High School Statistics — MGU Mathematics Series

Part IV: Random Variables, Distributions, and Inference

Random Variables · Distributions · Expected Value · Binomial · Geometric

Chapter 10 of 14

MGU Mathematics Series · High School Statistics · Library Textbook Edition

Chapter 10:
Random Variables
and Probability Distributions

Assigning numbers to uncertain outcomes — expected value, variability, the binomial distribution, and the foundations of probabilistic modeling

■ Math Level: High School Statistics ■

■ 14 Sections • Part IV of 5 ■

Follows Chapters 8 & 9: Probability & Conditional Probability · Prepares students for Chapter 11: Normal Distributions

Chapter 10 — Front Matter

Purpose and Overview

Probability, as studied in Chapters 8 and 9, describes the likelihood of events in terms of outcomes, sample spaces, and rules. Those methods work well when events have clear categorical outcomes: the coin lands heads, the card is a heart, the selected student prefers biology. But many real situations involve outcomes that are numerical — the number of customers who arrive, the amount of money won in a game, the number of defective items in a shipment, the number of free throws a player makes. In these situations, it is not enough to ask whether an event happens. It is more useful to ask how much, how many, or how large, and then to study the entire pattern of possible values and their probabilities together.

This chapter introduces random variables and probability distributions as tools for studying numerical outcomes of uncertain processes. A random variable assigns a number to each outcome of a random experiment. A probability distribution describes all the possible values that a random variable can take and how likely each value is. Together, these ideas allow students to compute averages, measure variability, model realistic chance situations, and prepare for the normal distribution, sampling distributions, and inference that follow.

Opening Question

A player at a school carnival pays $2 to roll a standard six-sided die. If the die shows a 6, the player wins $5. Otherwise, the player wins nothing. Is this game a good deal for the player? Should the carnival expect to make money in the long run? To answer carefully, you need to assign values to the outcomes, attach probabilities to each value, and find the long-run average. That is exactly what this chapter will teach you to do.

14Sections

13Objectives

16Key Terms

12+Worked Examples

Learning Objectives

1Explain what a random variable is and how it differs from a fixed variable.
2Identify discrete and continuous random variables and give examples of each.
3Construct and interpret a probability distribution table.
4Verify that a probability distribution is valid.
5Compute and interpret the expected value of a discrete random variable.
6Compute and interpret the variance and standard deviation of a discrete random variable.
7Recognize and check the four BINS conditions for a binomial setting.
8Compute binomial probabilities using the binomial formula.
9Find and interpret the mean and standard deviation of a binomial random variable.
10Describe the geometric setting and compute basic geometric probabilities.
11Apply random variable concepts to games, finance, insurance, and risk.
12Identify and correct common mistakes in working with distributions.
13Describe how discrete distributions connect to continuous distributions and inference.

Chapter 10 — Contents

Sections at a Glance

10.1What a Random Variable IsNumerical outcomes; definition; types of random variables; discrete vs. continuous; notation
10.2Discrete Random VariablesListing values from sample spaces; assigning numbers to outcomes; notation X and x
10.3Probability DistributionsDistribution tables; two validity conditions; probability histograms; constructing examples
10.4Expected ValueLong-run average; formula E(X) = Σ x·P(X=x); computing from a table; game applications
10.5Interpreting Expected ValueNot the single-trial prediction; weighted average; interpretation requirements
10.6Variability of a Random VariableWhy spread matters; variance formula; computing from a distribution table
10.7Standard Deviation of a Random VariableSquare root of variance; same units as X; complete distribution analysis
10.8Binomial SettingsBINS conditions; binary, independent, fixed n, same p; examples and non-examples
10.9Binomial ProbabilitiesBinomial formula; binomial coefficient; full distribution tables; technology use
10.10Mean and Standard Deviation of a BinomialShortcuts μ = np and σ = √(np(1−p)); interpretation; public health context
10.11Geometric Settings as an IntroductionTrials until first success; geometric formula; expected value 1/p; comparison with binomial
10.12Random Variables in Games, Finance, and RiskGames and lotteries; insurance logic; investment risk; expected value in decisions
10.13Common Random Variable MistakesSeven recurring errors; explanations and corrections; prevention strategies
10.14Preparing for Normal DistributionsContinuous distributions; area as probability; bridge to Chapter 11; college preview

Section 10.1

What a Random Variable Is

Arithmetic is precise; probability is uncertain. In earlier chapters, probability described whether an event happens — whether a coin lands heads, whether a card is a heart. But many real situations produce numerical outcomes: the number of cars that pass through an intersection, the amount of rainfall in a week, the number of correct answers on a quiz. When an experiment produces numerical outcomes, it is useful to define a rule that assigns a number to each outcome. That rule is called a random variable.

A random variable is typically denoted by a capital letter such as X, Y, or Z. A specific value it might take is denoted by a lowercase letter such as x. So if X is the number of heads in three coin flips, then X can be 0, 1, 2, or 3. The statement X = 2 means exactly two heads occurred.

