MGU Mathematics Series — Probability and Statistics
Part I: Probability Foundations and Random Variables
PMFs · CDFs · Expected Value · Variance · Named Distributions
Chapter 3 of 14
MGU Mathematics Series  ·  Probability and Statistics  ·  Library Textbook Edition

Chapter 3:
Discrete Random
Variables

Random variables as functions on sample spaces, probability mass functions, cumulative distribution functions, expected value, variance, and the six fundamental discrete distributions

■ Part I: Probability Foundations ■

■ 14 Sections  •  6 Named Distributions ■

Follows Chapter 2: Probability Rules and Conditional Probability  ·  Prepares students for Chapter 4: Continuous Random Variables

Chapter 3 — Front Matter

Purpose and Overview

"The purpose of a random variable is to translate uncertain outcomes into numbers we can work with mathematically."

In Chapters 1 and 2, probability was built from sample spaces, events, and rules. We assigned probabilities to outcomes and learned how to combine them using complements, unions, intersections, conditional probability, and Bayes' theorem. That framework is powerful, but it works directly with outcomes — and outcomes can be almost anything: heads or tails, a specific card drawn, a disease being present or absent.

Mathematics becomes most useful when it works with numbers. The language of calculus, algebra, and analysis is a language of functions, sums, integrals, and operations on numerical quantities. A random variable is the bridge that translates uncertain outcomes into numbers, making the full machinery of mathematics available for analyzing uncertainty.

This chapter introduces random variables, beginning with the discrete case — situations where the possible numerical values can be listed, at least in principle. We develop the probability mass function as the fundamental tool for describing a discrete random variable, the cumulative distribution function for accumulating probability, and the summary measures of expected value, variance, and standard deviation. We study the most important families of discrete distributions: Bernoulli, binomial, geometric, negative binomial, hypergeometric, and Poisson.

By the end of this chapter, you should be able to turn a word problem about uncertainty into a precise mathematical model, compute probabilities and summary measures from that model, and interpret what those numbers actually say about the real situation.

Opening Question

A quality inspector checks items on a production line. Each item is independently defective with probability 0.10. If she inspects 15 items, how likely is it that she finds at most one defective? How many defectives should she expect on average? What if she instead keeps inspecting until she finds the first defective — how long should she expect to wait? By the end of this chapter, you will have precise mathematical tools to answer all of these questions, and recognize the named distributions that model each situation.

14Sections
6Named Distributions
20+Key Terms
15+Worked Examples

Learning Objectives

  1. 1Define a random variable as a function from a sample space to the real numbers and explain what it means for a random variable to be discrete.
  2. 2Construct and verify a probability mass function; compute event probabilities by summing over the support.
  3. 3Define and compute the cumulative distribution function; recover the PMF from the CDF.
  4. 4Compute the expected value of a discrete random variable and of a function of a random variable using LOTUS.
  5. 5Compute variance and standard deviation; apply variance formulas under linear transformations.
  6. 6Standardize a random variable and interpret standardized values.
  7. 7Identify, apply, and compute probabilities for Bernoulli and binomial random variables.
  8. 8Identify, apply, and compute probabilities for geometric and negative binomial random variables; state the memoryless property.
  9. 9Identify, apply, and compute probabilities for hypergeometric random variables; explain when it differs from the binomial.
  10. 10Identify, apply, and compute probabilities for Poisson random variables; explain the Poisson as a limit of the binomial.
  11. 11Recognize and avoid the ten most common discrete random variable mistakes.
Chapter 3 — Contents

Sections at a Glance

Sections 3.1 – 3.4
Random Variables, PMFs, and CDFs
Definition · Discrete vs. Continuous · Probability Mass Functions · Cumulative Distribution Functions
Section 3.1

What a Random Variable Is

The Need for Numerical Outcomes

Suppose you roll two fair six-sided dice and want to analyze the sum. The sample space contains 36 equally likely outcomes: (1,1), (1,2), (1,3), and so on through (6,6). Working directly with pairs of numbers is awkward. What you actually care about — the sum — is a single number between 2 and 12. Similarly, if you monitor a hospital emergency room and want to analyze how many patients arrive in a given hour, each possible scenario is a complex, unlisted outcome, but the count of patients is a number you can work with directly. A random variable formalizes this translation from outcomes to numbers.

Definition — Random Variable

Let S be a sample space for a probability experiment. A random variable X is a function X : S → ℝ that assigns a real number X(s) to each outcome s ∈ S.

Random variables are conventionally denoted by uppercase letters near the end of the alphabet: X, Y, Z. The realized value after an experiment is denoted in lowercase: x, y, z.

This definition deserves patient reading. A random variable is a function, not a number and not an outcome. It takes outcomes as inputs and produces real numbers as outputs. The randomness comes from the fact that the experiment's outcome is uncertain; once the outcome is determined, the random variable's value is determined too.

Example 3.1.1
Sum of two dice
  • SetupRoll two fair dice. Let X be the sum of the two faces.
  • FunctionX maps each ordered pair to its sum: X((1,3)) = 4, X((5,6)) = 11, and so forth.
  • ValuesThe possible values of X are {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.
Example 3.1.2
Number of heads in three flips
  • SetupFlip a fair coin three times. Let Y be the number of heads.
  • FunctionY(HHT) = 2, Y(TTT) = 0. Each outcome is a sequence; Y counts the H's.
  • ValuesPossible values: {0, 1, 2, 3}.
Example 3.1.3
Indicator of a defective item
  • SetupA quality inspector selects one item from a production line. Let Z = 1 if defective, Z = 0 if not.
  • ValuesZ only takes two values: {0, 1}. Simple as it looks, this is one of the most important random variables in all of probability — the Bernoulli random variable.
Common Confusion

A random variable is not the same as the probability model on S. The probability model says how likely each outcome is. The random variable says how to convert outcomes to numbers. The two work together: once X is defined, the probability model on S induces a probability model on the values of X. Also, X is not a variable in the algebraic sense. Before the experiment, X is uncertain; its value depends on which outcome occurs.

Quick Check 3.1
  1. A drawer contains 4 red socks and 6 blue socks. Two socks are drawn without replacement.
  2. (a) Describe the sample space.   (b) Define a random variable X that counts the number of red socks drawn.   (c) List all possible values of X.
Section 3.2

Discrete Random Variables

Two Fundamental Types

A random variable X is called discrete if its set of possible values is countable — meaning the values can be listed in a sequence, possibly finite or possibly infinite but still enumerable.

