The Central Idea
Suppose a finite sample space
\[ S=\{s_1,s_2,\ldots,s_n\} \]contains \(n\) elementary outcomes and each outcome has the same probability.
Because the outcome probabilities must sum to \(1\),
\[ P(\{s_i\})=\frac{1}{n} \]for every \(i\).
Therefore, if an event \(A\) contains \(k\) of those outcomes,
\[ \boxed{P(A)=\frac{k}{n}=\frac{|A|}{|S|}}. \]What Does Equally Likely Mean?
Outcomes are equally likely when they are assigned the same probability.
If
\[ S=\{s_1,s_2,\ldots,s_n\}, \]then equal likelihood means
\[ P(\{s_1\}) = P(\{s_2\}) = \cdots = P(\{s_n\}). \]Since the probabilities sum to \(1\),
\[ \boxed{P(\{s_i\})=\frac1n}. \]Equal Likelihood Is an Assumption, Not an Axiom
The probability axioms do not require all outcomes to have equal probability.
For example, on
\[ S=\{A,B,C\}, \]the assignment
\[ P(A)=0.2,\qquad P(B)=0.3,\qquad P(C)=0.5 \]is perfectly valid.
Equal likelihood must come from additional structure, such as symmetry, design, or a justified modeling assumption.
The Classical Counting Formula
When the elementary outcomes in a finite sample space are equally likely, the probability of an event \(A\) is
\[ \boxed{ P(A)=\frac{|A|}{|S|} }. \]This is sometimes called the classical probability formula.
The numerator counts favorable outcomes, and the denominator counts all equally likely elementary outcomes.
Why the Counting Formula Works
Suppose \(S\) contains \(n\) equally likely outcomes. Each has probability
\[ \frac1n. \]If \(A\) contains \(k\) outcomes, then finite additivity gives
\[ P(A) = \underbrace{\frac1n+\frac1n+\cdots+\frac1n}_{k\text{ terms}} = \frac{k}{n}. \]Therefore,
\[ \boxed{P(A)=\frac{|A|}{|S|}}. \]Example: One Fair Coin
For one fair coin toss,
\[ S=\{H,T\}. \]The outcomes are modeled as equally likely, so
\[ P(H)=P(T)=\frac12. \]Thus,
\[ \boxed{P(H)=\frac{1}{2}}. \]Example: One Fair Die
For a fair six-sided die,
\[ S=\{1,2,3,4,5,6\}. \]Each face is assigned probability
\[ \frac16. \]If
\[ A=\{2,4,6\} \]is the event “roll an even number,” then
\[ P(A) = \frac{3}{6} = \boxed{\frac12}. \]Symmetry Often Supports Equal Likelihood
In idealized models, symmetry often provides the reason for assigning equal probabilities.
A geometrically symmetric, balanced coin is modeled so that neither face is favored.
A fair die is modeled so that no face is favored.
A thoroughly shuffled standard deck is often modeled so that each card is equally likely to occupy a given position.
Physical Symmetry and Mathematical Symmetry
Equal likelihood may be supported by physical design, but the probability model remains a mathematical idealization.
A real die can have small imperfections. A real coin can be slightly unbalanced. A real shuffle can be imperfect.
The mathematical model may still treat outcomes as equally likely when that approximation is appropriate for the purpose of the analysis.
Equal Labels Do Not Guarantee Equal Probabilities
Merely listing outcomes in a set does not make them equally likely.
If
\[ S=\{A,B,C\}, \]nothing about the notation itself implies
\[ P(A)=P(B)=P(C)=\frac13. \]The equality requires an independent justification.
Example: A Biased Coin
Suppose a coin is modeled by
\[ P(H)=0.7, \qquad P(T)=0.3. \]The sample space is still
\[ S=\{H,T\}, \]but the outcomes are not equally likely.
Therefore, the calculation
\[ P(H)=\frac12 \]would be invalid for this model.
Example: A Weighted Spinner
Suppose a spinner has three labeled regions \(A\), \(B\), and \(C\), but their areas are unequal.
