The Central Idea
If a finite sample space \(S\) consists of equally likely elementary outcomes, then for an event \(A\),
\[ \boxed{P(A)=\frac{|A|}{|S|}}. \]Counting-based probability therefore has two main tasks:
- count the total number of equally likely elementary outcomes;
- count the outcomes that belong to the event.
The probability is the ratio of those counts.
Counting Does Not Replace Probability Modeling
Counting methods are useful only after the probability model has been specified correctly.
Before counting, ask:
- What is the experiment?
- What are the elementary outcomes?
- Are those outcomes equally likely?
- Does order matter?
- Is repetition allowed?
- Are selections made with or without replacement?
A correct count built on the wrong sample space still produces the wrong probability.
The Basic Counting Probability Formula
Suppose
\[ |S|=N \]and
\[ |A|=K. \]If all \(N\) elementary outcomes are equally likely, then
\[ \boxed{P(A)=\frac{K}{N}}. \]The denominator counts all possible outcomes. The numerator counts only those satisfying the event.
Example: Selecting One Integer
One integer is selected uniformly from
\[ \{1,2,\ldots,30\}. \]Find the probability that the selected integer is divisible by \(5\).
The multiples of \(5\) are
\[ 5,10,15,20,25,30. \]Thus,
\[ |A|=6, \qquad |S|=30. \]Therefore,
\[ \boxed{ P(A)=\frac6{30}=\frac15 }. \]The Multiplication Principle for Sample Spaces
When an experiment is completed through several stages, the multiplication principle can count the possible outcomes.
If stage \(1\) has \(n_1\) possibilities, stage \(2\) has \(n_2\), and so on, then
\[ \boxed{ |S|=n_1n_2\cdots n_k }. \]This is especially useful for codes, sequences, repeated trials, and multistage selections.
Example: Four-Digit Codes
A four-digit code uses the digits \(0\) through \(9\), repetition is allowed, and leading zeroes are permitted.
Each of the four positions has \(10\) choices, so
\[ |S|=10^4=10000. \]Suppose the event \(A\) is “the first digit is \(7\).”
Then the remaining three positions are unrestricted:
\[ |A|=10^3=1000. \]Therefore,
\[ \boxed{ P(A)=\frac{1000}{10000}=\frac1{10} }. \]Counting Restrictions Position by Position
Some events are easiest to count by applying restrictions directly to each stage.
For the same four-digit code, suppose \(A\) is the event that every digit is even.
Each position has
\[ 5 \]even choices:
\[ 0,2,4,6,8. \]Therefore,
\[ |A|=5^4, \]so
\[ \boxed{ P(A)=\frac{5^4}{10^4}=\frac1{16} }. \]Order Matters in Some Experiments
When positions, ranks, or stages are distinguished, different orders usually represent different outcomes.
For example,
\[ (A,B,C) \]and
\[ (C,B,A) \]are different ordered selections.
Permutations are the natural counting tool when order matters.
Permutations Without Repetition
The number of ordered selections of \(r\) distinct objects from \(n\) distinct objects is
\[ \boxed{ P(n,r)=\frac{n!}{(n-r)!} }. \]This counts arrangements when an object cannot be selected more than once.
Example: Assigning Three Offices
Ten people are eligible for three distinct offices: president, vice president, and secretary.
The number of possible assignments is
\[ P(10,3) = 10\cdot9\cdot8 = 720. \]Suppose one particular person must be president.
Then only the remaining two offices need to be filled:
\[ 9\cdot8=72. \]Therefore,
\[ \boxed{ P(\text{that person is president}) = \frac{72}{720} = \frac1{10} }. \]Order Does Not Matter in Other Experiments
When only the selected group matters, not the order in which its members are chosen, combinations are appropriate.
The number of ways to choose \(r\) objects from \(n\) distinct objects is
\[ \boxed{ \binom nr=\frac{n!}{r!(n-r)!} }. \]Example: Choosing a Committee
A committee of \(4\) people is chosen uniformly from \(12\) people.
The total number of committees is
\[ |S|=\binom{12}{4}. \]Suppose a particular person must be on the committee.