A random variable X is called continuous if its set of possible values forms an interval (or union of intervals) in ℝ, so that no listing is possible.

A set is countable if its elements can be placed in one-to-one correspondence with a subset of the positive integers. The set {0, 1, 2, 3} is obviously countable. So is {1, 2, 3, ...}, the set of all positive integers — even though it is infinite, you can list it. Real intervals such as [0, 1] are not countable, because there are uncountably many real numbers in any interval.

For discrete random variables, the possible values are almost always non-negative integers or integers in practice. This chapter focuses entirely on the discrete case. Chapter 4 develops the continuous case, where calculus becomes essential.

Random VariableTypeReason
Number of defectives in a batch of 20DiscreteValues in {0, 1, 2, ..., 20} — finite list
Number of coin flips until first headDiscreteValues in {1, 2, 3, ...} — countably infinite list
Exact time for a radioactive atom to decayContinuousValues in (0, ∞) — an uncountable interval
Weight of a randomly selected appleContinuousAny positive real value — no listing possible
Number of emails in a spam folderDiscreteValues in {0, 1, 2, ...} — countable
Why the Distinction Matters

For a discrete random variable, we assign a specific probability to each possible value and sum probabilities. For a continuous random variable, individual values have probability zero, and probability is computed using integrals over intervals. The mathematical tools are different and the intuition differs too.

Quick Check 3.2
  1. Classify each as discrete or continuous: (a) number of text messages per day; (b) weight of an apple in grams; (c) machine breakdowns per month; (d) proportion of voters supporting a candidate; (e) number of emails in a spam folder.
Section 3.3

Probability Mass Functions

Definition — Probability Mass Function

The probability mass function (PMF) of a discrete random variable X is the function p : ℝ → [0,1] defined by

p(x) = P(X = x)

for each real number x. A function p is a valid PMF if and only if: (1) p(x) ≥ 0 for all x; (2) p(x) = 0 for all x not in the support; (3) Σ p(x) = 1 over all x in the support.

The term "mass function" is evocative: think of probability as mass sitting on the real line, concentrated at each possible value of X. A PMF tells you exactly how much mass sits at each point. The support of X is the set of all values x for which p(x) > 0.

Example 3.3.1
Uniform PMF — single fair die
  • SetupRoll one fair six-sided die. Let X be the face shown.
  • PMFp(x) = 1/6 for x ∈ {1, 2, 3, 4, 5, 6}, p(x) = 0 otherwise
  • VerifySum: 6 · (1/6) = 1. ✓ All values non-negative. ✓ This is a uniform distribution on six values.
Example 3.3.2
PMF from a two-dice sum
  • SetupRoll two fair dice. Let X be the sum. Count the outcomes producing each sum among the 36 equally likely pairs.
  • PMF Table k: 2 3 4 5 6 7 8 9 10 11 12 Count: 1 2 3 4 5 6 5 4 3 2 1 p(k): 1 2 3 4 5 6 5 4 3 2 1 ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ 36 36 36 36 36 36 36 36 36 36 36
  • VerifyTotal: (1+2+3+4+5+6+5+4+3+2+1)/36 = 36/36 = 1. ✓
Example 3.3.3
Computing event probabilities from the PMF
  • RuleP(X ∈ A) = Σ p(x) for all x in A. Sum the PMF over the relevant values.
  • P(X ≤ 4)P(X ≤ 4) = p(2) + p(3) + p(4) = 1/36 + 2/36 + 3/36 = 6/36 = 1/6
  • P(X ≥ 10)P(X ≥ 10) = p(10) + p(11) + p(12) = 3/36 + 2/36 + 1/36 = 6/36 = 1/6
  • NoteThe symmetry is not a coincidence. The two-dice sum distribution is symmetric around 7.
Quick Check 3.3
  1. (a) A random variable X has p(x) = cx for x ∈ {1, 2, 3, 4} and p(x) = 0 otherwise. Find c.
  2. (b) With the c you found, compute P(X ≥ 3).
Section 3.4

Cumulative Distribution Functions

Definition — Cumulative Distribution Function

The cumulative distribution function (CDF) of a random variable X is the function F : ℝ → [0,1] defined by

F(x) = P(X ≤ x)

for all real x. For a discrete random variable: F(x) = Σ p(t) for all t ≤ x. Every CDF is non-decreasing, right-continuous, with lim F(x) = 0 as x → −∞ and lim F(x) = 1 as x → +∞.

For a discrete random variable, the CDF is a step function: it stays flat between support values and jumps at each support value by the amount p(x). The CDF is defined for all real numbers, not just the support.

Example 3.4.1
Computing the CDF from a PMF
  • PMFLet X have p(1) = 0.1, p(2) = 0.4, p(3) = 0.3, p(4) = 0.2.
  • CDF F(x) = 0 for x < 1 F(x) = 0.1 for 1 ≤ x < 2 F(x) = 0.5 for 2 ≤ x < 3 F(x) = 0.8 for 3 ≤ x < 4 F(x) = 1.0 for x ≥ 4
  • CheckF(2) = P(X ≤ 2) = p(1) + p(2) = 0.1 + 0.4 = 0.5. ✓ Each jump equals the PMF value at that point.
Computing Probabilities Using the CDF

P(X ≤ b) = F(b)

P(X > b) = 1 − F(b)

P(a < X ≤ b) = F(b) − F(a)

P(a ≤ X ≤ b) = F(b) − F(a−)   where F(a−) denotes the left-hand limit at a.

Example 3.4.2
CDF calculations — two-dice sum
  • F(7)F(7) = Σ p(k) for k = 2 to 7 = (1+2+3+4+5+6)/36 = 21/36 = 7/12
  • F(9)F(9) = Σ p(k) for k = 2 to 9 = (1+2+3+4+5+6+5+4)/36 = 30/36 = 5/6
  • IntervalP(7 < X ≤ 9) = F(9) − F(7) = 5/6 − 7/12 = 10/12 − 7/12 = 3/12 = 1/4
Recovering the PMF from the CDF

For a discrete random variable, p(x) = F(x) − F(x−), the size of the jump in F at the point x. At any value between support points, the CDF is flat and the PMF is zero.