Even though the sample space can be written
\[ S=\{A,B,C\}, \]the three outcomes need not have probability \(1/3\).
If the spinner is designed so that landing probability is proportional to region angle, larger regions receive larger probability.
The Choice of Sample Space Matters
A counting method works only when the outcomes being counted are equally likely.
Sometimes a coarse sample space hides unequal multiplicities.
For example, if two fair coins are tossed and only the number of heads is recorded, then
\[ S=\{0,1,2\}. \]These three recorded outcomes are not equally likely.
Why the Number of Heads Is Not Uniform
For two fair coin tosses, the detailed ordered sample space is
\[ \{HH,HT,TH,TT\}. \]These four outcomes are equally likely.
The number of heads satisfies
\[ 0\leftrightarrow TT, \] \[ 1\leftrightarrow HT,TH, \]and
\[ 2\leftrightarrow HH. \]Therefore,
\[ P(0)=\frac14, \qquad P(1)=\frac24, \qquad P(2)=\frac14. \]Refining the Sample Space Can Restore Equal Likelihood
A coarse sample space may contain outcomes with unequal probabilities, while a more detailed sample space contains equally likely elementary outcomes.
For two fair coins,
\[ \{0,1,2\} \]is not an equally likely sample space, but
\[ \boxed{\{HH,HT,TH,TT\}} \]is equally likely under the standard fair-independent model.
Example: Sum of Two Fair Dice
If two fair dice are rolled and only the sum is recorded, then
\[ S=\{2,3,4,5,6,7,8,9,10,11,12\}. \]These eleven sums are not equally likely.
For example,
\[ 2 \]occurs only through
\[ (1,1), \]while
\[ 7 \]occurs through six ordered pairs.
The Better Two-Dice Sample Space
For two distinguishable fair dice, the ordered-pair sample space
\[ S=\{(i,j):i,j\in\{1,2,3,4,5,6\}\} \]contains
\[ 6\cdot6=36 \]equally likely elementary outcomes under the standard fair-independent model.
This makes counting-based probability appropriate.
Example: Probability of Sum Seven
The ordered pairs with sum \(7\) are
\[ (1,6),(2,5),(3,4),(4,3),(5,2),(6,1). \]There are \(6\) favorable outcomes among \(36\) equally likely outcomes.
\[ P(\text{sum }7) = \frac{6}{36} = \boxed{\frac16}. \]Example: Probability of Doubles
The doubles are
\[ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6). \]There are \(6\) favorable ordered outcomes among \(36\) equally likely outcomes.
\[ \boxed{ P(\text{doubles}) = \frac{6}{36} = \frac16 }. \]Equal Likelihood and Ordered Outcomes
When stages are distinguishable, ordered outcomes often provide the correct equally likely elementary structure.
For two fair dice,
\[ (2,5) \]and
\[ (5,2) \]are distinct outcomes and each has probability
\[ \frac1{36}. \]Example: Drawing One Card
Suppose one card is drawn from a thoroughly shuffled standard \(52\)-card deck.
Under the usual model, each card is equally likely.
The probability of drawing an ace is therefore
\[ P(\text{ace}) = \frac{4}{52} = \boxed{\frac1{13}}. \]Example: Drawing a Heart
A standard deck contains \(13\) hearts among \(52\) cards.
Under the equally likely card model,
\[ P(\text{heart}) = \frac{13}{52} = \boxed{\frac14}. \]Example: Drawing a Face Card
Each of the four suits contains a jack, queen, and king.
Therefore, there are
\[ 4\cdot3=12 \]face cards.
Thus,
\[ P(\text{face card}) = \frac{12}{52} = \boxed{\frac3{13}}. \]Equally Likely Does Not Mean Independent
Equal likelihood compares probabilities of outcomes or events.
Independence concerns whether one event changes the probability of another.
These concepts are different.
A collection of outcomes can be equally likely without raising any question about independence at all.
Repeated Fair Trials
Suppose a fair coin is tossed \(n\) times and the tosses are modeled as independent.