Then the remaining \(3\) members are selected from the other \(11\):
\[ |A|=\binom{11}{3}. \]Therefore,
\[ P(A) = \frac{\binom{11}{3}}{\binom{12}{4}} = \boxed{\frac13}. \]Choosing Between Permutations and Combinations
A useful decision rule is:
\[ \boxed{ \begin{aligned} \text{order matters} &\Rightarrow \text{permutations},\\ \text{order does not matter} &\Rightarrow \text{combinations}. \end{aligned} } \]Asking this question early prevents many counting errors.
Example: Five-Card Hands
A five-card hand from a standard \(52\)-card deck is an unordered selection.
Therefore, the total number of hands is
\[ \boxed{ |S|=\binom{52}{5} }. \]Permutations would overcount each hand by treating different orders of the same five cards as different outcomes.
Example: Exactly Two Aces in a Five-Card Hand
To build a hand with exactly two aces:
- choose \(2\) of the \(4\) aces;
- choose \(3\) of the \(48\) non-aces.
Thus,
\[ |A| = \binom42\binom{48}{3}. \]The probability is
\[ \boxed{ P(\text{exactly two aces}) = \frac{\binom42\binom{48}{3}}{\binom{52}{5}} }. \]Why the Product Appears in Combination Counts
The choice of aces and the choice of non-aces are separate stages.
There are
\[ \binom42 \]ways to choose the aces and
\[ \binom{48}{3} \]ways to choose the remaining cards.
By the multiplication principle,
\[ |A|=\binom42\binom{48}{3}. \]Example: At Least One Ace
A five-card hand contains at least one ace.
Direct counting would require several cases:
\[ 1,\ 2,\ 3,\ \text{or }4\text{ aces}. \]The complement is easier.
The number of hands with no aces is
\[ \binom{48}{5}. \]Therefore,
\[ \boxed{ P(\text{at least one ace}) = 1-\frac{\binom{48}{5}}{\binom{52}{5}} }. \]Complementary Counting
If an event is difficult to count directly but its complement is easy, use
\[ |A|=|S|-|A^c|. \]Then
\[ P(A) = 1-P(A^c). \]This is particularly effective for phrases such as:
- at least one;
- not all;
- not none;
- at least one repeated value.
Example: At Least One Repeated Birthday
In a simplified \(365\)-day model, suppose \(n\) people have independent uniformly distributed birthdays and leap day is ignored.
The direct event “at least two share a birthday” is complicated.
Its complement is “all birthdays are distinct.”
The total number of ordered birthday assignments is
\[ 365^n. \]For \(n\le365\), the number with all birthdays distinct is
\[ 365\cdot364\cdots(365-n+1). \]Therefore,
\[ \boxed{ P(\text{at least one shared birthday}) = 1- \frac{365\cdot364\cdots(365-n+1)}{365^n} }. \]Counting by Cases
Some events naturally break into mutually exclusive cases.
If
\[ A=A_1\cup A_2\cup\cdots\cup A_k \]and the cases are disjoint, then
\[ |A| = |A_1|+|A_2|+\cdots+|A_k|. \]Probability can then be computed from the total count.
Example: A Hand with Exactly One or Two Aces
The cases “exactly one ace” and “exactly two aces” are disjoint.
Their counts are
\[ \binom41\binom{48}{4} \]and
\[ \binom42\binom{48}{3}. \]Therefore,
\[ \boxed{ P(\text{one or two aces}) = \frac{ \binom41\binom{48}{4} + \binom42\binom{48}{3} }{ \binom{52}{5} } }. \]Restrictions Can Be Built into the Count
Instead of counting all outcomes and removing invalid ones afterward, it is often cleaner to build the restriction into the construction.
For example, to form a committee containing exactly \(2\) people from one group and \(3\) from another, count
\[ \binom{m}{2}\binom{n}{3}. \]The restriction becomes part of the counting structure.
Example: Mixed Committee
A committee of \(5\) is chosen uniformly from \(8\) engineers and \(7\) designers.
Find the probability that the committee contains exactly \(3\) engineers and \(2\) designers.
The total number of committees is
\[ \binom{15}{5}. \]The favorable count is
\[ \binom83\binom72. \]Therefore,
\[ \boxed{ P(\text{exactly }3\text{ engineers}) = \frac{\binom83\binom72}{\binom{15}{5}} }. \]At Least and At Most Conditions
Phrases such as at least and at most often create several allowable cases.