Quick Check 3.4
  1. A random variable W takes values {0, 1, 2, 3} with probabilities 0.2, 0.3, 0.4, 0.1.
  2. (a) Write out F(w) as a step function.   (b) Compute P(W > 1).   (c) Compute P(1 ≤ W ≤ 2).
Sections 3.5 – 3.7
Expected Value, Variance, and Transformations
Expectation · LOTUS · Linearity · Variance · Standard Deviation · Standardization
Section 3.5

Expected Value

Definition — Expected Value

The expected value (or mean) of a discrete random variable X is

E[X] = μ = Σ x · p(x)

where the sum is over all values in the support of X. Expected value is a weighted average of all possible values, weighted by their probabilities.

Expected value goes by many names: expectation, mean, mathematical expectation, or the first moment. In statistics it is often called the population mean. All these terms refer to the same mathematical object. The expected value is a long-run average over many repetitions of the experiment — it is a property of the distribution, not a guaranteed value for a single trial.

Example 3.5.1
Expected value of one fair die
  • SetupLet X be the outcome of rolling one fair die: p(x) = 1/6 for x ∈ {1,...,6}.
  • ComputeE[X] = 1·(1/6) + 2·(1/6) + 3·(1/6) + 4·(1/6) + 5·(1/6) + 6·(1/6) = 21/6 = 3.5
  • InterpretNote that 3.5 is not a possible outcome of a die roll. Expected value is a long-run average, not a guaranteed value.
E[X] = 3.5
Example 3.5.2
A game of chance
  • SetupYou pay $2 to play. You flip a fair coin: heads wins $5, tails wins nothing. Let X be your net winnings.
  • ValuesHeads: X = 5 − 2 = 3. Tails: X = 0 − 2 = −2. Each with probability 1/2.
  • ComputeE[X] = 3·(1/2) + (−2)·(1/2) = (3−2)/2 = 0.5
  • InterpretOn average, you gain $0.50 per game. Over many plays, you expect to come out ahead.
E[X] = $0.50 per play
LOTUS — Law of the Unconscious Statistician

If g is a function and X is a discrete random variable, then

E[g(X)] = Σ g(x) · p(x)

You do not need to find the distribution of g(X) first; sum g(x) · p(x) over the support of X.

Linearity of Expectation

For constants a and b and any random variable X with finite expected value:

E[aX + b] = aE[X] + b

A deeper result (proved in Chapter 5): for any two random variables X and Y (not necessarily independent), E[X + Y] = E[X] + E[Y].

Proof of Linearity
E[aX + b] = Σ (ax + b)p(x) = a Σ x·p(x) + b Σ p(x) = a·E[X] + b·1 = aE[X] + b. □
Example 3.5.3
Expected value of X² using LOTUS
  • SetupLet X be the roll of one die: p(x) = 1/6 for x ∈ {1,...,6}. Find E[X²].
  • ComputeE[X²] = 1²·(1/6) + 2²·(1/6) + 3²·(1/6) + 4²·(1/6) + 5²·(1/6) + 6²·(1/6) = (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6 ≈ 15.17
E[X²] = 91/6 ≈ 15.17
Quick Check 3.5
  1. (a) A random variable X has p(1) = 0.3, p(2) = 0.5, p(3) = 0.2. Compute E[X].
  2. (b) Compute E[2X + 1].
  3. (c) A game pays $10 if you roll a 6 on a fair die, and $0 otherwise. If the game costs $2 to play, what is your expected net gain?
Section 3.6

Variance and Standard Deviation

Definition — Variance and Standard Deviation

The variance of a discrete random variable X is

Var(X) = σ² = E[(X − μ)²] = Σ (x − μ)² · p(x)

The standard deviation is σ = √Var(X), in the same units as X.

The computing formula (almost always easier) is: Var(X) = E[X²] − (E[X])²

Derivation of the Computing Formula
E[(X−μ)²] = E[X² − 2μX + μ²] = E[X²] − 2μ·E[X] + μ² = E[X²] − 2μ² + μ² = E[X²] − μ². □
Example 3.6.1
Variance of one fair die
  • KnownE[X] = 3.5 and E[X²] = 91/6.
  • VarianceVar(X) = 91/6 − (3.5)² = 91/6 − 49/4 = 182/12 − 147/12 = 35/12 ≈ 2.917
  • SDσ = √(35/12) ≈ 1.708
Var(X) = 35/12 ≈ 2.917; σ ≈ 1.708
Variance Under Linear Transformations

For constants a and b:

Var(aX + b) = a² · Var(X)

Shifting by a constant b does not change spread. Rescaling by a multiplies variance by a² and standard deviation by |a|.

Quick Check 3.6
  1. (a) Let X have E[X] = 4 and E[X²] = 20. Find Var(X) and σ.
  2. (b) For the same X, find Var(3X − 2) and SD(3X − 2).
Section 3.7

Linear Transformations of Random Variables

Summary — Linear Transformations

If Y = aX + b for constants a and b, then:

E[Y] = aE[X] + b

Var(Y) = a² Var(X)

SD(Y) = |a| · SD(X)

Standardization

If μ = E[X] and σ = SD(X) > 0, the standardized form is:

Z = (X − μ) / σ

Then E[Z] = 0 and Var(Z) = 1, regardless of the original units or scale of X. Standardization is used throughout statistics for comparing quantities on different scales and for computing probabilities from standard reference distributions.

