Compound Events

Many probability situations involve more than one event.

Instead of asking:

"What is the probability of rolling a 6?"

we might ask:

"What is the probability of rolling a 6 and then rolling an even number?"

Instead of asking:

"What is the probability of flipping heads?"

we might ask:

"What is the probability of flipping two heads in a row?"

These are examples of compound events.

A compound event combines two or more events, and we can use organized lists, outcome diagrams, probability trees, and multiplication rules to determine their probabilities.

Compound probability is important because many real situations involve sequences or combinations of events rather than a single outcome.

What Is a Compound Event?

A compound event consists of two or more simple events.

For example:

Event 1: Flip a coin.

Event 2: Roll a die.

A possible compound outcome is:

Heads and 4

Another example is flipping a coin twice.

Possible compound outcomes are:

HH, HT, TH, TT

Each result contains outcomes from two events.

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Simple Events and Compound Events

A simple event involves one outcome or one stage.

Example:

Rolling a 5

A compound event combines events or outcomes.

Examples:

• rolling a 5 and then a 3
• flipping heads twice
• selecting a red counter and then a blue counter
• rolling an even number or a number greater than 4
• selecting two aces from a deck

Compound events can involve:

AND

or:

OR

Understanding the wording is extremely important.

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Sample Spaces for Compound Events

When several events occur, the sample space includes all possible combined outcomes.

Suppose a coin is flipped twice.

First flip:

H or T

Second flip:

H or T

The complete sample space is:

{HH, HT, TH, TT}

There are:

4 possible outcomes

If the coin is fair, each outcome has probability:

1/4

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Organized Outcome Diagrams

One way to display compound outcomes is with an outcome diagram.

For two coin flips:

An organized diagram helps prevent us from:

• forgetting outcomes
• counting outcomes twice
• incorrectly assuming how many possibilities exist

Worked Example: Two Coin Flips

Sample space:

{HH, HT, TH, TT}

Probability of two heads

Only:

HH

is favourable.

Therefore:

P(two heads) = 1/4

= 25%

Probability of exactly one head

Favourable outcomes:

HT, TH

Therefore:

P(exactly one head) = 2/4

= 1/2

= 50%

Probability of at least one head

Favourable outcomes:

HH, HT, TH

Therefore:

P(at least one head) = 3/4

= 75%

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"Exactly" and "At Least"

Probability questions often use important words.

Exactly one head

Means:

one head and no more

For two flips:

HT, TH

At least one head

Means:

one or more heads

For two flips:

HH, HT, TH

At most one head

Means:

zero or one head

For two flips:

HT, TH, TT

Carefully reading these words can completely change the answer.

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Probability Trees

A probability tree is a diagram showing possible outcomes at each stage of a compound event.

For a fair coin flipped twice:

First flip:

• H → 1/2
• T → 1/2

From each branch, the second flip has:

• H → 1/2
• T → 1/2

This creates four paths:

H → H

H → T

T → H

T → T

Each complete path represents one possible compound outcome.

Multiplying Along a Probability Tree

To calculate the probability of following a particular path:

Multiply the probabilities along the branches.

For:

Heads followed by Heads

P(HH) = 1/2 × 1/2

= 1/4

For:

Heads followed by Tails

P(HT) = 1/2 × 1/2

= 1/4

The same applies to TH and TT.

This gives:

Check:

1/4 + 1/4 + 1/4 + 1/4 = 1

All possible outcomes together must have a total probability of 1.

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The Multiplication Rule

When calculating the probability that one event occurs AND another event occurs:

Multiply probabilities along the path.

For independent events:

P(A and B) = P(A) × P(B)

For example:

P(heads and then rolling 6)

= 1/2 × 1/6

= 1/12

This rule becomes especially useful when many outcomes are possible.

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Independent Events

Two events are independent when the outcome of one event does not affect the probability of the other.

Examples include:

• flipping a coin and rolling a die
• rolling a die twice
• flipping a coin several times
• spinning a spinner twice

Suppose you flip heads on the first coin toss.

Does this change the probability of heads on the second toss?

No.

The second flip still has:

P(heads) = 1/2

The events are independent.