For example,
\[ \text{at least }2 \]may mean
\[ 2,\ 3,\ 4,\ldots \]depending on the maximum possible value.
Count the cases separately or use a complement if that is shorter.
Example: At Least Three Engineers
Continue with \(8\) engineers and \(7\) designers, choosing a committee of \(5\).
At least three engineers means:
\[ 3,\ 4,\text{ or }5\text{ engineers}. \]The favorable count is
\[ \binom83\binom72 + \binom84\binom71 + \binom85\binom70. \]Therefore,
\[ \boxed{ P(\text{at least }3\text{ engineers}) = \frac{ \binom83\binom72+ \binom84\binom71+ \binom85 }{ \binom{15}{5} } }. \]Counting with Repetition
When repeated selections are allowed and positions are distinguished, the multiplication principle often gives powers.
If each of \(r\) positions has \(n\) choices, then
\[ \boxed{n^r} \]ordered strings are possible.
Example: Passwords
A password consists of six lowercase letters, repetition is allowed, and all \(26^6\) strings are equally likely.
Find the probability that every character is the same letter.
There are \(26\) favorable strings:
\[ aaaaaa,\ bbbbbb,\ \ldots \]Therefore,
\[ \boxed{ P(\text{all same}) = \frac{26}{26^6} = \frac1{26^5} }. \]Example: No Repeated Digits
A four-digit code is selected uniformly from all \(10^4\) strings of decimal digits, with leading zeroes allowed.
Find the probability that no digit repeats.
The favorable count is
\[ 10\cdot9\cdot8\cdot7. \]Therefore,
\[ \boxed{ P(\text{no repeated digit}) = \frac{10\cdot9\cdot8\cdot7}{10^4} }. \]Example: At Least One Repeated Digit
Using the same four-digit code model, the event “at least one repeated digit” is the complement of “all digits distinct.”
Thus,
\[ \boxed{ P(\text{at least one repeated digit}) = 1- \frac{10\cdot9\cdot8\cdot7}{10^4} }. \]Inclusion–Exclusion in Counting-Based Probability
If favorable outcomes are described by overlapping conditions, the counts must be corrected for overlap.
For two events,
\[ |A\cup B| = |A|+|B|-|A\cap B|. \]If all sample-space outcomes are equally likely, then dividing by \(|S|\) gives the probability addition rule.
Example: Divisible by Two or Three
An integer is selected uniformly from
\[ \{1,2,\ldots,60\}. \]Let \(A\) be “divisible by \(2\)” and \(B\) be “divisible by \(3\).”
Then
\[ |A|=30, \qquad |B|=20. \]Their intersection consists of multiples of \(6\):
\[ |A\cap B|=10. \]Therefore,
\[ |A\cup B| = 30+20-10 = 40. \]Hence,
\[ \boxed{ P(A\cup B) = \frac{40}{60} = \frac23 }. \]Counting Sequential Draws Directly
Some without-replacement probabilities can be solved either by conditional multiplication or by counting equally likely complete selections.
Both approaches should agree when the model is constructed correctly.
Example: Two Aces by Counting
Two cards are selected uniformly as an unordered pair from a standard deck.
The total number of pairs is
\[ \binom{52}{2}. \]The favorable pairs consisting of two aces are
\[ \binom42. \]Therefore,
\[ \boxed{ P(\text{two aces}) = \frac{\binom42}{\binom{52}{2}} = \frac1{221} }. \]This agrees with
\[ \frac4{52}\cdot\frac3{51}. \]Different Correct Sample Spaces Can Give the Same Probability
A problem may admit more than one valid equally likely sample space.
For two-card selection without replacement, one can use:
- ordered pairs of distinct cards;
- unordered two-card subsets.
If the event is translated consistently, both models produce the same probability.
Ordered Counting for Two Aces
In the ordered model, the total number of outcomes is
\[ 52\cdot51. \]The favorable count is
\[ 4\cdot3. \]Thus,
\[ \frac{4\cdot3}{52\cdot51} = \boxed{\frac1{221}}. \]The same result appears because both numerator and denominator count ordered outcomes consistently.
