1. What is the classical probability model?
A classical probability model applies when:
- The sample space contains a finite number of elementary outcomes.
- Those elementary outcomes are equally likely.
If there are n equally likely outcomes and event A contains m of them:
P(A) = m/n
Equivalently:
P(A) = number of favorable outcomes ÷ total number of equally likely outcomes
“Favorable” means satisfying the event’s condition. It does not necessarily mean desirable.
Example
A fair six-sided die is rolled once.
Ω = {1, 2, 3, 4, 5, 6}
Let A be the event “the number is greater than 4.”
A = {5, 6}
There are two favorable outcomes among six equally likely outcomes.
P(A) = 2/6 = 1/3
Why the assumptions matter
The formula does not apply by simply counting outcomes that have unequal probabilities.
For example, two color labels do not imply probability 1/2 for each color.
The correct sample space must represent outcomes at a level where equal likelihood is justified.
2. Building a suitable sample space
Before calculating, state exactly what the experiment records.
A suitable sample space must:
- Include every possible outcome.
- Avoid counting the same elementary outcome twice.
- Distinguish outcomes when the experiment distinguishes them.
- Support the equal-likelihood assumption.
A practical method
- Describe one complete outcome.
- List or organize all such outcomes.
- Check whether they are equally likely.
- Translate the event into a condition on those outcomes.
- Count the total and favorable outcomes.
- Form and simplify the fraction.
Worked example 1: Numbered cards
One card is selected uniformly from cards numbered 1 to 12.
Find the probability of selecting a multiple of 3.
Sample space:
Ω = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
“Uniformly” means each card has the same probability of selection.
Favorable outcomes:
A = {3, 6, 9, 12}
Therefore:
P(A) = 4/12 = 1/3
Verification: Four of the twelve cards satisfy the condition.
3. Choosing the correct elementary outcomes
A recorded category may combine several elementary outcomes.
Worked example 2: Colored balls
A bag contains:
- Three red balls.
- Two blue balls.
Each individual ball is equally likely to be selected.
Find the probability of red.
Label the balls:
Ω = {R₁, R₂, R₃, B₁, B₂}
There are five equally likely elementary outcomes.
The event “red” is:
A = {R₁, R₂, R₃}
Therefore:
P(red) = 3/5
Why {Red, Blue} cannot be counted equally
The recorded colors have probabilities:
- Red: 3/5.
- Blue: 2/5.
The color sample space is valid as a description, but its outcomes are not equally likely.
Key lesson: Equal likelihood belongs to the model, not to the number of labels you write down.
4. Ordered outcomes
In an experiment with several stages, one outcome must describe every stage.
An ordered pair records a first result and a second result:
(first result, second result)
The outcomes (1, 2) and (2, 1) are different when order or identity matters.
Examples
- Two successive draws: first card, then second card.
- Two distinguishable dice: red die, then blue die.
- A coin toss followed by a die roll: coin result, then die result.
Counting two-stage outcomes
If:
- The first stage has m possibilities.
- Each first-stage result allows n second-stage possibilities.
Then there are:
m × n complete outcomes
This counts possible outcomes. To use the classical probability formula, we must also justify that the complete outcomes are equally likely.
Worked example 3: A coin and a die
A fair coin is tossed and a fair six-sided die is rolled independently.
“Independently” means the coin result does not change the die’s probabilities.
The outcomes are:
Ω = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6),
(T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
There are:
2 × 6 = 12 equally likely outcomes
Find the probability of heads and an even number.
Favorable outcomes:
A = {(H,2), (H,4), (H,6)}
Therefore:
P(A) = 3/12 = 1/4
5. Systematic listing for repeated coin tosses
Worked example 4: Three coin tosses
A fair coin is tossed three times independently. Find the probability of exactly two heads.
