Problem structure	Useful method	Main check
Few possible outcomes	Systematic listing or a table	No omissions or duplicates
Several stages	Multiplication counting principle	Correct number of choices at each stage
Separate, non-overlapping cases	Addition counting principle	Cases do not overlap
Either condition, with possible overlap	Add the counts and subtract the overlap	Each outcome counted once
At least one occurrence	Count the complement when easier	Complement means no occurrences
Select two distinct items without roles	Count ordered selections, then divide by 2	Every unordered pair has exactly two orders

Important: Counting outcomes and calculating their probabilities are different tasks. A correct outcome count supports the classical formula only when those outcomes are equally likely.

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3. Systematic listing

Listing is a reliable method when the sample space is small.

Organize the list so that a reader can see why it is complete.

Worked example 1: Sum of two selected numbers

Two numbers are drawn successively from {1, 2, 3, 4}, with replacement. Each draw is uniform and independent.

Find the probability that their sum is 6.

Step 1: Count all ordered outcomes

Each draw has four possibilities:

Total = 4 × 4 = 16

Step 2: List favorable outcomes

For a sum of 6:

A = {(2,4), (3,3), (4,2)}

Step 3: Calculate

P(A) = 3/16

Verification

The outcomes (2,4) and (4,2) are distinct because draw order is recorded.

The repeated number in (3,3) is allowed because replacement is used.

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4. The multiplication counting principle

Suppose a process has several stages.

If:

• Stage 1 has m choices.
• For each first-stage choice, stage 2 has n choices.

Then the number of complete outcomes is:

m × n

For three stages with fixed numbers of choices:

Total = m × n × r

Example: Clothing combinations

A person chooses one of three shirts and one of two pairs of trousers.

Number of outfits = 3 × 2 = 6

This is a counting result. It does not say the outfits are equally likely to be chosen.

Worked example 2: A three-position code

A code has three positions. Each position contains one of the digits:

1, 2, 3, 4

Repetition is allowed.

All possible codes are equally likely.

Find the probability that the first and last digits are equal.

Step 1: Count all codes

Total = 4 × 4 × 4 = 64

Step 2: Count favorable codes

• First digit: four choices.
• Middle digit: four choices.
• Last digit: one choice, because it must equal the first.

Favorable = 4 × 4 × 1 = 16

Step 3: Calculate

P = 16/64 = 1/4

Key lesson: A restriction changes the number of choices at the relevant stage.

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5. Counting without repetition

If an item cannot be reused, the number of choices decreases.

For three ordered selections from five distinct items without repetition:

• First position: five choices.
• Second position: four choices.
• Third position: three choices.

Total = 5 × 4 × 3 = 60

Worked example 3: A code with distinct digits

A three-digit code uses the digits 1, 2, 3, 4, and 5 without repetition.

All valid codes are equally likely.

Find the probability that the code begins with an even digit.

Step 1: Count all valid codes

Total = 5 × 4 × 3 = 60

Step 2: Count favorable codes

The first digit can be 2 or 4:

• First digit: two choices.
• Second digit: four remaining choices.
• Third digit: three remaining choices.

Favorable = 2 × 4 × 3 = 24

Step 3: Calculate

P = 24/60 = 2/5

Codes versus numbers

A code may permit an initial zero if the rules allow it.

An ordinary three-digit positive integer cannot begin with zero.

Always read the wording before counting.

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6. The addition counting principle

If outcomes are divided into separate cases that do not overlap, add the case counts.

If one case contains m outcomes and another non-overlapping case contains n outcomes:

Combined count = m + n

Worked example 4: Exactly one six

Two fair dice are rolled independently.

Find the probability of exactly one six.

Step 1: Count all outcomes

Total = 36

Step 2: Divide the event into two cases

Case 1: The first die is six and the second is not six.

Count = 1 × 5 = 5

Case 2: The second die is six and the first is not six.

Count = 5 × 1 = 5

The cases do not overlap.

Step 3: Add and calculate

Favorable = 5 + 5 = 10

P(exactly one six) = 10/36 = 5/18

The outcome (6,6) is excluded because it contains two sixes.

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7. Overlapping events: Avoiding double-counting

Let:

• A be one event.
• B be another event.

We use:

• A ∪ B: A or B, including both.
• A ∩ B: A and B.

When we add the counts of A and B, the outcomes in both events are counted twice.