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Independent Does Not Mean Different

Suppose you roll a die twice.

The first roll is:

6

The probability of rolling another 6 is still:

1/6

The die does not "remember" the previous result.

Similarly, if a coin produces five heads in a row, then for a fair coin:

P(heads on next flip) = 1/2

and:

P(tails on next flip) = 1/2

The previous results do not change the probabilities.

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Worked Example: Coin and Die

A fair coin is flipped and a fair six-sided die is rolled.

What is the probability of:

Heads AND 6?

P(heads) = 1/2

P(6) = 1/6

The events are independent.

Therefore:

P(heads and 6)

= 1/2 × 1/6

= 1/12

≈ 8.3%

Outcome Diagram: Coin and Die

We can also list the complete sample space.

There are:

12 equally likely outcomes

Only one is:

H6

Therefore:

P(H and 6) = 1/12

The outcome diagram and multiplication method produce the same result.

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Worked Example: Heads and an Even Number

Using the same experiment:

P(heads) = 1/2

P(even number) = 3/6 = 1/2

Therefore:

P(heads and even)

= 1/2 × 1/2

= 1/4

Looking at the outcome diagram, the favourable outcomes are:

H2, H4, H6

There are:

3 favourable outcomes out of 12.

Therefore:

3/12 = 1/4

Again, both methods agree.

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Three Independent Events

The multiplication rule can be extended.

Suppose a fair coin is flipped three times.

What is the probability of:

HHH?

P(HHH)

= 1/2 × 1/2 × 1/2

= 1/8

The complete sample space is:

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

There are eight equally likely outcomes.

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Probability of a Specific Sequence

Suppose a fair coin is flipped four times.

What is:

P(HTTH)?

Multiply:

1/2 × 1/2 × 1/2 × 1/2

= 1/16

Any particular sequence of four fair coin flips has probability:

1/16

This includes:

HHHH
HTTH
TTTT
THHT

Each specific sequence is equally likely.

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Dependent Events

Two events are dependent when the outcome of one event changes the probability of another.

A common example involves selecting objects without replacement.

Suppose a bag contains:

3 red counters

2 blue counters

Total:

5 counters

We select one counter and do not replace it.

If the first counter is red, the bag changes.

Before the first selection:

P(red) = 3/5

After selecting red:

There are now:

2 red

2 blue

Total:

4 counters

Therefore:

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

= 1/2

The first event changed the probability of the second event.

The events are dependent.

With Replacement vs Without Replacement

This distinction is extremely important.

With Replacement

The selected object is returned before the next selection.

The original probabilities are restored.

Events are often independent.

Without Replacement

The selected object is not returned.

The contents change.

The probabilities for later selections change.

Events are dependent.

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Worked Example: Without Replacement

A bag contains:

4 red counters

3 blue counters

Two counters are selected without replacement.

Find:

P(red then red)

First selection:

P(red) = 4/7

After selecting red:

3 red counters remain.

Total counters remaining:

6

Therefore:

P(second red | first red) = 3/6

Now multiply:

P(red then red)

= 4/7 × 3/6

= 12/42

= 2/7

≈ 28.6%

Notice that we did not calculate:

4/7 × 4/7

because the first selection changed the contents of the bag.

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Worked Example: Red Then Blue

Using the same bag:

4 red
3 blue

Total = 7

Find:

P(red then blue)

First:

P(red) = 4/7

After selecting red:

There are still:

3 blue

but only:

6 counters total

Therefore:

P(blue second | red first) = 3/6

Multiply:

P(red then blue)

= 4/7 × 3/6

= 12/42

= 2/7

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The Probability Tree Changes

For dependent events, the probabilities on the second set of branches depend on what happened first.

Suppose a bag contains:

2 red counters
1 blue counter

Two counters are selected without replacement.

First draw:

P(R) = 2/3

P(B) = 1/3

If the first counter is red:

Remaining:

1 red, 1 blue

Therefore:

P(R second) = 1/2

P(B second) = 1/2

If the first counter is blue:

Remaining:

2 red, 0 blue

Therefore:

P(R second) = 1

P(B second) = 0

This changing probability is what makes the events dependent.