Step 1: List all ordered outcomes
Ω = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
There are:
2 × 2 × 2 = 8 equally likely outcomes
Step 2: Identify exactly two heads
A = {HHT, HTH, THH}
Step 3: Calculate
P(A) = 3/8
A common incorrect model
A student lists only the number of heads:
{0, 1, 2, 3}
and concludes that exactly two heads has probability 1/4.
This is incorrect because those four counts are not equally likely.
- Zero heads occurs in one ordered outcome.
- One head occurs in three ordered outcomes.
- Two heads occurs in three ordered outcomes.
- Three heads occurs in one ordered outcome.
Key lesson: Count the equally likely sequences before grouping them into categories.
6. Two-dice sample spaces
Roll two fair six-sided dice independently.
To distinguish them, call them the red die and the blue die.
A complete outcome is:
(red result, blue result)
There are:
6 × 6 = 36 equally likely outcomes
A table of sums
Each cell below corresponds to one ordered outcome. The number in the cell is its sum.
| Red / Blue | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Worked example 5: Sum equal to 8
The favorable ordered outcomes are:
{(2,6), (3,5), (4,4), (5,3), (6,2)}
There are five.
P(sum = 8) = 5/36
Why sums are not equally likely
The possible sums are:
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
There are eleven possible sums, but they are not equally likely.
- Sum 2 comes only from (1,1).
- Sum 7 comes from six ordered outcomes.
Therefore, the probability of a particular sum is not automatically 1/11.
7. Translating “and,” “or,” and “not”
These words describe which outcomes belong to an event.
“And”
An outcome must satisfy both conditions.
For one die:
“Even and greater than 3” = {4, 6}
“Or”
In mathematics, “or” normally includes outcomes satisfying either condition or both.
For one die:
“Even or greater than 4” = {2, 4, 5, 6}
The outcome 6 satisfies both conditions but is listed only once.
“Not”
The event “not A” contains all outcomes in the sample space that do not belong to A.
It is called the complement of A.
We write it as Aᶜ.
For one die:
A = {5, 6}
Aᶜ = {1, 2, 3, 4}
An event and its complement:
- Cannot occur together.
- Together cover the entire sample space.
Therefore:
P(Aᶜ) = 1 − P(A)
Worked example 6: Avoiding double-counting
A fair die is rolled once.
Find the probability that the result is even or greater than 4.
The event is:
A = {2, 4, 5, 6}
Therefore:
P(A) = 4/6 = 2/3
An incorrect count would add three even outcomes and two outcomes greater than 4 to get five.
That counts 6 twice.
Key lesson: In an event set, each elementary outcome is counted once.
8. “At least one” and complementary events
An event can sometimes be counted more easily by counting its complement.
Worked example 7: At least one six
Two fair dice are rolled independently.
Find the probability of at least one six.
Step 1: Identify the complement
The complement of “at least one six” is:
“Neither die shows six.”
Step 2: Count all outcomes
6 × 6 = 36
Step 3: Count outcomes with no six
Each die can show 1, 2, 3, 4, or 5.
5 × 5 = 25
Step 4: Subtract
Favorable outcomes = 36 − 25 = 11
Therefore:
P(at least one six) = 11/36
Direct-count verification
- First die is six: six outcomes.
- Second die is six: six outcomes.
- The outcome (6,6) belongs to both lists.
Therefore:
6 + 6 − 1 = 11
Distinguish the wording
- At least one six: One six or two sixes.
- Exactly one six: One six, but not two.
For exactly one six:
Favorable outcomes = 5 + 5 = 10
P(exactly one six) = 10/36 = 5/18
9. Sampling with replacement
With replacement means the selected item is returned before the next selection.
The same item can therefore be selected again.
To obtain equally likely ordered outcomes, assume each draw selects uniformly from the available items and the selection mechanism does not favor earlier results.
Worked example 8: Repeated numbered draws
A bag contains three balls numbered 1, 2, and 3.
Two balls are selected successively with replacement, uniformly on each draw.
Find the probability that the two numbers are equal.