Therefore:

Count of A or B = count of A + count of B − count of both

The probability version is:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

If the events are mutually exclusive, the overlap is empty and its probability is zero.

Worked example 5: Multiples of 2 or 3

One integer is selected uniformly from 1 to 20.

Find the probability that it is divisible by 2 or by 3.

Step 1: Count multiples of 2

{2, 4, 6, 8, 10, 12, 14, 16, 18, 20}

Count: 10.

Step 2: Count multiples of 3

{3, 6, 9, 12, 15, 18}

Count: 6.

Step 3: Count the overlap

Numbers divisible by both 2 and 3 are multiples of 6:

{6, 12, 18}

Count: 3.

Step 4: Remove double-counting

Favorable = 10 + 6 − 3 = 13

P = 13/20

Check: Simply using 16/20 would count 6, 12, and 18 twice.

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8. Complementary-event calculations

The complement of A is written Aᶜ and means “A does not occur.”

P(Aᶜ) = 1 − P(A)

Equivalently:

P(A) = 1 − P(Aᶜ)

For counting:

Favorable count = total count − unfavorable count

Worked example 6: At least one head

A fair coin is tossed four times independently.

Find the probability of at least one head.

Step 1: Count all outcomes

Total = 2 × 2 × 2 × 2 = 16

Step 2: Identify the complement

The complement of “at least one head” is:

“No heads.”

Only one sequence satisfies this:

TTTT

Step 3: Calculate

P(at least one head) = 1 − 1/16 = 15/16

Important distinction

The complement of “exactly one head” is not “no heads.”

It includes:

• Zero heads.
• Two heads.
• Three heads.
• Four heads.

Translate the entire condition before taking its complement.

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9. Ordered versus unordered selections

Ordered selection

The positions or roles matter.

For example, selecting a captain and an assistant from five students:

Total = 5 × 4 = 20

Choosing A as captain and B as assistant is different from choosing B as captain and A as assistant.

Unordered selection

Only the selected group matters.

For example, choosing two students as a team:

{A, B} = {B, A}

Each unordered pair appears twice in the ordered count.

Therefore, the number of unordered pairs from n distinct items is:

n(n − 1)/2

This applies when selecting two distinct items without replacement and without assigning roles.

Worked example 7: A two-student team

A class has four girls and two boys. A two-student team is selected uniformly from all possible pairs.

Find the probability that the team contains one girl and one boy.

Step 1: Count all teams

There are six students:

Total = 6 × 5 ÷ 2 = 15

Step 2: Count mixed teams

Choose one of four girls and one of two boys:

Favorable = 4 × 2 = 8

There is no need to divide this count by 2. Each mixed team is already counted once by choosing its girl and its boy.

Step 3: Calculate

P = 8/15

Ordered-count verification

If students were selected successively without replacement:

• Girl then boy: 4 × 2 = 8.
• Boy then girl: 2 × 4 = 8.

P = 16/30 = 8/15

Both consistent methods agree.

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10. Replacement changes the sample space

With replacement

The selected item is returned before the next selection.

For two uniform independent draws from N distinct items:

Total ordered outcomes = N × N

Without replacement

The selected item is not returned.

For two uniform successive draws from N distinct items:

Total ordered outcomes = N × (N − 1)

Worked example 8: Comparing two models

A bag contains two red balls and three blue balls.

Find the probability of drawing two red balls:

1. With replacement and independent uniform draws.
2. Without replacement and uniform selection from the remaining balls.

Case 1: With replacement

Label the individual balls.

Total ordered outcomes:

5 × 5 = 25

Both red:

2 × 2 = 4

P(both red) = 4/25

The same physical red ball may be selected twice.

Case 2: Without replacement

Total ordered outcomes:

5 × 4 = 20

Both red:

2 × 1 = 2

P(both red) = 2/20 = 1/10

Interpretation

The answers differ because the experiment differs.

Without replacement, selecting a red ball first leaves only one red ball available.

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11. Counting outcomes that satisfy an “and” condition

“And” tells us that both conditions must hold.

It does not automatically tell us to multiply the probabilities of the two events.

For this lesson, use the sample space to count outcomes satisfying both conditions.

Worked example 9: Even sum and different dice

Two fair dice are rolled independently.

Find the probability that the sum is even and the two numbers are different.