Conditional Probability

When one event affects another, we often describe the probability using the word given.

For example:

P(red second given red first)

means:

the probability of selecting red on the second draw given that red was selected first.

If a bag originally contains:

4 red and 3 blue

and a red counter has already been removed:

P(red second | red first)

= 3/6

= 1/2

The vertical line:

|

can be read as:

"given that"

This idea is called conditional probability.

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Adding Probabilities

Probability trees also help with situations where several different paths satisfy the same event.

Suppose a fair coin is flipped twice.

What is the probability of getting:

exactly one head?

There are two possible paths:

HT

or:

TH

Each has probability:

1/2 × 1/2 = 1/4

Now add the probabilities:

P(exactly one head)

= 1/4 + 1/4

= 1/2

This gives us another important rule:

Multiply along branches.

Add between separate favourable paths.

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AND Usually Means Multiply

Consider:

P(heads AND heads)

We follow one path:

H → H

Therefore:

1/2 × 1/2 = 1/4

The word AND often indicates that several events must occur together.

This usually involves multiplication.

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OR Often Means Add

Consider:

P(HH OR TT)

P(HH) = 1/4

P(TT) = 1/4

These are separate outcomes.

Therefore:

P(HH or TT)

= 1/4 + 1/4

= 1/2

The word OR often involves adding probabilities of different favourable outcomes.

However, care is needed when events overlap.

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Mutually Exclusive Events

Events are mutually exclusive when they cannot occur at the same time in a single trial.

For one die roll:

rolling a 2

and:

rolling a 5

are mutually exclusive.

You cannot roll both on the same die at the same time.

Therefore:

P(2 or 5)

= P(2) + P(5)

= 1/6 + 1/6

= 2/6

= 1/3

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Events That Overlap

Some events are not mutually exclusive.

Consider rolling one die.

Event A:

rolling an even number

= {2, 4, 6}

Event B:

rolling a number greater than 3

= {4, 5, 6}

Notice:

4 and 6

belong to both events.

If we simply add:

3/6 + 3/6

we would count 4 and 6 twice.

The correct favourable outcomes are:

{2, 4, 5, 6}

Therefore:

P(even OR greater than 3)

= 4/6

= 2/3

For overlapping events, we must avoid double counting.

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General Addition Rule

For two events A and B:

P(A or B) = P(A) + P(B) − P(A and B)

We subtract the overlap because it was counted twice.

Using the previous example:

P(even) = 3/6

P(greater than 3) = 3/6

P(even and greater than 3) = 2/6

Therefore:

P(even or greater than 3)

= 3/6 + 3/6 − 2/6

= 4/6

= 2/3

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"At Least One"

Compound probability questions frequently ask for:

at least one

For example:

What is the probability of getting at least one head in three coin flips?

We could count:

HHH
HHT
HTH
HTT
THH
THT
TTH

That works, but there is an easier method.

The only outcome that does not contain a head is:

TTT

P(TTT)

= 1/2 × 1/2 × 1/2

= 1/8

Therefore:

P(at least one head)

= 1 − 1/8

= 7/8

This is called using the complement.

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The Complement Strategy

Sometimes it is easier to calculate what we do not want.

The general rule is:

P(at least one success) = 1 − P(no successes)

For example, suppose a die is rolled twice.

What is the probability of rolling at least one 6?

Instead of listing every favourable outcome, calculate:

P(no 6 on one roll) = 5/6

P(no 6 on either roll)

= 5/6 × 5/6

= 25/36

Therefore:

P(at least one 6)

= 1 − 25/36

= 11/36

≈ 30.6%

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Worked Example: At Least One Head

A fair coin is flipped four times.

Find:

P(at least one head)

Probability of tails on one flip:

1/2

Probability of four tails:

P(TTTT)

= (1/2)⁴

= 1/16

Therefore:

P(at least one head)

= 1 − 1/16

= 15/16

= 93.75%

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Compound Events with Spinners

Suppose a spinner has four equal sections:

Red, Blue, Green, Yellow

It is spun twice.

What is:

P(red then blue)?

P(red) = 1/4

P(blue) = 1/4

The spins are independent.

