P(defective)

= 0.014 + 0.018

= 0.032

or:

3.2%

A probability tree clearly shows both paths leading to a defective item.

---

Communicating Probability Reasoning

A correct numerical answer is important, but good probability work should also explain why the calculation is appropriate.

Instead of writing only:

3/5 × 2/4 = 3/10

write:

The first red marble has probability 3/5. Because the marble is not replaced, only 2 red marbles remain among 4 total marbles. Therefore, the probability of two red marbles is 3/5 × 2/4 = 3/10.

Clear reasoning should identify:

• the events
• whether they are independent or dependent
• why probabilities are multiplied
• why different paths are added
• whether replacement occurs
• what the final probability represents

---

Checking a Tree Diagram

Before accepting a tree diagram, check:

Do Branches from Each Point Add to 1?

They should.

Have the Probabilities Changed Correctly?

For without-replacement problems, totals usually change.

Have You Included Every Possible Outcome?

Missing branches lead to incorrect answers.

Did You Multiply Along Each Path?

A complete sequence usually requires multiplication.

Did You Add All Relevant Paths?

An event may occur in more than one way.

Is the Final Answer Between 0 and 1?

Every probability must satisfy:

0 ≤ P ≤ 1

---

Checking a Probability Table

For a table, check:

• row totals
• column totals
• grand total
• missing entries
• whether probabilities sum to 1 where appropriate
• whether the correct row or column is being used for conditional probability

A well-organized table should allow totals to be checked in more than one direction.

---

Common Mistakes

Adding Along a Tree Path

Along a path, probabilities are usually multiplied.

Multiplying Alternative Paths

If an event can occur through different mutually exclusive paths, those path probabilities are added.

Forgetting to Change Probabilities Without Replacement

After an item is removed, both the total and possibly the favourable number change.

Changing Probabilities When There Is Replacement

Replacement restores the original situation.

Forgetting a Possible Path

For "one of each," both AB and BA may need to be included.

Assuming Order Does Not Matter

A sequential experiment may distinguish AB from BA.

Using the Grand Total for Conditional Probability

The denominator should be the total for the given condition.

Reversing Conditional Probabilities

P(A | B) and P(B | A) usually use different row or column totals.

Reading the Wrong Cell

Always identify the row and column representing the desired intersection.

Ignoring Table Totals

Row and column totals are valuable checks.

Assuming Every Tree Has Equal Branches

Real probabilities are often unequal.

Rounding Too Early

Keep fractions or several decimal places until the final step.

Giving Only a Number

For multi-step problems, show enough reasoning to explain which paths or table entries were used.

---

Key Terms

Tree diagram — A branching representation of possible sequences of events.

Probability tree — A tree diagram with probabilities assigned to its branches.

Branch — One possible outcome at a stage of a tree.

Node — A point where a tree splits into possible outcomes.

Path — A complete sequence of branches through a tree.

Path probability — The probability of following a particular complete path.

Sequential event — An event occurring as part of an ordered sequence.

Independent events — Events where one result does not change the probability of another.

Dependent events — Events where one result changes the probability of another.

Conditional probability — The probability of an event given that another event has occurred.

With replacement — Returning a selected object before the next selection.

Without replacement — Keeping a selected object out before the next selection.

Two-way table — A table organizing observations according to two categorical variables.

Contingency table — Another term for a two-way table.

Frequency table — A table showing how often different outcomes occur.

Probability table — A table showing outcomes and their probabilities.

Probability distribution — A description of the probabilities associated with possible values of a variable.

Joint probability — The probability of two conditions occurring together.

Marginal probability — The probability of one event without conditioning on another.

Conditional frequency — The frequency of an outcome within a particular subgroup.

Row total — The total of the entries across a row.

Column total — The total of the entries down a column.

Grand total — The total number of observations in a table.

Complement — All outcomes in which a specified event does not occur.

Sample space — The complete set of possible outcomes.

Mutually exclusive paths — Alternative complete paths that cannot occur simultaneously in a single trial.