Step 1: List the ordered outcomes
Ω = {(1,1), (1,2), (1,3),
(2,1), (2,2), (2,3),
(3,1), (3,2), (3,3)}
There are:
3 × 3 = 9 equally likely outcomes
Step 2: Identify matching numbers
A = {(1,1), (2,2), (3,3)}
Step 3: Calculate
P(A) = 3/9 = 1/3
10. Sampling without replacement
Without replacement means the first item is not returned before the second selection.
The same physical item cannot be selected twice.
Worked example 9: Different numbered balls
Use the same three balls numbered 1, 2, and 3.
Two balls are selected successively without replacement, uniformly from those remaining.
The sample space is:
Ω = {(1,2), (1,3), (2,1), (2,3), (3,1), (3,2)}
There are:
3 × 2 = 6 equally likely outcomes
Find the probability that the first number is smaller than the second.
Favorable outcomes:
A = {(1,2), (1,3), (2,3)}
Therefore:
P(A) = 3/6 = 1/2
Replacement comparison
| Feature | With replacement | Without replacement |
|---|---|---|
| First item returned? | Yes | No |
| Same physical item can appear twice? | Yes | No |
| Second-draw choices from N distinct items | N | N − 1 |
| Two-draw ordered outcome count | N × N | N × (N − 1), for N ≥ 2 |
Important: Without replacement prevents repeating the same physical item. It does not necessarily prevent repeating a color.
11. Colored selections without replacement
Worked example 10
A bag contains three red balls and two blue balls.
Two balls are selected successively without replacement. At each stage, every remaining ball is equally likely to be selected.
Find the probability that the colors are different.
Step 1: Label the physical balls
R₁, R₂, R₃, B₁, B₂
Step 2: Count the total ordered outcomes
There are five choices first and four choices second:
Total = 5 × 4 = 20
Step 3: Count different-color outcomes
Red first, blue second:
3 × 2 = 6
Blue first, red second:
2 × 3 = 6
These two groups do not overlap.
Favorable total = 6 + 6 = 12
Step 4: Calculate
P(different colors) = 12/20 = 3/5
Verification using the complement
Both red:
3 × 2 = 6 outcomes
Both blue:
2 × 1 = 2 outcomes
Same-color total:
6 + 2 = 8
Different-color total:
20 − 8 = 12
The counts agree.
12. Ordered and unordered counting must agree
Sometimes the experiment asks only which items were selected, not their order.
You can use ordered or unordered outcomes, provided:
- The chosen outcomes are equally likely.
- The numerator and denominator use the same counting convention.
Worked example 11: Selecting two cards
Two cards are selected uniformly as a pair from cards numbered 1, 2, 3, and 4.
Find the probability that their sum is 5.
Method 1: Unordered pairs
List each pair once:
{1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}
There are six equally likely pairs.
Favorable pairs:
{1,4}, {2,3}
Therefore:
P(sum = 5) = 2/6 = 1/3
Method 2: Ordered draws without replacement
There are:
4 × 3 = 12 ordered outcomes
Favorable outcomes:
(1,4), (4,1), (2,3), (3,2)
Therefore:
P(sum = 5) = 4/12 = 1/3
Both methods give the same result.
Do not mix methods
Using two favorable unordered pairs over twelve ordered outcomes would be incorrect.
The numerator and denominator must describe the same kind of outcome.
13. Common mistakes and verification
Mistake 1: Counting sums as equally likely
Two dice have eleven possible sums, but thirty-six equally likely ordered outcomes.
Mistake 2: Omitting reversed outcomes
For distinguishable dice, (2,5) and (5,2) are different.
Mistake 3: Double-counting overlap
In an “or” event, an outcome satisfying both conditions is still counted once.
Mistake 4: Ignoring replacement
With replacement and without replacement usually have different sample spaces.
Mistake 5: Treating colors as individual equally likely items
A color category can contain several physical balls.
Mistake 6: Confusing “at least one” with “exactly one”
“At least one” includes cases with two or more.