Therefore:

P(red then blue)

= 1/4 × 1/4

= 1/16

Worked Example: Same Colour Twice

Suppose a spinner has three equal sections:

Red, Blue, Green

It is spun twice.

Possible same-colour outcomes:

RR

BB

GG

Each has probability:

1/3 × 1/3 = 1/9

There are three favourable paths.

Therefore:

P(same colour)

= 1/9 + 1/9 + 1/9

= 3/9

= 1/3

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Compound Events with Cards

Suppose a standard 52-card deck is used.

Two cards are drawn without replacement.

Find the probability that both cards are aces.

First card:

P(ace) = 4/52

After drawing an ace:

3 aces remain.

51 cards remain.

Therefore:

P(second ace | first ace) = 3/51

Multiply:

P(two aces)

= 4/52 × 3/51

= 12/2652

= 1/221

≈ 0.45%

The events are dependent because the first card is not replaced.

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What If the Card Is Replaced?

Suppose the first card is returned to the deck and the deck is shuffled.

Then:

P(ace first) = 4/52

and:

P(ace second) = 4/52

The probabilities remain the same.

Therefore:

P(two aces with replacement)

= 4/52 × 4/52

= 1/13 × 1/13

= 1/169

The replacement changes the probability structure.

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Reading a Probability Tree

A probability tree can tell us:

• possible outcomes
• probability at each stage
• probability of each complete path
• whether probabilities change
• whether events appear independent or dependent

When reading a tree:

Step 1: Start at the beginning.

Step 2: Follow one complete path.

Step 3: Multiply probabilities along that path.

Step 4: Repeat for other favourable paths.

Step 5: Add the probabilities of the favourable paths.

This process works for many compound probability problems.

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Checking a Probability Tree

At every branching point, probabilities should total:

1

For example:

P(H) + P(T)

= 1/2 + 1/2

= 1

For a bag:

P(red) = 3/5

P(blue) = 2/5

Then:

3/5 + 2/5 = 1

If branch probabilities do not total 1, check your tree.

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Independent or Dependent?

Ask one simple question:

Does the first event change the probability of the second event?

If no:

The events are independent.

If yes:

The events are dependent.

Examples:

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Worked Example: Identifying Dependence

A bag contains:

5 red counters
3 blue counters

A counter is selected.

Situation A

The counter is replaced.

Second selection:

P(red) = 5/8

The probability has not changed.

Independent

Situation B

The counter is not replaced and the first counter was red.

Second selection:

P(red) = 4/7

The probability changed.

Dependent

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Real-World Compound Events

Compound probabilities occur in many situations.

Genetics

Several genetic traits may be inherited together.

Manufacturing

A product may need several components to function correctly.

Reliability Engineering

A system may depend on several components working successfully.

Medicine

Several risk factors or test outcomes may be considered together.

Weather

Sequences or combinations of weather events may be analyzed.

Sports

Analysts may examine sequences of shots, passes, or scoring events.

Games

Cards, dice, spinners, and other random processes frequently involve compound events.

Probability trees help organize these more complicated situations.

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Compound Probability and Experimental Results

Compound probabilities can also be investigated experimentally.

Suppose two coins are flipped 100 times.

The theoretical probability of HH is:

1/2 × 1/2 = 1/4

Therefore, expected HH outcomes:

100 × 1/4 = 25

Suppose the experiment produces:

28 HH outcomes

Experimental probability:

28/100 = 0.28

Theoretical probability:

0.25

The values are not identical, but the difference may simply be due to random variation.

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Probability Trees as Models

A probability tree is a model.

It depends on correct assumptions.

For example, when constructing a tree for two coin flips, we may assume:

• the coin is fair
• the flips are independent

When constructing a tree for counters, we must know:

• how many counters of each colour exist
• whether selection is random
• whether counters are replaced
• whether all counters are equally likely to be selected

Incorrect assumptions produce incorrect probabilities.

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Worked Example: Full Probability Tree Problem

A bag contains:

3 red counters

2 blue counters

Two counters are selected without replacement.

Find the probability of selecting one counter of each colour.