Each ordered \(H/T\) sequence has probability
\[ \left(\frac12\right)^n. \]Since there are
\[ 2^n \]sequences, each has probability
\[ \boxed{\frac{1}{2^n}}. \]Example: Three Fair Coin Tosses
The sample space is
\[ S= \{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}. \]There are \(8\) equally likely sequences.
The event “exactly two heads” is
\[ \{HHT,HTH,THH\}. \]Therefore,
\[ \boxed{ P(\text{exactly two heads}) = \frac38 }. \]Example: At Least One Head
For three fair coin tosses, the outcomes with at least one head are all outcomes except
\[ TTT. \]Therefore, there are \(7\) favorable outcomes among \(8\) equally likely sequences.
\[ \boxed{ P(\text{at least one head}) = \frac78 }. \]Counting Can Replace Listing
When a sample space is large but its elementary outcomes are equally likely, counting methods can replace explicit listing.
If there are
\[ |S|=N \]equally likely outcomes and an event contains
\[ |A|=K, \]then
\[ \boxed{P(A)=\frac{K}{N}}. \]This connects probability directly to combinatorics.
Example: Random Selection from Distinct Objects
Suppose one object is chosen uniformly at random from \(20\) distinct objects.
If \(7\) of the objects are red, then
\[ P(\text{red}) = \frac{7}{20}. \]The phrase uniformly at random signals that the \(20\) individual objects are equally likely to be selected.
Uniformly at Random
In a finite sample space, the phrase chosen uniformly at random generally means that each elementary outcome has the same probability.
If
\[ |S|=n, \]then
\[ \boxed{P(\{s\})=\frac1n} \]for every \(s\in S\).
Uniform Does Not Mean the Same Thing in Every Setting
In a finite sample space, uniform probability assigns equal probability to each outcome.
In a continuous interval, a uniform model distributes probability evenly by length rather than by giving positive equal probability to individual points.
Thus, the phrase uniform must be interpreted in the context of the sample space.
Why Individual Points Cannot All Have Positive Equal Probability on an Interval
A continuous interval contains infinitely many points.
In a standard uniform continuous model, each individual point has probability
\[ 0, \]while intervals of positive length can have positive probability.
Therefore, the finite counting formula
\[ \frac{|A|}{|S|} \]is not the right tool for ordinary continuous uniform probability.
Example: Uniform Selection on an Interval
Suppose \(X\) is selected uniformly from
\[ [0,10]. \]The event
\[ A=[2,5] \]occupies length
\[ 5-2=3. \]The entire interval has length
\[ 10. \]Thus,
\[ \boxed{P(A)=\frac{3}{10}}. \]Geometric Uniformity
In geometric probability, equal likelihood is often modeled through equal length, area, volume, or another geometric measure.
The corresponding probability ratio may take the form
\[ \frac{\text{favorable length}}{\text{total length}}, \] \[ \frac{\text{favorable area}}{\text{total area}}, \]or
\[ \frac{\text{favorable volume}}{\text{total volume}}. \]A later lesson develops geometric probability directly.
Equal Likelihood and Replacement
The rules of an experiment can affect whether elementary outcomes remain equally likely.
Suppose two objects are drawn from a collection.
With replacement, the same set of choices may be available at each stage.
Without replacement, the second-stage possibilities depend on the first selection.
A careful sample-space model must reflect that change.
Example: Ordered Draws Without Replacement
Suppose two distinct objects are selected in order from \(5\) distinct objects without replacement, with all ordered selections equally likely.
There are
\[ 5\cdot4=20 \]ordered outcomes.
Therefore, each ordered pair has probability
\[ \boxed{\frac1{20}}. \]Unordered Versus Ordered Models
The same physical procedure can sometimes be represented using ordered or unordered elementary outcomes.
If all unordered pairs are equally likely, then one may count unordered pairs directly.
If the natural experiment produces equally likely ordered sequences, then counting should begin with ordered sequences unless a valid grouping argument shows that all grouped outcomes receive equal multiplicity.
Grouping Equally Likely Outcomes
Suppose a detailed sample space contains equally likely elementary outcomes.
If broader categories each contain the same number of elementary outcomes, then those broader categories are also equally likely.