Do Not Mix Ordered and Unordered Counts
A common error is to count the denominator using ordered outcomes and the numerator using unordered outcomes, or vice versa.
For a valid ratio,
\[ \boxed{ \text{numerator and denominator must use the same type of elementary outcome} }. \]Symmetry Can Simplify Counting
Sometimes a symmetry argument avoids detailed enumeration.
If \(n\) equally likely positions are symmetric and a distinguished object occupies exactly one of them, then the probability it occupies any specified position is
\[ \frac1n. \]Counting can verify this, but symmetry may reveal it immediately.
Example: First Position in a Random Ordering
Ten distinct people are placed in a uniformly random order.
The total number of orderings is
\[ 10!. \]If a particular person must be first, the remaining people can be arranged in
\[ 9! \]ways.
Therefore,
\[ P(\text{particular person first}) = \frac{9!}{10!} = \boxed{\frac1{10}}. \]Example: Two Particular People Adjacent
Ten distinct people are arranged uniformly in a line.
Find the probability that two particular people stand next to one another.
Treat the two people as a block. The block and the other eight people form \(9\) units, which can be arranged in
\[ 9! \]ways.
The two people inside the block can switch order in
\[ 2 \]ways.
Thus,
\[ |A|=2\cdot9!, \qquad |S|=10!. \]Therefore,
\[ \boxed{ P(\text{adjacent}) = \frac{2\cdot9!}{10!} = \frac15 }. \]Example: No Two Particular People Adjacent
Using the previous arrangement problem,
\[ P(\text{adjacent})=\frac15. \]Therefore,
\[ \boxed{ P(\text{not adjacent}) = 1-\frac15 = \frac45 }. \]A counting result and the complement rule work together naturally.
Counting with Identical Categories
Some probability questions concern arrangements of repeated categories rather than fully distinct objects.
If \(n\) positions contain category counts
\[ n_1,n_2,\ldots,n_k \]with
\[ n_1+\cdots+n_k=n, \]then the number of distinct category strings is
\[ \boxed{ \frac{n!}{n_1!n_2!\cdots n_k!} }. \]Example: Five Tosses with Exactly Three Heads
A fair coin is tossed \(5\) times.
Every \(H/T\) string has probability
\[ \frac1{2^5}. \]The number of strings with exactly \(3\) heads is
\[ \binom53. \]Therefore,
\[ \boxed{ P(\text{exactly }3\text{ heads}) = \frac{\binom53}{2^5} = \frac{10}{32} = \frac5{16} }. \]Why \(\binom nr\) Appears in Repeated Binary Trials
To obtain exactly \(r\) successes in \(n\) positions, choose which \(r\) positions contain the successes.
There are
\[ \binom nr \]such position sets.
For equally likely fair binary sequences, this immediately gives the probability count.
Counting and Unequal Probabilities
If elementary outcomes are not equally likely, the ratio
\[ \frac{|A|}{|S|} \]is generally invalid.
Counting may still help organize the event, but the probability must weight outcomes according to their actual probabilities.
Example: Biased Coin Strings
Suppose
\[ P(H)=0.7, \qquad P(T)=0.3. \]In three independent tosses, the eight \(H/T\) sequences are not equally likely.
For example,
\[ P(HHH)=0.7^3 \]while
\[ P(TTT)=0.3^3. \]Counting three-head strings versus all strings does not produce the probability.
Counting Still Helps with a Biased Coin
For exactly two heads in three independent biased tosses, there are
\[ \binom32=3 \]position patterns.
Each such pattern has probability
\[ 0.7^2(0.3). \]Therefore,
\[ \boxed{ P(\text{exactly two heads}) = \binom32(0.7)^2(0.3) }. \]Here counting determines the multiplicity, while probability weights each pattern.
Counting Methods Can Support Conditional Probability
Conditional probability restricts attention to a smaller sample space.
When the remaining outcomes are equally likely within that condition, counting can be used again:
\[ P(A\mid B) = \frac{|A\cap B|}{|B|}. \]This interpretation will be developed in the dedicated conditional probability lesson.