Mistake 7: Mixing ordered and unordered counts
Use one convention consistently in both numerator and denominator.
Mistake 8: Counting possibilities without justifying probabilities
The fact that there are n possible outcomes does not, by itself, give probability 1/n for each.
Verification checklist
- Is the sample space finite?
- Are its elementary outcomes equally likely?
- Is order recorded or deliberately ignored?
- Is replacement specified?
- Are all valid outcomes included?
- Is each outcome counted once?
- Does the event match the wording?
- Are numerator and denominator counted consistently?
- Does the answer lie between 0 and 1?
- Can a complement or second method verify the count?
14. Independent practice
Complete these before reading the solutions.
Practice 1
One card is selected uniformly from cards numbered 1 to 15.
Find the probability that its number is a multiple of 4.
Practice 2
A bag contains five white balls and three black balls. Each ball is equally likely to be selected.
Find the probability of black.
Practice 3
A fair coin is tossed twice independently.
List the sample space and find the probability that both results are the same.
Practice 4
A fair coin is tossed three times independently.
Find the probability of exactly one head.
Practice 5
Two fair dice are rolled independently.
Find the probability that their sum is 9.
Practice 6
Two fair dice are rolled independently.
Find the probability that both show the same number.
Practice 7
Two fair dice are rolled independently.
Find the probability of at least one 1.
Practice 8
A fair die is rolled once.
Find the probability that the result is odd or greater than 4.
Practice 9
Two successive draws are made uniformly from balls numbered 1, 2, and 3, with replacement.
Find the probability that the first number is greater than the second.
Practice 10
Two successive draws are made uniformly from balls numbered 1, 2, 3, and 4, without replacement.
Find the probability that the sum is 5.
Practice 11
A bag contains two red balls and three blue balls.
Two balls are drawn successively without replacement, uniformly from those remaining.
Find the probability that both are blue.
Practice 12
Two cards are selected uniformly as an unordered pair from cards numbered 1, 2, 3, 4, and 5.
List the favorable pairs and find the probability that both numbers are odd.
15. Practice solutions
Solution 1
Multiples of 4:
{4, 8, 12}
There are three favorable outcomes among fifteen.
P = 3/15 = 1/5
Solution 2
There are eight equally likely individual balls, three of which are black.
P(black) = 3/8
Solution 3
Ω = {HH, HT, TH, TT}
Both results are the same in:
{HH, TT}
P = 2/4 = 1/2
Solution 4
Exactly one head occurs in:
{HTT, THT, TTH}
There are eight equally likely ordered outcomes.
P = 3/8
Solution 5
Sum 9 occurs in:
{(3,6), (4,5), (5,4), (6,3)}
P = 4/36 = 1/9
Solution 6
Matching outcomes:
{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
P = 6/36 = 1/6
Solution 7
There are thirty-six ordered outcomes.
With no 1, each die has five choices:
5 × 5 = 25
At least one 1:
36 − 25 = 11
P = 11/36
Solution 8
Odd outcomes:
{1, 3, 5}
Outcomes greater than 4:
{5, 6}
Combined event, counting 5 once:
{1, 3, 5, 6}
P = 4/6 = 2/3
Solution 9
With replacement:
Total = 3 × 3 = 9
First greater than second:
{(2,1), (3,1), (3,2)}
P = 3/9 = 1/3
Solution 10
Without replacement:
Total = 4 × 3 = 12
Sum 5:
{(1,4), (4,1), (2,3), (3,2)}
P = 4/12 = 1/3
Solution 11
Label the five physical balls.
Total ordered outcomes:
5 × 4 = 20
Both blue:
3 × 2 = 6
P = 6/20 = 3/10
Solution 12
There are ten equally likely unordered pairs:
{1,2}, {1,3}, {1,4}, {1,5},
{2,3}, {2,4}, {2,5},
{3,4}, {3,5}, {4,5}
Favorable pairs:
{1,3}, {1,5}, {3,5}
P = 3/10
16. Lesson self-check
This is a formative check within 04.2, not a replacement for the scheduled probability quiz.