If the categories contain different numbers of elementary outcomes, then the broader categories generally are not equally likely.
Example: Equal-Size Grouping
A fair die has six equally likely faces.
Group the outcomes into
\[ A=\{1,2\}, \qquad B=\{3,4\}, \qquad C=\{5,6\}. \]Each event contains \(2\) of the \(6\) equally likely outcomes, so
\[ P(A)=P(B)=P(C)=\frac{2}{6}=\boxed{\frac13}. \]Example: Unequal-Size Grouping
For a fair die, define
\[ A=\{1\}, \qquad B=\{2,3\}, \qquad C=\{4,5,6\}. \]Then
\[ P(A)=\frac16, \qquad P(B)=\frac26, \qquad P(C)=\frac36. \]The three grouped outcomes are not equally likely.
Symmetry Must Match the Actual Experiment
A symmetric-looking answer set does not guarantee a symmetric process.
For example, the outcomes
\[ \{\text{win},\text{lose}\} \]contain two labels, but that does not imply
\[ P(\text{win})=P(\text{lose})=\frac12. \]Equal likelihood depends on the mechanism producing the outcomes.
“Random” Does Not Automatically Mean “Equally Likely”
A random experiment has uncertain outcomes, but those outcomes can have unequal probabilities.
Therefore, the word random alone does not justify
\[ P(A)=\frac{|A|}{|S|}. \]Look for wording such as fair, uniformly selected, or a stated symmetry assumption.
Empirical Evidence Can Challenge Equal-Likelihood Assumptions
Equal likelihood may begin as a theoretical model.
Repeated observations can later provide evidence about whether the model is reasonable.
If a supposedly fair coin produces a persistent imbalance over a large number of trials, the equal-likelihood assumption may need to be examined.
Simulation and empirical probability will address this connection later.
Equal Likelihood and Counting-Based Probability
Counting-based probability has two separate tasks:
- justify that the elementary outcomes are equally likely;
- count the favorable and total elementary outcomes correctly.
An error in either task can invalidate the result.
A Reliable Counting Workflow
- State the experiment precisely.
- Choose elementary outcomes that preserve the relevant distinctions.
- Determine whether those elementary outcomes are equally likely.
- Count the total number of elementary outcomes.
- Describe the event of interest.
- Count the favorable elementary outcomes.
- Compute \(P(A)=|A|/|S|\).
- Check that the result lies between \(0\) and \(1\).
Example: Full Counting Workflow with Two Dice
Find the probability that the sum of two fair distinguishable dice is at least \(10\).
The equally likely sample space has
\[ |S|=36 \]ordered pairs.
The favorable outcomes are
\[ (4,6),(5,5),(5,6),(6,4),(6,5),(6,6). \]There are \(6\) favorable outcomes, so
\[ \boxed{ P(\text{sum at least }10)=\frac6{36}=\frac16 }. \]Example: A Card Event with Overlap
One card is drawn uniformly from a standard \(52\)-card deck.
Find the probability that it is a heart or a king.
There are \(13\) hearts and \(4\) kings, but the king of hearts belongs to both groups.
Thus, the number of favorable cards is
\[ 13+4-1=16. \]Therefore,
\[ \boxed{ P(\text{heart or king}) = \frac{16}{52} = \frac4{13} }. \]Counting Still Requires Correct Event Operations
Equal likelihood makes probability proportional to outcome count, but counting must still respect unions, intersections, and complements.
Overlapping favorable categories must not be counted twice.
This is why the next lesson develops complement, addition, and multiplication rules systematically.
Complement Counting
Sometimes it is easier to count the outcomes that do not satisfy an event.
If the sample space is equally likely, then
\[ P(A^c)=\frac{|A^c|}{|S|}. \]Since
\[ P(A)=1-P(A^c), \]counting the complement can simplify many “at least one” problems.
Example: At Least One Six
Two fair dice are rolled.
Instead of counting outcomes containing at least one \(6\), count the complement: neither die shows \(6\).
There are
\[ 5\cdot5=25 \]outcomes with no \(6\) among \(36\) total outcomes.
