Analytical Statistics (T), NYU | Chapter 1, Sections 1.2–1.2.6 | Source: Hogg & Tanis, Probability and Statistical Inference, 9th ed.
Difficulty: Introductory to Intermediate
Prerequisites: Probability Basics and Set Theory study notes (Ch. 1, Sections 1.1–1.1.3). You need to be comfortable with sample spaces, events and the probability axioms before tackling counting methods.
When every outcome in a sample space is equally likely, computing a probability reduces to counting: favourable outcomes divided by total outcomes. This section equips you with the counting tools to do that. The multiplication principle, permutations and combinations appear throughout the rest of the course and are essential for working with discrete distributions (binomial, hypergeometric) later on. The hardest part of most counting problems is recognising which method applies, and the only reliable way to get better at that is to work through many problems.
When outcomes are equally likely, P(A) = N(A)/N(S). To count N(A) and N(S) you need three tools: the multiplication principle for sequential experiments, permutations (nPr) for ordered selections without replacement, and combinations (nCr, "n choose r") for unordered selections without replacement. The section ends by combining these to solve "how many of type 1 and type 2" problems (the pattern behind the hypergeometric distribution).
Equally likely outcomes
A sample space where every individual outcome has the same probability. When this holds, P(A) = N(A)/N(S). In simple terms, every result is just as likely as every other.
Multiplication principle
If experiment E₁ has n₁ outcomes and, for each of those, experiment E₂ has n₂ outcomes, the composite experiment E₁E₂ has n₁ x n₂ outcomes. Think of it as multiplying the number of choices at each stage.
Factorial, n!
n! = n x (n-1) x (n-2) x ... x 2 x 1. It counts the number of ways to arrange n distinct objects in a line. 0! is defined as 1.
Permutation (r-permutation), nPr
nPr = n! / (n-r)! . The number of ordered sequences of r objects chosen from n distinct objects without replacement. Also called a "sequence without repetition."
Combination (r-combination), nCr or "n choose r"
nCr = n! / [r!(n-r)!] . The number of unordered selections of r objects from n distinct objects without replacement. Think of it as "how many groups of r can I pick, ignoring order?"
With replacement
After each draw, the object is returned to the pool before the next draw. The pool stays the same size throughout.
Without replacement
Once drawn, an object is removed from the pool. The pool shrinks by one after each draw.
Ordered sample
A selection where the sequence matters: drawing (1, 5, 7) is different from drawing (7, 1, 5).
Unordered sample
A selection where only the identities of the chosen objects matter, not the sequence: {1, 5, 7} is the same regardless of the order drawn.
When every outcome in S has the same probability, P(A) = N(A) / N(S), where N(A) is the number of outcomes favourable to A.
The entire challenge reduces to counting N(A) and N(S) correctly.
Example: rolling a die, A = {2, 3}. N(A) = 2, N(S) = 6, so P(A) = 2/6.
If experiment E₁ has n₁ possible outcomes and experiment E₂ has n₂ possible outcomes for each outcome of E₁, then doing E₁ followed by E₂ gives n₁ x n₂ total outcomes.
Extends to any number of stages: E₁, E₂, ..., Eₖ gives n₁ x n₂ x ... x nₖ outcomes.
Example: roll a die then flip a coin. Total outcomes = 6 x 2 = 12.
Drawing all n objects from a pool of n, caring about order, gives n! arrangements.
Example: 10 numbered balls drawn one at a time without replacement. Total ordered arrangements = 10! = 3,628,800.
Probability of one specific ordering (e.g. 10, 9, 8, ..., 1) = 1/n! = 1/10!.
Drawing n times from a pool of n with replacement gives nⁿ ordered outcomes.
Example: 10 balls with replacement, drawing 10 times. Total = 10¹⁰.
Choose r objects from n, order matters, no replacement.
The first draw has n choices, the second has n-1, ..., the r-th has n-r+1.
nPr = n! / (n-r)!
Example: draw 3 balls from 10 without replacement. ₁₀P₃ = 10! / 7! = 10 x 9 x 8 = 720.
Probability of one specific sequence (e.g. 1, 2, 3 in that order) = 1 / nPr.
Choose r objects from n, order matters, with replacement. Each draw has n choices independently.
Total outcomes = nʳ.
Example: draw 3 balls from 10 with replacement. Total = 10³ = 1,000.
Choose r objects from n, order does not matter, no replacement.
Every unordered group of r objects corresponds to r! ordered arrangements, so the number of unordered groups is nPr / r! = nCr.
nCr = n! / [r!(n-r)!]
Example: draw 3 balls from 10, order does not matter. ₁₀C₃ = 10! / (3! x 7!) = 120.
Probability that you draw the set {1, 2, 3} (unordered) = 1 / nCr = 1/120.
This is r! times larger than the probability of the specific sequence (1, 2, 3), because there are r! ways to arrange those same three numbers.