Step 1: Count outcomes with an even sum

An even sum occurs when:

• Both numbers are odd.
• Both numbers are even.

Each die has three odd and three even faces.

Even-sum count = 3 × 3 + 3 × 3 = 18

Step 2: Remove equal-number outcomes

The six doubles are:

(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)

Every double has an even sum.

Even sum and different numbers = 18 − 6 = 12

Step 3: Calculate

P = 12/36 = 1/3

Key lesson: Translate the conditions first. Count their intersection rather than choosing a formula from the word “and” alone.

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12. A combined counting example

Worked example 10: At least one blue ball

A bag contains three red balls and two blue balls.

Two balls are drawn successively without replacement, uniformly from the remaining balls.

Find the probability of at least one blue ball.

Step 1: Count all ordered outcomes

Total = 5 × 4 = 20

Step 2: Count the complement

“No blue balls” means both balls are red.

Both-red count = 3 × 2 = 6

Step 3: Subtract

Favorable = 20 − 6 = 14

P(at least one blue) = 14/20 = 7/10

Verification by separate cases

Exactly one blue:

• Blue then red: 2 × 3 = 6.
• Red then blue: 3 × 2 = 6.

Two blue:

• 2 × 1 = 2.

Total favorable = 6 + 6 + 2 = 14

The results agree.

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13. Common mistakes and solution checks

Mistake 1: Adding choices across stages

Three shirts and two trousers produce 3 × 2 outfits, not 3 + 2.

Mistake 2: Multiplying overlapping cases

Separate alternative cases are usually added after ensuring they do not overlap.

Mistake 3: Forgetting overlap in an “or” event

Add both event counts, then subtract the count satisfying both.

Mistake 4: Dividing every pair count by 2

Divide by 2 only when each desired unordered pair has been counted exactly twice.

For a girl-boy team counted as “choose the girl, then choose the boy,” each team is counted once.

Mistake 5: Ignoring replacement

The second-stage number of available items changes without replacement.

Mistake 6: Multiplying probabilities merely because the question says “and”

The word “and” describes an intersection. Multiplying event probabilities requires additional conditions; counting the intersection directly avoids an unjustified assumption.

Mistake 7: Using inconsistent outcomes

Do not place an unordered favorable count over an ordered total count.

Mistake 8: Treating all categories as equally likely

Individual labeled balls may be equally likely while color patterns are not.

Verification checklist

• Have I justified equal likelihood?
• Do the numerator and denominator use the same kind of outcome?
• Does order matter?
• Is repetition allowed?
• Have I used the correct number of choices at each stage?
• Are separate cases disjoint?
• Have I removed overlap?
• Is the complement correct?
• Is the probability between 0 and 1?
• Can another method verify the count?

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14. Independent practice

Show the counting method, not just the final fraction.

Practice 1

One integer is selected uniformly from 1 to 24.

Find the probability that it is divisible by 5.

Practice 2

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

Find the probability of tails and a number greater than 4.

Practice 3

Two fair dice are rolled independently.

Find the probability that the sum is 10.

Practice 4

Two fair dice are rolled independently.

Find the probability of exactly one 2.

Practice 5

Two fair dice are rolled independently.

Find the probability of at least one 2.

Practice 6

One integer is selected uniformly from 1 to 30.

Find the probability that it is divisible by 3 or by 5.

Practice 7

A three-position code uses digits 1, 2, 3, and 4. Repetition is allowed, and all codes are equally likely.

Find the probability that the last digit is 4.

Practice 8

A three-position code uses digits 1, 2, 3, 4, and 5 without repetition. All valid codes are equally likely.

Find the probability that the first digit is 1 and the last digit is even.

Practice 9

A two-person team is selected uniformly from five people.

How many possible teams are there?

Practice 10

A group contains three girls and four boys. A two-person team is selected uniformly from all possible pairs.

Find the probability of selecting two girls.

Practice 11

A bag contains three white balls and two black balls.

Two balls are drawn successively with replacement, uniformly and independently.

Find the probability of different colors.

Practice 12

Use the bag in Practice 11, but draw without replacement.

Find the probability of different colors.

Practice 13

A fair coin is tossed four times independently.

Find the probability of exactly one tail.

Practice 14

Two fair dice are rolled independently.

Find the probability that their sum is odd.