There are two favourable paths:

Red then Blue

or:

Blue then Red

Red then Blue

P(R then B)

= 3/5 × 2/4

= 6/20

= 3/10

Blue then Red

P(B then R)

= 2/5 × 3/4

= 6/20

= 3/10

Now add:

P(one of each)

= 3/10 + 3/10

= 6/10

= 3/5

= 60%

This demonstrates the key probability-tree strategy:

Multiply along each path.

Add the favourable paths.

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Worked Example: Both the Same Colour

Using the same bag:

3 red
2 blue

Find:

P(both counters are the same colour)

Possible paths:

RR

or:

BB

RR

3/5 × 2/4

= 6/20

= 3/10

BB

2/5 × 1/4

= 2/20

= 1/10

Add:

3/10 + 1/10

= 4/10

= 2/5

= 40%

Notice:

P(one of each) + P(same colour)

= 3/5 + 2/5

= 1

These two events account for every possible result.

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A Strategy for Compound Probability

When solving a compound probability problem:

Identify the events

What happens first, second, third, and so on?

Determine whether they are independent or dependent

Ask:

Does one event change the probability of another?

Organize the outcomes

Use:

• a list
• an outcome diagram
• a table
• a probability tree

Label probabilities

Check that probabilities at each branching point total 1.

Multiply along paths

Use multiplication when events occur together along one route.

Add favourable paths

If several paths satisfy the question, calculate each path and add them.

Consider the complement

For "at least one" questions, calculating the opposite event may be easier.

Check your answer

Probability must satisfy:

0 ≤ P ≤ 1

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Common Mistakes

Adding when you should multiply

For events occurring along one path:

multiply

Multiplying when you should add

For separate favourable paths:

add

Assuming every event is independent

Without replacement often creates dependent events.

Forgetting to change the denominator

If an object is not replaced, the total number of objects decreases.

Forgetting to change the numerator

The number of favourable objects may also decrease.

Treating replacement and no replacement as identical

They produce different probabilities.

Assuming previous coin flips affect the next flip

Independent events do not "remember" previous outcomes.

Confusing exactly with at least

"Exactly one" and "at least one" describe different events.

Forgetting multiple possible orders

"One red and one blue" includes:

RB and BR

Double counting overlapping events

When events overlap, do not count shared outcomes twice.

Forgetting the complement strategy

"At least one" questions are often easier using:

1 − P(none)

Drawing incomplete probability trees

Every possible outcome at each stage should be represented.

Forgetting to check branch totals

Branches from the same point should total 1.

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Key Terms

Compound event — An event involving two or more simple events or conditions.

Simple event — An event involving a single outcome or stage.

Sample space — The complete set of possible outcomes.

Outcome diagram — An organized representation of possible combined outcomes.

Probability tree — A branching diagram showing possible outcomes and their probabilities across multiple stages.

Branch — One path segment on a probability tree representing an outcome.

Path — A sequence of branches representing a complete compound outcome.

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

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

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

Replacement — Returning a selected object before the next selection.

Without replacement — Keeping the selected object out, causing later probabilities to change.

Multiplication rule — A rule used to calculate the probability of events occurring together along a path.

Addition rule — A rule used to combine probabilities of different favourable outcomes.

Mutually exclusive events — Events that cannot occur simultaneously in a single trial.

Overlapping events — Events that share one or more outcomes.

Complement — The event that the original event does not occur.

At least — A phrase meaning the stated amount or more.

At most — A phrase meaning the stated amount or fewer.

Exactly — A phrase indicating precisely the stated number of occurrences.

Favourable outcome — An outcome satisfying the event being investigated.

Fair — Describes a probability process in which equivalent outcomes have equal probabilities.

Random — Describes an outcome that cannot be predicted with certainty beforehand.

Probability model — A mathematical representation of a random process.

Tree diagram — Another name commonly used for a probability tree.

Conditional notation — Notation such as P(B | A), meaning the probability of B given that A has occurred.