Example 3.7.1
Unit conversion
  • SetupBolt length X (in mm): E[X] = 50, Var(X) = 4. Measured in cm: Y = X/10.
  • MeanE[Y] = 50/10 = 5 cm
  • VarianceVar(Y) = (1/10)² · 4 = 0.04 cm²
  • SDSD(Y) = 0.2 cm
Example 3.7.2
Standardizing a quiz score
  • SetupA score X has E[X] = 70 and SD(X) = 10. Standardize a score of 85.
  • ComputeZ = (85 − 70)/10 = 15/10 = 1.5
  • InterpretA score of 85 is 1.5 standard deviations above the mean.
Z = 1.5
Quick Check 3.7
  1. A random variable X has μ = 3 and σ = 2. Define Y = −X + 10.
  2. (a) Find E[Y] and SD(Y).   (b) Write the standardized form Z = (X − μ)/σ and verify E[Z] = 0 and Var(Z) = 1.
Sections 3.8 – 3.13
The Six Fundamental Discrete Distributions
Bernoulli · Binomial · Geometric · Negative Binomial · Hypergeometric · Poisson
Section 3.8

Bernoulli Random Variables

Definition — Bernoulli Distribution

A random variable X has a Bernoulli distribution with parameter p ∈ (0,1), written X ~ Bernoulli(p), if

P(X = 1) = p and P(X = 0) = 1 − p = q

The value 1 is called "success" and 0 is called "failure." Compact PMF: p(x) = pˣ(1−p)^(1−x) for x ∈ {0, 1}.

Mean and variance: μ = p    σ² = p(1−p) = pq

The Bernoulli distribution is the building block for several other distributions. A sequence of independent Bernoulli trials — each with the same success probability p — forms the Bernoulli process, and counting successes in that process leads to the binomial, geometric, and negative binomial distributions.

Notice that variance is maximized when p = 0.5 (maximum uncertainty) and equals zero when p = 0 or p = 1 (no uncertainty).

Example 3.8.1
Defective item indicator
  • SetupA machine produces a defective item with probability 0.03. Let X = 1 if the next item is defective, X = 0 otherwise.
  • ModelX ~ Bernoulli(0.03)
  • SummaryE[X] = 0.03 Var(X) = (0.03)(0.97) = 0.0291
Section 3.9

Binomial Random Variables

Definition — Binomial Distribution

A random variable X has a Binomial distribution with parameters n and p, written X ~ Binomial(n, p), if X counts the number of successes in n independent Bernoulli(p) trials. Support: {0, 1, 2, ..., n}.

PMF: P(X = k) = C(n,k) · pᵏ · (1−p)^(n−k)   for k = 0, 1, ..., n

Mean and variance: μ = np    σ² = np(1−p)

Any specific arrangement of k successes and n−k failures has probability p^k (1−p)^(n−k) by independence. The binomial coefficient C(n,k) = n!/[k!(n−k)!] counts the number of distinct arrangements. The PMF sums to 1 by the binomial theorem: Σ C(n,k) p^k (1−p)^(n−k) = (p + (1−p))^n = 1.

The mean and variance follow by writing X = X₁ + X₂ + ··· + Xₙ where each Xᵢ is an independent Bernoulli(p). Linearity gives E[X] = np; independence gives Var(X) = np(1−p).

Example 3.9.1
Quality inspection
  • SetupA 10% defect rate. Inspector checks 15 items. Let X be the number of defectives.
  • ModelX ~ Binomial(15, 0.1)
  • SummaryE[X] = 15(0.1) = 1.5 Var(X) = 15(0.1)(0.9) = 1.35 σ ≈ 1.16
  • P(X = 0)P(X = 0) = C(15,0)(0.1)⁰(0.9)¹⁵ = (0.9)¹⁵ ≈ 0.206
  • P(X ≤ 1)P(X = 1) = C(15,1)(0.1)¹(0.9)¹⁴ = 15(0.1)(0.9)¹⁴ ≈ 0.343 P(X ≤ 1) ≈ 0.206 + 0.343 = 0.549
  • InterpretAbout 54.9% of the time, the inspector finds at most one defective item.
P(X ≤ 1) ≈ 0.549
Binomial Assumptions

The binomial requires: (1) fixed number of trials n; (2) each trial is Bernoulli with outcomes success/failure; (3) constant probability p of success; (4) trials are independent. When drawing without replacement from a finite population, use the hypergeometric distribution (Section 3.12).

Quick Check 3.9
  1. A fair coin is flipped 8 times. Let X be the number of heads.
  2. (a) Identify the distribution and parameters.   (b) Find P(X = 4).   (c) Find E[X] and SD(X).
Section 3.10

Geometric Random Variables

Definition — Geometric Distribution

A random variable X has a Geometric distribution with parameter p ∈ (0,1), written X ~ Geometric(p), if X is the number of independent Bernoulli(p) trials required to obtain the first success. Support: {1, 2, 3, ...}.

PMF: P(X = k) = (1−p)^(k−1) · p   for k = 1, 2, 3, ...

Mean and variance: μ = 1/p    σ² = (1−p)/p²

The PMF sums to 1 via the geometric series: Σ (1−p)^(k−1) p = p · [1/(1−(1−p))] = 1. The mean 1/p is intuitive: if the probability of success is p, you expect to wait 1/p trials on average.

The Memoryless Property

If X ~ Geometric(p), then for any positive integers m and n:

P(X > m + n | X > m) = P(X > n)

Given that the first success has not occurred in the first m trials, the distribution of additional trials required is still Geometric(p). Past failures carry no information about the future. This is the defining property of the geometric distribution.

Proof of Memoryless Property
P(X > k) = (1−p)^k (the first k trials are all failures). Then:
P(X > m+n | X > m) = P(X > m+n) / P(X > m) = (1−p)^(m+n) / (1−p)^m = (1−p)^n = P(X > n). □
Example 3.10.1
First miss by a free throw shooter
  • SetupA player makes each throw with probability 0.75. Let X be the number of throws until the first miss.
  • ModelX ~ Geometric(0.25), since p = 0.25 is the probability of a miss (success in our counting).
  • MeanE[X] = 1/0.25 = 4 throws on average until first miss
  • P(X ≤ 3)P(X ≤ 3) = 1 − P(X > 3) = 1 − (0.75)³ = 1 − 0.4219 = 0.5781
E[X] = 4; P(first miss within 3 throws) ≈ 0.578
Two Parameterizations

Some textbooks define the geometric as the number of failures before the first success, giving support {0, 1, 2, ...} and mean (1−p)/p. Always confirm which definition is in use before computing.

Section 3.11

Negative Binomial Random Variables: An Introduction

Definition — Negative Binomial Distribution

A random variable X has a Negative Binomial distribution with parameters r ∈ {1, 2, 3, ...} and p ∈ (0,1), written X ~ NegBin(r, p), if X is the number of independent Bernoulli(p) trials required to obtain the r-th success. Support: {r, r+1, r+2, ...}.