Example: Conditional Counting
One integer is selected uniformly from
\[ \{1,2,\ldots,20\}. \]Given that the integer is even, find the probability that it is divisible by \(4\).
There are \(10\) even integers.
Among them, the multiples of \(4\) are
\[ 4,8,12,16,20, \]giving \(5\) favorable outcomes.
Therefore,
\[ \boxed{ P(\text{divisible by }4\mid\text{even}) = \frac5{10} = \frac12 }. \]Common Error: Using the Counting Formula Without Equal Likelihood
The formula
\[ P(A)=\frac{|A|}{|S|} \]requires equally likely elementary outcomes.
Counting labels alone is not enough.
Common Error: Using a Combination When Order Matters
If outcomes distinguish first, second, and third positions, then combinations discard information that the experiment retains.
Use permutations or stage-by-stage multiplication instead.
Common Error: Using a Permutation When Order Does Not Matter
If the outcome is a committee, hand, or subset, different orders of the same selected objects do not represent different outcomes.
Using permutations then counts every unordered outcome multiple times.
Common Error: Mixing Counting Conventions
If the denominator counts ordered outcomes, the numerator must also count ordered outcomes.
If the denominator counts unordered outcomes, the numerator must use the same convention.
Consistency is essential.
Common Error: Counting Overlapping Cases as Disjoint
If two favorable categories overlap, adding their counts directly counts shared outcomes twice.
Use inclusion–exclusion:
\[ |A\cup B| = |A|+|B|-|A\cap B|. \]Common Error: Forgetting the Complement Strategy
Some “at least one” events have many direct cases but a simple complement.
Before launching into a long case count, ask whether
\[ |A|=|S|-|A^c| \]is easier.
Common Error: Forgetting Impossible Cases
Restrictions can make some nominal arrangements impossible.
Without replacement, for example, the same physical object cannot appear twice in one selection.
The sample-space count must reflect the rules of the experiment.
Common Error: Double Counting the Same Outcome
A counting construction should generate each elementary outcome exactly once.
If the same outcome can be produced through several construction paths, the count must be corrected or redesigned.
A Reliable Procedure for Counting-Based Probability
- State the experiment and event precisely.
- Choose elementary outcomes that are equally likely.
- Decide whether order matters.
- Decide whether repetition or replacement is allowed.
- Count the full sample space.
- Count the favorable outcomes using the simplest valid method.
- Use cases, complements, or inclusion–exclusion when appropriate.
- Keep numerator and denominator counting conventions consistent.
- Compute \(P(A)=|A|/|S|\).
- Check that the result lies in \([0,1]\).
Example: Full Workflow
A committee of \(5\) is selected uniformly from \(9\) managers and \(6\) analysts.
Find the probability that the committee contains exactly \(2\) managers.
Order does not matter, so combinations are appropriate.
The total number of committees is
\[ |S|=\binom{15}{5}. \]To obtain exactly \(2\) managers, choose \(2\) managers and \(3\) analysts:
\[ |A|=\binom92\binom63. \]Therefore,
\[ \boxed{ P(A) = \frac{\binom92\binom63}{\binom{15}{5}} }. \]Practice 1: Uniform Integer Selection
One integer is selected uniformly from
\[ \{1,2,\ldots,40\}. \]Find the probability that it is divisible by \(8\).
Practice 2: Code Restriction
A three-digit code uses digits \(0\) through \(9\), repetition is allowed, and leading zeroes are permitted.
Find the probability that all three digits are odd.
Practice 3: Random Ordering
Eight distinct people are placed in a uniformly random order.
Find the probability that a particular person is first.
Practice 4: Committee Membership
A committee of \(3\) is chosen uniformly from \(10\) people.
Find the probability that a particular person is included.
Practice 5: Exactly One Ace
A five-card hand is selected uniformly from a standard deck.
Write a counting expression for the probability that the hand contains exactly one ace.
Practice 6: At Least One Ace
A five-card hand is selected uniformly from a standard deck.
Write a complement-based expression for the probability that it contains at least one ace.
Practice 7: Mixed Committee
A committee of \(4\) is selected uniformly from \(6\) engineers and \(5\) designers.
Find the probability that the committee contains exactly \(2\) engineers and \(2\) designers.