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Key Takeaways

• A compound event combines two or more events.
• Compound outcomes can be organized using lists, tables, outcome diagrams, and probability trees.
• Probability trees show possible outcomes across several stages.
• Each complete route through a probability tree represents a possible compound outcome.
• Multiply probabilities along a path.
• Add probabilities of separate favourable paths.
• Independent events do not change each other's probabilities.
• Dependent events do change later probabilities.
• Repeated coin flips and die rolls are usually independent.
• Selecting objects without replacement creates dependent events.
• Replacement usually restores the original probabilities.
• For independent events:

P(A and B) = P(A) × P(B)

• Conditional probability describes the probability of one event given that another has already occurred.
• Probability trees for dependent events have changing branch probabilities.
• The word AND often indicates multiplication along a path.
• The word OR often involves adding probabilities of different outcomes.
• Mutually exclusive events cannot occur simultaneously.
• Overlapping events must not be counted twice.
• For overlapping events:

P(A or B) = P(A) + P(B) − P(A and B)

• "Exactly one" means one and only one.
• "At least one" means one or more.
• "At most one" means zero or one.
• "At least one" problems are often easiest using a complement.
• A useful rule is:

P(at least one success) = 1 − P(no successes)

• Branch probabilities leaving the same point should total 1.
• All complete outcomes together should have total probability 1.
• Probability trees depend on assumptions such as fairness, random selection, independence, and replacement.
• Compound theoretical probabilities can be compared with experimental results.
• Experimental results may differ from theoretical predictions because of random variation.

For probability trees, remember:

Multiply along. Add across.

For independence, ask:

Did the first event change the probability of the next event?

And for "at least one," ask:

Would it be easier to calculate none and subtract from 1?

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Check Your Understanding

1. What is a compound event?

2. Explain the difference between independent and dependent events.

3. A fair coin is flipped twice. List the complete sample space.

4. Calculate the probability of:

a. HH
b. HT
c. exactly one head
d. at least one head
e. no heads

5. A fair coin is flipped three times. Calculate P(HHH).

6. Calculate the probability of exactly three tails in three flips.

7. Calculate the probability of at least one head in three flips.

8. Explain why using a complement makes Question 7 easier.

9. A fair coin is flipped and a fair die is rolled. Calculate:

a. P(heads and 6)
b. P(tails and an even number)
c. P(heads and a number greater than 4)

10. Are the coin and die events in Question 9 independent or dependent? Explain.

11. A bag contains 4 red and 3 blue counters. Two counters are selected without replacement. Calculate:

a. P(red then red)
b. P(red then blue)
c. P(blue then red)
d. P(blue then blue)

12. Check that the probabilities in Question 11 total 1.

13. Why do the probabilities change after the first selection in Question 11?

14. Repeat Question 11 assuming that the first counter is replaced before the second selection.

15. Explain why replacement changes the calculations.

16. A bag contains 3 red and 2 blue counters. Two are selected without replacement. Calculate the probability of selecting one of each colour.

17. Why must both RB and BR be included in Question 16?

18. A fair die is rolled twice. Calculate the probability of rolling two sixes.

19. Calculate the probability of rolling at least one 6 in two rolls.

20. A spinner has four equal sections. It is spun twice. Calculate the probability of landing on the same section both times.

21. Explain the difference between "exactly one success" and "at least one success."

22. A standard deck contains 52 cards. Two cards are drawn without replacement. Calculate the probability that both are aces.

23. Are the card selections in Question 22 independent or dependent? Explain.

24. How would the calculation change if the first card were replaced?

25. A die is rolled once. Find:

P(even OR greater than 3)

Explain why simply adding P(even) and P(greater than 3) would give an incorrect answer.

26. Explain the rule:

Multiply along branches and add favourable paths.

27. What should the probabilities leaving each branching point on a probability tree total?

28. Give two real-world situations involving compound events.

29. A fair coin is flipped four times. Calculate the probability of:

a. four tails
b. at least one head
c. the specific sequence HHTT

30. Challenge: A bag contains 5 red, 3 blue, and 2 green counters. Two counters are selected without replacement.

Construct a probability tree and use it to determine:

a. P(red then red)
b. P(red then blue)
c. P(blue then red)
d. P(two green counters)
e. P(one red and one blue, in either order)
f. P(at least one red counter)

Explain why the events are dependent and how the probabilities on the second branches change after the first counter is selected.