With replacement | Without replacement | |
|---|---|---|
Ordered | nʳ | nPr = n!/(n-r)! |
Unordered | (not covered in this course) | nCr = n!/[r!(n-r)!] |
When a sequence of n items contains exactly r of type 1 and n-r of type 2, the number of distinct orderings is nCr.
Example: Madrid and Barcelona play 10 matches. Madrid wins exactly 3. The number of distinct win/loss sequences = ₁₀C₃ = 120.
An urn holds n balls: s of type 1, n-s of type 2. You draw r without replacement. The probability of getting exactly k of type 1 and r-k of type 2:
P(A) = [sCk x (n-s)C(r-k)] / nCr
The numerator counts the ways to pick k from the s type-1 balls multiplied by the ways to pick the remaining r-k from the n-s type-2 balls. The denominator counts all possible unordered draws of r from n.
Example: 10 balls, 4 red, 6 black. Draw 5. Probability of exactly 3 red and 2 black = [₄C₃ x ₆C₂] / ₁₀C₅ = (4 x 15) / 252 = 60/252 ≈ 0.238.
Equally likely probability:
P(A) = N(A) / N(S)
Multiplication principle:
n₁ x n₂ x ... x nₖ total outcomes for k sequential experiments
Factorial:
n! = n x (n-1) x ... x 2 x 1, with 0! = 1
Permutations (ordered, without replacement):
nPr = n! / (n-r)!
Combinations (unordered, without replacement):
nCr = n! / [r!(n-r)!]
Ordered with replacement:
nʳ
Hypergeometric counting (k of type 1, r-k of type 2):
P = [sCk x (n-s)C(r-k)] / nCr
Combinations underpin lottery odds, poker-hand probabilities, and quality-control sampling plans. The hypergeometric counting pattern (how many defective items in a sample from a mixed lot) is used routinely in manufacturing acceptance testing. The birthday problem, a classic application of ordered sampling with replacement, shows that in a group of just 23 people there is roughly a 50% chance two share a birthday.
Students frequently confuse nPr and nCr. If the problem says "how many ways to choose" or "how many groups," order does not matter and you want nCr. If it says "how many sequences" or "how many arrangements," order matters and you want nPr.
Forgetting whether sampling is with or without replacement changes the answer dramatically. Always check the problem statement.
A common error in the hypergeometric setup is to forget to count the ways of choosing the type-2 objects. The numerator is the product of the two binomial coefficients, not just one of them.
Students sometimes try to use nCr when the objects are not all distinct, or when sampling is with replacement. The standard nCr formula assumes distinct objects and no replacement.
⚠️ Know how to decide which counting method applies: ordered vs unordered, with vs without replacement. This decision is the crux of most exam questions.
⚠️ The hypergeometric counting formula (sCk x (n-s)C(r-k) / nCr) is a favourite exam problem. Practise setting up the numerator carefully.
⚠️ Be comfortable computing nCr and nPr by hand for small numbers, including edge cases like nC0 = 1 and nCn = 1.
⚠️ The birthday problem is a standard exam application. Know how to set it up: the complement is that all birthdays are different (ordered, without replacement from 365 days), divided by the total (ordered, with replacement).
True or false: 0! = 1.
Fill in the blank: the number of ways to arrange 5 objects in a line is ___.
True or false: nCr = nPr / r!.
Fill in the blank: drawing 4 items from 10 with replacement, ordered, gives ___ outcomes.
True or false: ₅C₂ = ₅C₃.
Q: An urn contains 10 balls numbered 1 to 10. You draw 3 without replacement. How many ordered sequences are possible?
A: ₁₀P₃ = 10 x 9 x 8 = 720.
Q: Same urn, same draw. How many unordered selections are possible?
A: ₁₀C₃ = 10! / (3! x 7!) = 120.
Q: You roll a die 5 times. What is the probability of getting the exact sequence 1, 2, 3, 4, 5?
A: Each roll has 6 outcomes, so N(S) = 6⁵ = 7,776. There is exactly 1 favourable outcome. P = 1/7,776.
Q: An urn has 4 red and 6 black balls. You draw 5 without replacement. What is the probability that exactly 3 are red?
A: P = [₄C₃ x ₆C₂] / ₁₀C₅ = (4 x 15) / 252 = 60/252 ≈ 0.238.
Q: In a class of r students, what is the probability that at least two share a birthday? Set up the calculation (assume 365 days, equally likely).
A: It is easier to compute the complement (all birthdays different). P(all different) = ₃₆₅Pr / 365ʳ. Then P(at least one match) = 1 - ₃₆₅Pr / 365ʳ. For r = 23, this is approximately 0.507.
The hypergeometric counting pattern introduced here is the basis of the hypergeometric distribution covered later in the course. Combinations also appear directly in the binomial distribution (choosing which k out of n trials are successes). The multiplication principle is used again when computing joint probabilities for independent events. If counting methods feel shaky, they will trip you up repeatedly in later chapters, so it is worth drilling them now.
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