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Key Takeaways

• Tree diagrams organize sequential probability events.
• Tables organize probability information into rows and columns.
• Each branch of a probability tree represents a possible outcome.
• Probabilities leaving the same node must add to 1.
• To find the probability of a complete path:

multiply along the branches

• If an event can occur through several mutually exclusive paths:

add the relevant path probabilities

• Independent events keep the same probabilities at later stages.
• Dependent events have probabilities that change according to earlier outcomes.
• Drawing without replacement usually produces dependent events.
• Drawing with replacement often produces independent events.
• Probability trees can represent unequal as well as equal probabilities.
• Trees can contain two, three, or many stages.
• Complement reasoning can simplify complicated tree problems.
• Two-way tables are especially useful for categorical and survey data.
• Interior table cells often represent joint frequencies.
• Row and column totals can be used to calculate marginal probabilities.
• Conditional probability uses the total of the given group as its denominator.
• P(A | B) and P(B | A) are generally different.
• Probability tables can also represent complete sample spaces such as two-dice experiments.
• Probability distribution tables must have probabilities that sum to 1.
• Missing table probabilities can often be found using this total.
• Trees are usually best when events occur sequentially.
• Tables are usually best when categories are being compared.
• A strong probability solution should explain the reasoning, not only provide the final number.

The central strategy for probability trees is:

Follow a path → Multiply

Combine successful paths → Add

And for tables:

Identify the condition → Find the correct row or column → Use the appropriate total

---

Check Your Understanding

1. What information does a probability tree diagram show?

2. What must the probabilities leaving a single node add to?

3. When should probabilities along a tree path be multiplied?

4. When should different path probabilities be added?

5. A fair coin is tossed twice. Construct or describe the probability tree.

6. Use the tree to find:

P(HH)

7. Find:

P(exactly one head)

8. A fair coin is tossed three times. Find:

P(HHH)

9. Find the probability of exactly two heads.

10. Find the probability of at least one head using a complement.

11. A bag contains 5 red and 3 blue marbles. Two are selected without replacement. Find:

P(RR)

12. Find:

P(RB)

13. Find the probability of one marble of each colour.

14. Repeat Question 11 if the first marble is replaced.

15. Explain why the answers to Questions 11 and 14 differ.

Use the following table for Questions 16–21:

16. Find P(reads daily).

17. Find P(Grade 9).

18. Find P(Grade 9 AND reads daily).

19. Find P(reads daily | Grade 9).

20. Find P(Grade 9 | reads daily).

21. Explain why the answers to Questions 19 and 20 are different.

22. A probability distribution is:

Find the missing probability.

23. Explain the difference between a joint probability and a marginal probability.

24. Explain why a tree diagram is particularly useful for events occurring without replacement.

25. Explain when a table might be more useful than a tree diagram.

26. A box contains 6 white balls and 4 black balls. Two balls are selected without replacement. Find the probability of:

a. WW
b. WB
c. BW
d. BB

27. Check that the four probabilities in Question 26 add to 1. Explain why they should.

28. A student has a 0.8 probability of completing homework on any given day. Assuming days are independent, calculate the probability that the student completes homework on three consecutive days.

29. Explain how you would clearly communicate the reasoning used in Question 28.

30. Challenge: A company receives electronic components from two suppliers.

Supplier A provides 65% of the components.

Supplier B provides 35%.

Of Supplier A's components, 3% are defective.

Of Supplier B's components, 7% are defective.

a. Construct or describe a probability tree for this situation.
b. Find P(A and defective).
c. Find P(B and defective).
d. Calculate the overall probability that a component is defective.
e. Calculate the probability that a component is not defective.
f. A component is known to be defective. Calculate P(A | defective).
g. Calculate P(B | defective).
h. Explain why the supplier probabilities on the first branches add to 1.
i. Explain why the defective and non-defective probabilities leaving each supplier branch add to 1.
j. Explain where multiplication is used in the tree.
k. Explain where addition is used.
l. Organize the same information into a two-way table for a hypothetical shipment of 10,000 components.
m. Use your table to check the probabilities calculated from the tree.
n. Explain which representation—the tree or the table—you find more useful for this problem and justify your answer.