PMF: P(X = k) = C(k−1, r−1) · pʳ · (1−p)^(k−r)   for k = r, r+1, ...

Mean and variance: μ = r/p    σ² = r(1−p)/p²

When r = 1, the negative binomial reduces to the geometric distribution. The PMF has a clear structure: for the r-th success to occur on trial k, exactly r−1 successes must occur in the first k−1 trials (counted by C(k−1, r−1)), and then a success occurs on trial k.

The binomial counts successes in a fixed number of trials. The negative binomial has a fixed number of successes and a random number of trials. These are complementary perspectives on the same Bernoulli process.

Example 3.11.1
Recruiting eligible participants
  • SetupEach candidate meets eligibility criteria with probability 0.3. Let X be the number of candidates interviewed until the 5th eligible participant is found.
  • ModelX ~ NegBin(5, 0.3)
  • SummaryE[X] = 5/0.3 ≈ 16.7 candidates interviewed Var(X) = 5(0.7)/(0.3)² ≈ 38.9
  • P(X = 10)P(X = 10) = C(9,4)(0.3)⁵(0.7)⁵ = 126 · 0.00243 · 0.16807 ≈ 0.0514
Section 3.12

Hypergeometric Random Variables

Definition — Hypergeometric Distribution

Setup: A population of N objects contains M successes (type 1) and N−M failures (type 2). We select n objects without replacement. Let X be the number of successes in the sample.

X ~ Hyper(N, M, n). PMF:

P(X = k) = [C(M,k) · C(N−M, n−k)] / C(N,n)

Mean and variance: μ = nM/N    σ² = n·(M/N)·((N−M)/N)·((N−n)/(N−1))

The mean has an intuitive form: M/N is the proportion of successes in the population, and we expect that proportion in our sample. The variance formula includes the factor (N−n)/(N−1), called the finite population correction factor. As N → ∞ with M/N → p, this factor approaches 1 and the hypergeometric variance approaches np(1−p), recovering the binomial.

Example 3.12.1
Committee selection
  • SetupA committee of 5 is selected from 10 people: 6 women and 4 men. Let X be the number of women selected.
  • ModelX ~ Hyper(10, 6, 5)
  • MeanE[X] = 5 · (6/10) = 3
  • P(X = 3)P(X = 3) = [C(6,3) · C(4,2)] / C(10,5) = (20 · 6) / 252 = 120/252 = 10/21 ≈ 0.476
P(X = 3) ≈ 0.476; E[X] = 3
Use Hypergeometric when...Use Binomial when...
Sampling without replacementSampling with replacement
Population is finite and sample is a significant fractionPopulation is large (sample < 5–10% of population)
Successive draws are dependentTrials are independent
Quick Check 3.12
  1. A deck of 52 cards contains 13 hearts. Five cards are dealt. Let X be the number of hearts.
  2. (a) Identify the distribution and parameters.   (b) Find E[X].   (c) Find P(X = 2).
Section 3.13

Poisson Random Variables

Definition — Poisson Distribution

A random variable X has a Poisson distribution with parameter λ > 0, written X ~ Poisson(λ), if

P(X = k) = e^(−λ) · λᵏ / k! for k = 0, 1, 2, 3, ...

λ is the rate — the expected number of events in the interval. A remarkable feature: E[X] = Var(X) = λ.

The PMF sums to 1 because Σ e^(−λ) λ^k / k! = e^(−λ) · e^λ = 1, using the Maclaurin series for e^λ. The equal mean and variance is a signature property that can be used to check whether a Poisson model is plausible for a given dataset.

Poisson as a Limit of Binomials

If X ~ Binomial(n, p) with p = λ/n, then as n → ∞ with λ held fixed, P(X = k) → e^(−λ)λᵏ/k!. The Poisson approximation is accurate when n ≥ 20 and p ≤ 0.05.

Example 3.13.1
Call center arrivals
  • SetupA call center receives an average of 8 calls per hour. Assume a Poisson process. Let X be the number of calls in a given hour.
  • ModelX ~ Poisson(8). E[X] = Var(X) = 8; σ = √8 ≈ 2.83.
  • P(X = 0)P(X = 0) = e⁻⁸ ≈ 0.000335 (very unlikely to have no calls)
  • P(X = 5)P(X = 5) = e⁻⁸ · 8⁵ / 5! = e⁻⁸ · 32768 / 120 ≈ 0.0916
Example 3.13.2
Poisson approximation to the binomial
  • Setup1000 bolts produced per hour; each independently defective with probability 0.002. Let X be the number of defectives.
  • ExactX ~ Binomial(1000, 0.002). But n is large and p is small, so approximate with X ≈ Poisson(λ = 1000 × 0.002 = 2).
  • P(X = 0)P(X = 0) ≈ e⁻² ≈ 0.135
  • P(X ≥ 3)P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − e⁻²(1 + 2 + 2) = 1 − 5e⁻² ≈ 0.323
P(X = 0) ≈ 0.135; P(X ≥ 3) ≈ 0.323
Example 3.13.3
Rescaling the Poisson rate
  • SetupOn average, 3 customers per minute enter a store (Poisson process). Find P(X = 10) in a 4-minute period.
  • Rateλ = 3 per minute; t = 4 minutes, so X ~ Poisson(12).
  • ComputeP(X = 10) = e⁻¹² · 12¹⁰ / 10! ≈ 0.1048
P(X = 10) ≈ 0.105
Rate Units Must Match Interval Length

A Poisson rate λ defined per hour cannot be applied directly to a 30-minute interval. Always adjust: if events arrive at rate λ per unit time and you want probabilities over t units, use Poisson(λt).

Quick Check 3.13
  1. On average, 2 typographical errors appear per page.
  2. (a) P(exactly 0 errors on a page)?   (b) P(3 or more errors on a page)?   (c) On a 10-page chapter, what distribution governs total errors, and what is its mean?
Section 3.14

Common Discrete Random Variable Mistakes

Mistake 1 — Treating Expected Value as a Guaranteed Value
"E[X] = 3.5 means X will usually equal 3.5."

Correction: E[X] is a long-run average over many repetitions. For a fair die, 3.5 is a value the die can never show. Expected value is a property of the distribution, not a prediction for a single trial.

Mistake 2 — Forgetting to Check PMF Requirements
Accepting any non-negative function as a valid PMF without verifying the sum equals 1.