Practice 8: No Repeated Digits
A four-digit code is chosen uniformly from all \(10^4\) decimal strings.
Find the probability that all digits are distinct.
Practice 9: Divisible by Two or Five
One integer is selected uniformly from
\[ \{1,2,\ldots,100\}. \]Find the probability that it is divisible by \(2\) or by \(5\).
Practice 10: Exactly Three Heads
A fair coin is tossed \(6\) times.
Find the probability of exactly \(3\) heads.
Practice 11: Adjacent People
Seven distinct people are arranged uniformly in a line.
Find the probability that two particular people are adjacent.
Practice 12: Identify the Error
A student computes the probability of drawing two aces from a deck without replacement by writing
\[ \frac{\binom42}{52\cdot51}. \]Explain the counting inconsistency and give one correct counting ratio.
Practice Answers
Practice 1
The multiples of \(8\) are
\[ 8,16,24,32,40. \]Therefore,
\[ \boxed{ P=\frac5{40}=\frac18 }. \]Practice 2
There are \(5\) odd digits for each position:
\[ 1,3,5,7,9. \]Thus,
\[ \boxed{ P=\frac{5^3}{10^3}=\frac18 }. \]Practice 3
\[ P = \frac{7!}{8!} = \boxed{\frac18}. \]Practice 4
The total number of committees is
\[ \binom{10}{3}. \]If the particular person is included, choose the remaining \(2\) members from \(9\):
\[ \binom92. \]Therefore,
\[ \boxed{ P = \frac{\binom92}{\binom{10}{3}} = \frac3{10} }. \]Practice 5
\[ \boxed{ P(\text{exactly one ace}) = \frac{\binom41\binom{48}{4}}{\binom{52}{5}} }. \]Practice 6
\[ \boxed{ P(\text{at least one ace}) = 1-\frac{\binom{48}{5}}{\binom{52}{5}} }. \]Practice 7
\[ \boxed{ P = \frac{\binom62\binom52}{\binom{11}{4}} }. \]Practice 8
\[ \boxed{ P = \frac{10\cdot9\cdot8\cdot7}{10^4} }. \]Practice 9
There are \(50\) multiples of \(2\), \(20\) multiples of \(5\), and \(10\) multiples of \(10\).
By inclusion–exclusion,
\[ |A\cup B| = 50+20-10 = 60. \]Therefore,
\[ \boxed{ P=\frac{60}{100}=\frac35 }. \]Practice 10
There are \(2^6=64\) equally likely toss sequences and
\[ \binom63=20 \]sequences with exactly three heads.
Therefore,
\[ \boxed{ P=\frac{\binom63}{2^6}=\frac5{16} }. \]Practice 11
Treat the two particular people as one block. The block and the remaining five people form \(6\) units:
\[ |A|=2\cdot6!. \]Since
\[ |S|=7!, \]we obtain
\[ \boxed{ P=\frac{2\cdot6!}{7!}=\frac27 }. \]Practice 12
The numerator counts unordered pairs while the denominator counts ordered pairs, so the elementary outcomes are inconsistent.
One correct unordered ratio is
\[ \boxed{ \frac{\binom42}{\binom{52}{2}} }. \]An equivalent ordered ratio is
\[ \boxed{ \frac{4\cdot3}{52\cdot51} }. \]What to Remember
For a finite sample space of equally likely elementary outcomes,
\[ \boxed{ P(A)=\frac{|A|}{|S|} }. \]Useful counting tools include:
- the multiplication principle for multistage outcomes;
- permutations when order matters;
- combinations when order does not matter;
- counting by cases for disjoint alternatives;
- complementary counting for difficult “at least one” events;
- inclusion–exclusion for overlapping conditions;
- restriction-based counting for constrained selections.
Always keep the numerator and denominator based on the same elementary outcome convention.
The most important habit is: build the correct equally likely sample space before choosing a counting formula.
Where We Go Next
Counting-based probability works naturally when outcomes can be represented as finite equally likely cases. Other probability models are built from continuous regions rather than discrete outcome counts.
The next lesson, Geometric Probability, replaces outcome counts with lengths, areas, volumes, and other geometric measures to model uniformly distributed continuous outcomes.