Check 1
Which condition is required to calculate probability by simply counting favorable outcomes and dividing by the total?
A. The event contains exactly one outcome.
B. The finite elementary outcomes being counted are equally likely.
C. The experiment has only two outcomes.
D. Every event has probability 1/2.
Check 2
Two fair dice are rolled independently. What is the probability that their sum is 7?
A. 1/11
B. 1/12
C. 1/6
D. 1/2
Check 3
A fair coin is tossed three times independently. What is the probability of at least one head?
A. 7/8
B. 3/8
C. 1/2
D. 1/8
Check 4
Two balls are drawn successively without replacement from five distinct balls.
How many ordered outcomes are possible?
A. 10
B. 25
C. 5
D. 20
Check 5
A bag contains three red balls and one blue ball.
Two balls are drawn successively without replacement, uniformly from those remaining.
What is the probability that their colors are different?
A. 1/4
B. 1/2
C. 3/4
D. 1/3
Check 6
A fair die is rolled once.
What is the probability that the result is even or at least 5?
A. 5/6
B. 1/2
C. 2/3
D. 1/3
17. Self-check answers and explanations
Check 1: B
The counting formula requires a finite sample space of equally likely elementary outcomes.
Check 2: C
Sum 7 occurs in:
(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
P = 6/36 = 1/6
The eleven possible sums are not equally likely.
Check 3: A
There are eight equally likely sequences.
The complement of “at least one head” is TTT, which is one outcome.
P(at least one head) = 1 − 1/8 = 7/8
Check 4: D
There are five choices first and four choices second.
Total = 5 × 4 = 20
Check 5: B
Total ordered outcomes:
4 × 3 = 12
Different colors:
- Red then blue: 3 × 1 = 3.
- Blue then red: 1 × 3 = 3.
P = 6/12 = 1/2
Check 6: C
Even outcomes:
{2, 4, 6}
Outcomes at least 5:
{5, 6}
Their combined event is:
{2, 4, 5, 6}
P = 4/6 = 2/3
The outcome 6 is counted once.
18. Revision guidance
- Difficulty with Practices 1–2: Review equally likely individual outcomes.
- Difficulty with Practices 3–4: Review complete ordered sequences.
- Difficulty with Practices 5–7: Use the two-dice table and count cells.
- Difficulty with Practice 8: Review inclusive “or” and double-counting.
- Difficulty with Practices 9–11: Review replacement and ordered draws.
- Difficulty with Practice 12: Review unordered pairs and consistent counting.
For each incorrect or guessed answer, write:
- The sample space or counting structure.
- Why its elementary outcomes are equally likely.
- The favorable outcomes or favorable count.
- The probability.
- A verification using a second count or complement where practical.
19. Lesson review
You should now be able to:
- Recognize when a classical probability model applies.
- Construct finite sample spaces systematically.
- Distinguish elementary outcomes from grouped categories.
- Count ordered outcomes in simple multistage experiments.
- Use a two-dice table.
- Interpret “and,” inclusive “or,” and “not.”
- Avoid counting overlapping outcomes twice.
- Use a complement for “at least one” problems.
- Distinguish sampling with and without replacement.
- Keep ordered and unordered counting consistent.
- Calculate and simplify classical probabilities.
- Verify that a result lies between 0 and 1.
Essential rules
P(A) = favorable equally likely outcomes ÷ total equally likely outcomes
P(Aᶜ) = 1 − P(A)
For two draws from N distinct items:
- With replacement: N × N ordered outcomes.
- Without replacement: N × (N − 1) ordered outcomes.
These outcome counts support classical probability calculations when the selection assumptions make the counted outcomes equally likely.
Completion target: Construct the sample space correctly, justify equal likelihood, and independently correct every missed practice or self-check question.