Correction: A PMF is valid only if (1) all values are non-negative, and (2) the values sum to exactly 1. To find a constant c that makes a function a valid PMF, set the sum equal to 1 and solve.

Mistake 3 — Confusing p(k) with F(k)
Using CDF value F(k) when the problem asks for the PMF value p(k), or vice versa.

Correction: p(k) = P(X = k) is the mass at exactly k. F(k) = P(X ≤ k) is the accumulated mass up to and including k. For discrete distributions, F(k) = Σ p(j) for j ≤ k.

Mistake 4 — Using Binomial When Hypergeometric Applies
Applying the binomial formula when sampling without replacement from a small population.

Correction: If sampling is without replacement and the sample is a non-trivial fraction of the population, use the hypergeometric. The binomial is appropriate when sampling with replacement, or when the population is large enough that removing sampled items does not change the success probability meaningfully.

Mistake 5 — Applying the Geometric Formula Incorrectly
P(X = k) = p^(k−1)(1−p) instead of (1−p)^(k−1)p

Correction: Confirm which parameterization is in use. If X is the number of trials until the first success, P(X = k) = (1−p)^(k−1)p. If X is the number of failures before the first success, P(X = k) = (1−p)^k · p.

Mistake 6 — Variance Rescaling Error
Var(aX + b) = a · Var(X)

Correction: Var(aX + b) = a² · Var(X). Rescaling multiplies standard deviation by |a| and multiplies variance by a². Shifting by a constant does not change variance or standard deviation.

Mistake 7 — Assuming the Poisson Applies to Any Count
Automatically applying the Poisson distribution to any integer-valued random variable.

Correction: The Poisson is appropriate when events occur independently at a constant rate and are rare relative to the interval. Bounded counts (like items in a batch of fixed size) need different models.

Mistake 8 — Generalizing Memorylessness
Applying the memoryless property to binomial or hypergeometric distributions.

Correction: Memorylessness is specific to the geometric (and, for continuous variables, the exponential). It holds because of trial independence. Do not generalize this to other distributions.

Mistake 9 — Misreading Poisson Rate Units
Using λ defined per hour for a 30-minute interval without adjusting.

Correction: The Poisson rate must match the interval length. If events arrive at rate λ per unit time and you want probabilities over t units of time, use Poisson(λt).

Mistake 10 — Computing E[X²] as (E[X])²
E[X²] = (E[X])² → concluding Var(X) = 0

Correction: In general E[X²] ≠ (E[X])², and Var(X) = E[X²] − (E[X])² ≥ 0, with equality only when X is constant. Always compute E[X²] directly as Σ x² p(x).

Chapter 3 — Back Matter

Chapter Summary

Distribution Reference

Bernoulli
Bernoulli(p)
  • Support: {0, 1}
  • PMF: p(x) = p^x (1−p)^(1−x)
  • Mean: p
  • Variance: p(1−p)
  • Use when: single trial, two outcomes
Binomial
Binomial(n, p)
  • Support: {0, 1, ..., n}
  • PMF: C(n,k) pᵏ(1−p)^(n−k)
  • Mean: np
  • Variance: np(1−p)
  • Use when: count successes in n independent trials
Geometric
Geometric(p)
  • Support: {1, 2, 3, ...}
  • PMF: (1−p)^(k−1) p
  • Mean: 1/p
  • Variance: (1−p)/p²
  • Use when: trials until first success; memoryless
Negative Binomial
NegBin(r, p)
  • Support: {r, r+1, r+2, ...}
  • PMF: C(k−1,r−1) pʳ(1−p)^(k−r)
  • Mean: r/p
  • Variance: r(1−p)/p²
  • Use when: trials until r-th success
Hypergeometric
Hyper(N, M, n)
  • Support: {max(0,n−(N−M)),...,min(n,M)}
  • PMF: C(M,k)C(N−M,n−k)/C(N,n)
  • Mean: nM/N
  • Variance: includes (N−n)/(N−1)
  • Use when: sampling without replacement, finite population
Poisson
Poisson(λ)
  • Support: {0, 1, 2, ...}
  • PMF: e^(−λ) λᵏ / k!
  • Mean: λ
  • Variance: λ   (mean = variance)
  • Use when: rare, independent events; rate × time

Key Terms Review

random variable
A function X : S → ℝ assigning a real number to each outcome in a sample space.
discrete random variable
A random variable whose range is a countable set; values can be listed.
support
The set of values x for which p(x) > 0.
PMF
Probability mass function: p(x) = P(X = x). Non-negative, sums to 1 over the support.
CDF
Cumulative distribution function: F(x) = P(X ≤ x). Non-decreasing step function for discrete variables.
expected value
E[X] = Σ x·p(x). The long-run average; a weighted average of possible values.
LOTUS
Law of the unconscious statistician: E[g(X)] = Σ g(x)p(x). No need to find the distribution of g(X) first.
linearity of expectation
E[aX+b] = aE[X]+b and E[X+Y] = E[X]+E[Y] for any X, Y.
variance
Var(X) = E[(X−μ)²] = E[X²]−(E[X])². Measures spread around the mean.
standard deviation
σ = √Var(X). In the same units as X.
standardization
Z = (X−μ)/σ. The standardized form has E[Z] = 0 and Var(Z) = 1.
Bernoulli distribution
A discrete distribution on {0,1} with P(X=1) = p. Mean p, variance p(1−p).
binomial distribution
Counts successes in n independent Bernoulli(p) trials. Mean np, variance np(1−p).
geometric distribution
Trials until first success. Mean 1/p. Has the memoryless property.
memoryless property
P(X>m+n | X>m) = P(X>n). Unique to the geometric among discrete distributions.
negative binomial
Trials until r-th success. Generalizes geometric (r=1). Mean r/p.
hypergeometric distribution
Counts successes when sampling without replacement from a finite population. Mean nM/N.
finite population correction
(N−n)/(N−1) in the hypergeometric variance. Accounts for sampling without replacement.
Poisson distribution
Models rare, independent event counts. Mean equals variance equals λ. Limit of Binomial(n, λ/n).
Poisson process
A model for events occurring independently at a constant rate λ. Counts over interval t follow Poisson(λt).