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15. Practice solutions

Solution 1

Multiples of 5:

{5, 10, 15, 20}

P = 4/24 = 1/6

Solution 2

Total outcomes:

2 × 6 = 12

Favorable outcomes:

(T,5), (T,6)

P = 2/12 = 1/6

Solution 3

Sum 10:

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

P = 3/36 = 1/12

Solution 4

First die is 2, second is not:

1 × 5 = 5

Second die is 2, first is not:

5 × 1 = 5

P = 10/36 = 5/18

Solution 5

No 2:

5 × 5 = 25 outcomes

At least one 2:

36 − 25 = 11 outcomes

P = 11/36

Solution 6

Multiples of 3: 10.

Multiples of 5: 6.

Multiples of both, meaning multiples of 15: 2.

Favorable = 10 + 6 − 2 = 14

P = 14/30 = 7/15

Solution 7

Total codes:

4 × 4 × 4 = 64

Codes ending in 4:

4 × 4 × 1 = 16

P = 16/64 = 1/4

Solution 8

Total valid codes:

5 × 4 × 3 = 60

For favorable codes:

• First digit is fixed as 1.
• Last digit has two choices: 2 or 4.
• Middle digit has three choices after the first and last digits are fixed.

Favorable = 1 × 2 × 3 = 6

P = 6/60 = 1/10

Solution 9

Teams = 5 × 4 ÷ 2 = 10

Each unordered team has two possible selection orders.

Solution 10

Total teams:

7 × 6 ÷ 2 = 21

Two-girl teams:

3 × 2 ÷ 2 = 3

P = 3/21 = 1/7

Solution 11

Total ordered outcomes:

5 × 5 = 25

Different colors:

3 × 2 + 2 × 3 = 12

P = 12/25

Solution 12

Total ordered outcomes:

5 × 4 = 20

Different colors:

3 × 2 + 2 × 3 = 12

P = 12/20 = 3/5

The favorable count is unchanged, but the total count changes.

Solution 13

Total sequences:

2 × 2 × 2 × 2 = 16

Exactly one tail:

THHH, HTHH, HHTH, HHHT

P = 4/16 = 1/4

Solution 14

An odd sum requires one odd and one even result.

• Odd then even: 3 × 3 = 9.
• Even then odd: 3 × 3 = 9.

P = 18/36 = 1/2

16. Probability calculation assessment

Suggested time: 12–15 minutes
Questions: 8 single-answer multiple-choice questions
Marks: 1 per question
Suggested course mastery target: 7/8

These settings are internal course guidance.

Question 1

A three-position code uses digits 1, 2, 3, 4, and 5. Repetition is allowed.

How many codes are possible?

A. 15
B. 60
C. 125
D. 25

Question 2

Two fair dice are rolled independently.

What is the probability that their sum is 6?

A. 5/36
B. 1/11
C. 1/6
D. 6/36

Question 3

One integer is selected uniformly from 1 to 12.

What is the probability that it is even or divisible by 3?

A. 10/12
B. 1/2
C. 1/3
D. 2/3

Question 4

A fair coin is tossed three times independently.

What is the probability of at least one tail?

A. 3/8
B. 7/8
C. 1/2
D. 1/8

Question 5

Two distinct students are selected from six students to form a team with no assigned roles.

How many teams are possible?

A. 30
B. 12
C. 15
D. 36

Question 6

A bag contains two red balls and three blue balls.

Two balls are drawn successively without replacement, uniformly from those remaining.

What is the probability that both are blue?

A. 3/10
B. 9/25
C. 3/5
D. 1/5

Question 7

A three-position code uses digits 1, 2, 3, and 4 without repetition. All valid codes are equally likely.

What is the probability that its first digit is even?

A. 1/4
B. 1/3
C. 3/4
D. 1/2

Question 8

Two fair dice are rolled independently.

What is the probability of exactly one six?

A. 11/36
B. 5/18
C. 1/3
D. 1/36

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17. Instructor answer key

Tutor LMS placement: Put the assessment in a Quiz item if it will be graded. Keep the answer key and explanations separate from student-facing questions until submission.

 

Question	Answer	Skill assessed
1	C	Multiplication counting principle
2	A	Ordered dice outcomes
3	D	Overlapping events
4	B	Complementary events
5	C	Unordered pairs
6	A	Sampling without replacement
7	D	Restricted ordered counting
8	B	Disjoint cases and exactly one