Therefore,
\[ P(\text{at least one }6) = 1-\frac{25}{36} = \boxed{\frac{11}{36}}. \]Common Error: Counting Unequally Likely Outcomes
For the sum of two fair dice, there are \(11\) possible sums:
\[ 2,3,\ldots,12. \]It is incorrect to conclude that
\[ P(\text{sum }7)=\frac1{11}. \]The sums are not equally likely.
The correct equally likely sample space contains \(36\) ordered die pairs.
Common Error: Assuming Fairness from Appearance
A die with six labeled faces is not automatically fair.
A spinner with four labeled regions is not automatically uniform.
A coin with two sides is not automatically balanced.
Equal likelihood is a model assumption that must be stated, implied by the setup, or otherwise justified.
Common Error: Confusing Equal Counts with Equal Probability
Two events can contain the same number of outcomes and still have different probabilities if the elementary outcomes are weighted unequally.
Therefore,
\[ |A|=|B| \]does not imply
\[ P(A)=P(B) \]unless the underlying outcomes are equally likely.
Common Error: Using a Coarse Sample Space
If two fair coins are tossed and the recorded outcome is the number of heads,
\[ S=\{0,1,2\}. \]Dividing by \(3\) is incorrect because those three outcomes are not equally likely.
Refine the experiment to
\[ \{HH,HT,TH,TT\} \]before using direct counting.
Common Error: Counting Physical Objects Instead of Elementary Outcomes
The denominator must count the elementary outcomes of the probability model, not whatever objects happen to be easiest to count.
In a card problem, the elementary outcomes may be individual cards.
In a two-dice problem, they may be ordered pairs.
In a sequence problem, they may be complete strings.
Common Error: Forgetting Restrictions
If objects are selected without replacement, outcomes containing repeated copies of the same physical object may be impossible.
Counting them in the denominator would enlarge the sample space incorrectly and distort the probability.
The equally likely outcomes must satisfy every rule of the experiment.
Common Error: Forgetting Order
If the experiment distinguishes first and second positions, then
\[ (A,B) \]and
\[ (B,A) \]are different outcomes.
Combining them without justification can change both the numerator and denominator.
Common Error: Treating Continuous Uniform Models as Finite Counting Problems
In a continuous uniform interval, individual points do not each receive a positive probability of the form
\[ \frac1n. \]Probability is instead assigned according to interval length or another appropriate measure.
Finite outcome counting and continuous geometric probability are related ideas but not identical methods.
A Reliable Test for the Classical Formula
Before using
\[ P(A)=\frac{|A|}{|S|}, \]ask:
- Is the sample space finite?
- Are the outcomes I am counting elementary enough?
- Are those outcomes equally likely?
- Does the event correspond to a subset of those outcomes?
- Have I counted both the event and the sample space without omissions or duplicates?
If any answer is uncertain, the counting formula needs further justification.
Practice 1: Fair Die
A fair six-sided die is rolled.
Find the probability of rolling a number greater than \(4\).
Practice 2: Fair Coin
A fair coin is tossed three times.
Find the probability of getting exactly one head.
Practice 3: Two Dice
Two fair distinguishable six-sided dice are rolled.
Find the probability that the sum is \(8\).
Practice 4: Card Draw
One card is drawn uniformly from a standard \(52\)-card deck.
Find the probability of drawing a queen.
Practice 5: Card Color
One card is drawn uniformly from a standard deck.
Find the probability of drawing a red card.
Practice 6: Coarse Sample Space
Two fair coins are tossed, and only the number of heads is recorded.
Are the outcomes
\[ \{0,1,2\} \]equally likely? Explain.
Practice 7: Loaded Outcome Model
A spinner has three outcomes with probabilities
\[ P(A)=0.2,\qquad P(B)=0.3,\qquad P(C)=0.5. \]Can the classical formula \(P(A)=|A|/|S|\) be applied directly to the three labels?
Practice 8: At Least One Head
Four fair coins are tossed.
Find the probability of obtaining at least one head.