PMF Practice

3.PMFProbability Mass Functions
  1. A random variable X has PMF p(x) = c/x² for x ∈ {1, 2, 3}, zero otherwise. (a) Find c. (b) Construct a PMF table. (c) Find P(X ≤ 2).
  2. Which of the following could be a valid PMF? Justify each answer. (a) p(x) = 0.3 for x ∈ {1,2,3}, zero otherwise. (b) p(x) = (0.5)^x for x ∈ {1,2,3,...}. (c) p(x) = x/10 for x ∈ {1,2,3,4}. (d) p(1) = −0.2, p(2) = 1.2.
  3. Roll two fair dice. Let Y = |D₁ − D₂|. (a) List all possible values of Y. (b) Construct the PMF. (c) What is the most probable value of Y?
  4. A random variable X takes values {−2, −1, 0, 1, 2} with PMF p(x) = c(1+x²). Find c and sketch the PMF as a bar chart.

CDF Practice

3.CDFCumulative Distribution Functions
  1. A random variable X has PMF: p(1) = 0.05, p(2) = 0.20, p(3) = 0.40, p(4) = 0.25, p(5) = 0.10. (a) Construct the CDF table. (b) Sketch the step function. (c) Find P(2 ≤ X ≤ 4). (d) Find P(X > 3).
  2. A CDF is given by F(x) = 0 for x < 0; F(x) = 0.15 for 0 ≤ x < 2; F(x) = 0.55 for 2 ≤ x < 4; F(x) = 0.90 for 4 ≤ x < 6; F(x) = 1 for x ≥ 6. (a) Identify the support of X. (b) Find the PMF. (c) Find P(X ≥ 4).
  3. For the two-dice sum X, compute F(5), F(8), and P(5 < X ≤ 8).
  4. True or False? (a) If F(3) = 0.7, then P(X = 3) = 0.7. Explain. (b) If F(3) = 0.7 and F(2) = 0.4, then P(X = 3) = 0.3. Explain.

Expected Value Practice

3.EVExpected Value
  1. Compute E[X] for: (a) p(0)=0.2, p(1)=0.5, p(2)=0.3. (b) p(x) = 1/4 for x ∈ {1,2,3,4}. (c) p(x) = (3/4)(1/4)^(x−1) for x = 1,2,3,... (Hint: geometric distribution.)
  2. For the PMF in (a), compute E[X²] and E[3X − 1].
  3. A lottery ticket costs $5. You win $100 with probability 0.01 and nothing otherwise. What is your expected net gain per ticket?
  4. Roll a fair die: roll 1 → lose $3; roll 6 → win $6; otherwise win $1. Find the expected value of winnings.
  5. An insurance company receives 0 to 4 claims per day with probabilities 0.10, 0.30, 0.30, 0.20, 0.10. (a) Compute E[X]. (b) If each claim costs $2000, find the expected daily payout.

Variance Practice

3.VVariance and Standard Deviation
  1. Compute Var(X) and SD(X) for each PMF in Exercise 3.EV.1.
  2. A random variable X has E[X] = 5 and E[X²] = 30. Find Var(X), SD(X), and Var(2X − 3).
  3. Let X ~ Binomial(20, 0.4). Without computing the full distribution, find E[X], Var(X), and SD(X).
  4. Two independent random variables X and Y each have variance 9. Let Z = X + 3Y − 2. Find Var(Z). (Use: for independent X, Y: Var(X+Y) = Var(X)+Var(Y).)
  5. PMF of W: p(w) = w/10 for w ∈ {1,2,3,4}. Compute E[W], E[W²], and Var(W).
  6. Show that if P(X = c) = 1, then Var(X) = 0. Interpret this result.

Distribution Identification

For each scenario: identify the distribution family, state its parameters, and explain why that distribution applies.

3.DIDistribution Identification
  1. An unfair coin (P(heads) = 0.6) is flipped 25 times. Let X be the number of heads.
  2. A server receives 50 requests per minute on average; requests arrive independently. Let X be the number in a 3-second interval.
  3. A box has 8 working and 4 burned-out bulbs. Five are selected without replacement. Let X be the number of working bulbs.
  4. A factory has a 2% defect rate. Defects are independent. Let X be the number tested until the first defective.
  5. Continuing from 4: Let Y be the number tested until the 5th defective.
  6. A student randomly guesses on a 20-question true/false quiz. Let X be the number correct.
  7. Type O-negative blood: 7% of people have it. Donors arrive one at a time. Let X be the number until the first O-negative donor.
  8. Average 1.5 coding errors per 100 patient records. Let X be the number of errors in the next 100 records.

Applications

3.APPApplied Problems
  1. (Quality Control) A production run of 500 chips has a 1% defect rate. A batch of 15 chips is sampled. (a) Let X be defective chips. Find P(X = 0) and P(X ≥ 2). (b) Find the probability of rejection if any defective is found. (c) If instead sampling without replacement from 5 defectives among 500, which distribution is more appropriate? Explain.
  2. (Genetics) Offspring inherit a recessive trait with probability 1/4, independently. A botanist studies 12 offspring. (a) Find E[X] and Var(X). (b) Find P(X ≥ 5). (c) If the botanist waits until finding 3 with the recessive trait, what is the expected number of plants studied?
  3. (Epidemiology) A rare disease occurs at 0.8 cases per 1000 residents per year. A neighborhood has 2000 residents. (a) Identify the distribution for X, cases next year. (b) Find P(X = 0), P(X = 1), P(X ≥ 3). (c) If 5 cases are observed, discuss whether this suggests an elevated rate.
  4. (Finance) An insurer covers 1000 policyholders, each independently filing a claim with probability 0.04. (a) Identify the distribution; find E[X] and Var(X). (b) Use the Poisson approximation to set up P(X ≥ 50). (c) Find P(needing additional staff) if the limit is 60 claims.
  5. (Networks) A router drops each packet with probability 0.001. In 2000 packets, let X be the number dropped. (a) Give the exact and approximating distribution. (b) Find P(X ≤ 1). (c) Find E[X] and SD(X).
  6. (Biology) DNA replication introduces mutations at 1.2 per million base pairs. Consider 2 million base pairs. (a) Identify the distribution. (b) Find P(X = 0) and P(X ≥ 4).