Practice 9: Uniform Integer Selection
One integer is selected uniformly from
\[ \{1,2,\ldots,20\}. \]Find the probability that the selected integer is divisible by \(4\).
Practice 10: Without Replacement
Two distinct objects are selected in order from \(6\) distinct objects without replacement, with every ordered selection equally likely.
How many elementary outcomes are there, and what is the probability of one particular ordered pair?
Practice 11: Uniform Interval
A number is selected uniformly from
\[ [0,12]. \]Find the probability that it lies in
\[ [3,7]. \]Practice 12: Identify the Error
A student says:
“The possible sums of two dice are \(2\) through \(12\), so there are eleven outcomes and each has probability \(1/11\).”
Explain the error and give the correct probability of a sum of \(7\).
Practice Answers
Practice 1
The favorable outcomes are
\[ \{5,6\}. \]Therefore,
\[ \boxed{ P(\text{greater than }4) = \frac26 = \frac13 }. \]Practice 2
There are \(2^3=8\) equally likely sequences. Exactly one head occurs in
\[ HTT,\quad THT,\quad TTH. \]Therefore,
\[ \boxed{\frac38}. \]Practice 3
The ordered pairs with sum \(8\) are
\[ (2,6),(3,5),(4,4),(5,3),(6,2). \]Therefore,
\[ \boxed{ P(\text{sum }8)=\frac5{36} }. \]Practice 4
There are \(4\) queens among \(52\) equally likely cards:
\[ \boxed{ P(\text{queen})=\frac4{52}=\frac1{13} }. \]Practice 5
There are \(26\) red cards:
\[ \boxed{ P(\text{red})=\frac{26}{52}=\frac12 }. \]Practice 6
No. The detailed equally likely outcomes are
\[ HH,\ HT,\ TH,\ TT. \]Thus,
\[ P(0)=\frac14,\qquad P(1)=\frac12,\qquad P(2)=\frac14. \]Practice 7
No. The three labels have unequal probabilities, so the ratio of event count to total label count is not valid.
Practice 8
There are
\[ 2^4=16 \]equally likely sequences. Only one sequence,
\[ TTTT, \]has no heads. Therefore,
\[ \boxed{ P(\text{at least one head}) = 1-\frac1{16} = \frac{15}{16} }. \]Practice 9
The multiples of \(4\) are
\[ 4,8,12,16,20. \]Therefore,
\[ \boxed{ P(\text{divisible by }4)=\frac5{20}=\frac14 }. \]Practice 10
There are
\[ 6\cdot5=30 \]ordered outcomes. Therefore, one particular ordered pair has probability
\[ \boxed{\frac1{30}}. \]Practice 11
The favorable interval has length
\[ 7-3=4, \]while the total interval has length \(12\). Therefore,
\[ \boxed{ P([3,7])=\frac4{12}=\frac13 }. \]Practice 12
The eleven sums are not equally likely. The correct equally likely sample space consists of \(36\) ordered die pairs.
Six of those pairs have sum \(7\), so
\[ \boxed{ P(\text{sum }7)=\frac6{36}=\frac16 }. \]What to Remember
If a finite sample space contains equally likely elementary outcomes, then
\[ \boxed{ P(A)=\frac{|A|}{|S|} }. \]This formula works because each elementary outcome has probability
\[ \frac1{|S|}. \]- Equal likelihood is an assumption or modeling conclusion, not a probability axiom.
- Symmetry often supports equal-likelihood models.
- A coarse sample space can hide unequal probabilities.
- Ordered outcomes are often the correct elementary outcomes in multistage experiments.
- Counting is valid only when the outcomes being counted are equally likely.
- Uniform continuous models use geometric measure rather than finite outcome counts.
- “Random” does not automatically mean “equally likely.”
The most important habit is: justify equal likelihood before dividing favorable outcomes by total outcomes.
Where We Go Next
Equal likelihood makes many probability calculations reducible to counting, but probability problems also require rules for combining events.
The next lesson, Complement, Addition, and Multiplication Rules, develops the principal formulas for probabilities involving “not,” “or,” and sequential or joint occurrence.