Error Analysis

Each item contains one or more errors. Find and correct each error, and explain the correct reasoning.

Problem 3.EA.1

Problem: Let X ~ Binomial(10, 0.3). Compute P(X = 3).

P(X = 3) = (0.3)³(0.7)⁷ = (0.027)(0.0824) = 0.00222

Find the error. What is the correct computation?

Problem 3.EA.2

Problem: Let X have PMF p(1) = 0.4, p(2) = 0.4, p(3) = 0.3. Compute E[X].

E[X] = (1 + 2 + 3) / 3 = 2

Find the error(s). What is the correct computation?

Problem 3.EA.3

Problem: A geometric random variable has p = 0.4. Compute P(X ≤ 2).

P(X ≤ 2) = P(X=1) + P(X=2) = 0.4 + 0.4 = 0.8

Find the error. What is the correct answer?

Problem 3.EA.4

Problem: Let X ~ Poisson(3). Find Var(X).

Var(X) = √3 ≈ 1.73 because the standard deviation of a Poisson equals √λ.

Find the error. What is the correct answer?

Problem 3.EA.5

Problem: X has E[X] = 4 and Var(X) = 2. Let Y = 3X + 1. Find Var(Y).

Var(Y) = 3 · Var(X) + 0 = 3(2) = 6

Find the error. What is the correct answer?

Problem 3.EA.6

Problem: 50 cards drawn without replacement from a 52-card deck. Let X = number of aces. Student uses Binomial(50, 4/52).

Identify the error and state the correct distribution with parameters.

Chapter 3 Checkpoint

Chapter 3 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 3 · Allow 60–75 minutes
Section A — Short Answer (Questions 1–8)
  1. Define a random variable in your own words. What makes it "random"?
  2. State the three requirements for a valid PMF.
  3. What is the relationship between the PMF and the CDF for a discrete random variable? How do you move from one to the other?
  4. State the definition of expected value and the computing formula for variance.
  5. Explain in one to two sentences why expected value is called a long-run average.
  6. What is the memoryless property, and which discrete distribution possesses it?
  7. State the four conditions required for the binomial distribution to be appropriate.
  8. Why does the Poisson distribution have equal mean and variance?
Section B — Computation (Questions 9–18)
  1. A random variable X has p(x) = k(3−x) for x ∈ {0, 1, 2} and 0 otherwise. Find k, E[X], and Var(X).
  2. Let X have CDF: F(x) = 0 for x < 1; F(x) = 0.3 for 1 ≤ x < 3; F(x) = 0.6 for 3 ≤ x < 5; F(x) = 1.0 for x ≥ 5. Find the PMF of X and compute P(1 < X ≤ 5).
  3. A fair coin is flipped 12 times. Let X be the number of heads. Compute P(X = 6), E[X], and SD(X).
  4. A student guesses randomly on a 20-question multiple-choice test, each question with 5 options. Let X be the number correct. Find P(X ≥ 8). Set up the sum and simplify.
  5. Let X ~ Geometric(0.3). Find P(X = 4), E[X], and P(X > 5).
  6. A Poisson random variable has E[X] = 5. Find P(X = 3) and P(X ≤ 2).
  7. A committee of 4 is selected from 7 engineers and 5 scientists. Let X be the number of engineers. Find P(X = 2), E[X], and identify the distribution.
  8. Let X ~ NegBin(3, 0.5). Find E[X] and P(X = 5).
  9. A random variable X has E[X] = 6 and Var(X) = 4. Define Y = 2X − 3. Find E[Y], Var(Y), and the standardized value of X = 8.
  10. Verify that Σ (e^(−4) · 4^k / k!) = 1 for k = 0, 1, 2, ... by using the Maclaurin series for e^x. Explain each step.
Section C — Conceptual and Applied (Questions 19–22)
  1. A game costs $3 to play. Roll a fair die: 5 or 6 wins $10; otherwise nothing. (a) Compute expected net gain per play. (b) If you play 1000 times, approximately how much do you expect to win or lose in total?
  2. Car accidents at an intersection average 0.4 per day (Poisson process). (a) P(no accidents on a given day)? (b) P(3 or more accidents in a given week)?
  3. A process yields 5% defectives. Batch of 200 items. (a) Justify the Poisson approximation. (b) Set up P(15 or fewer defectives) clearly. (c) Compute E[X] and SD(X) under Poisson.
  4. A recruiter needs 4 qualified applicants; each interview independently yields a qualified candidate with probability 0.25. Let X be the number of interviews needed. (a) Identify the distribution. (b) Find E[X] and Var(X). (c) Set up and simplify P(X ≤ 10).
Bridge Note: From Discrete to Continuous Random Variables
Chapter 4: Continuous Random Variables

This chapter studied random variables whose possible values can be listed. The entire analysis rested on sums: Σ p(x) = 1, E[X] = Σ x·p(x), F(x) = Σ p(t) for t ≤ x.

Chapter 4 crosses into new territory. When a random variable can take any value in an interval — a length, a weight, a temperature, an exact measurement — listing values is impossible. There is no individual point that carries positive probability. Instead, probability is spread continuously, and the sum Σ is replaced by the integral ∫.

The shift is profound but the concepts carry over. The PMF becomes a probability density function (PDF), and Σ p(x) = 1 becomes ∫ f(x) dx = 1. The CDF is still F(x) = P(X ≤ x), but it is now ∫ f(t) dt from −∞ to x. Expected value becomes E[X] = ∫ x f(x) dx. The named distributions — uniform, exponential, normal, gamma, beta — have their own beautiful properties.

Before moving on, review:

  • Definition and properties of definite integrals
  • Integration techniques: u-substitution and integration by parts
  • Area under a curve as a measure of accumulation
MGU Library Connection
  • Chapter 2: Probability Rules and Conditional Probability — prerequisite for this chapter
  • Chapter 4: Continuous Random Variables — the parallel development with integrals
  • Chapter 5: Joint Distributions — extending to multiple random variables simultaneously
  • Chapter 6: Moment Generating Functions — algebraic tools for characterizing distributions
  • Appendix F: Common Discrete Distributions Reference — PMF, mean, variance, and MGF for all six distributions
  • MGU Dictionary entries: random variable, PMF, CDF, expected value, variance, Bernoulli, binomial, geometric, Poisson