Definition — Random Variable

A random variable is a variable that takes numerical values determined by the outcomes of a random experiment. Its value cannot be known with certainty before the experiment is run, but probability describes the structure of its possible values.

Situation	Random Variable X	Type
Roll a six-sided die	Face value shown	Discrete
Count emails received in one hour	Number of emails	Discrete
Measure height of a student	Height in centimeters	Continuous
Flip a coin 10 times	Number of heads	Discrete
Choose a light bulb from production	Lifetime in hours	Continuous
Survey a voter on a ballot measure	1 = supports, 0 = opposes	Discrete

Discrete vs. Continuous

A discrete random variable takes a countable number of possible values — usually whole numbers. A continuous random variable can take any value in an interval, including fractions and decimals. This chapter focuses on discrete random variables. Continuous distributions are introduced in Chapter 11.

Student Note — What "Random" Means

A random variable is not arbitrary or unknowable. "Random" means the value is determined by a chance process. Before the experiment, the value is uncertain; after the experiment, it is known. Probability describes the structure of that uncertainty.

Section 10.2

Discrete Random Variables

To work with a discrete random variable, begin by listing every possible value it can take, then count which outcomes produce each value. Consider a fair coin flipped twice. The sample space is {HH, HT, TH, TT}. Let X be the number of heads. Assign the value of X to each outcome:

Outcome	X (number of heads)	Probability
HH	2	1/4
HT	1	1/4
TH	1	1/4
TT	0	1/4

The possible values of X are 0, 1, and 2. Two outcomes give X = 1, so P(X = 1) = 2/4 = 1/2. One outcome each gives X = 0 and X = 2, so P(X = 0) = P(X = 2) = 1/4. This is the foundation for building a probability distribution.

Section 10.3

Probability Distributions

A probability distribution is a complete description of all possible values of a random variable and the probability of each value. For a discrete random variable, it is displayed as a table with one column for values and one for probabilities. The lowercase x represents a specific value; the uppercase X represents the random variable itself.

Two Requirements for a Valid Distribution

Condition 1: Every probability must satisfy 0 ≤ P(X = x) ≤ 1.

Condition 2: The probabilities must sum to exactly 1: Σ P(X = x) = 1.

If either condition fails, the table does not represent a valid probability distribution.

Example 10.3.1

Checking a probability distribution

ProblemIs the following table a valid probability distribution? Values: 1, 2, 3, 4. Probabilities: 0.15, 0.30, 0.40, 0.20.
ThinkCheck that each probability is between 0 and 1, then check that the sum equals 1.
Calculate Each value: 0.15, 0.30, 0.40, 0.20 — all between 0 and 1. ✓ Sum: 0.15 + 0.30 + 0.40 + 0.20 = 1.05
CheckThe sum is 1.05, not 1.00.
InterpretThis is not a valid distribution. The probabilities sum to more than 1, which is impossible in any probability model.

Answer: Invalid — probabilities sum to 1.05, not 1.

Example 10.3.2

Constructing a probability distribution

ProblemA spinner has three equally likely sections labeled 1, 2, and 3. Construct the probability distribution for one spin.
ThinkEach outcome is equally likely, so each has probability 1/3.
Represent x | P(X = x) ------+--------- 1 | 1/3 2 | 1/3 3 | 1/3
CheckEach probability is between 0 and 1. Sum = 1/3 + 1/3 + 1/3 = 1. ✓
InterpretEach value is equally likely. No outcome is favored.

Answer: Valid uniform distribution with P(X = 1) = P(X = 2) = P(X = 3) = 1/3.

Probability Histogram

A probability distribution can be shown as a probability histogram: place the possible values on the horizontal axis, draw a bar over each value with height equal to its probability, and label the vertical axis "Probability." The total area of all bars equals 1. This is the geometric bridge to density curves in Chapter 11, where area under a continuous curve represents probability.

Section 10.4

Expected Value

If a game were played an extremely large number of times, what would the average result be? This long-run average is called the expected value of the random variable. It is not the most likely single outcome — it is the average value across a very large number of trials. The expected value is also called the mean of the random variable, written µ_X.

Expected Value Formula

E(X) = μ_X = Σ x · P(X = x)

Multiply each value by its probability and sum all products.

Example 10.4.1

Expected value of a spinner

ProblemA spinner has four sections: value 1 (probability 0.10), value 2 (probability 0.20), value 3 (probability 0.40), value 4 (probability 0.30). Find E(X).
Represent x | P(X=x) | x · P(X=x) ---+--------+----------- 1 | 0.10 | 0.10 2 | 0.20 | 0.40 3 | 0.40 | 1.20 4 | 0.30 | 1.20 | Sum | 2.90
CalculateE(X) = 0.10 + 0.40 + 1.20 + 1.20 = 2.90
CheckE(X) = 2.90 is between the minimum of 1 and maximum of 4. Higher values have higher probabilities, so an expected value above 2.5 makes sense. ✓
InterpretIf this spinner were spun many times, the average result per spin would be approximately 2.90.

Answer: E(X) = 2.90

Example 10.4.2

Expected value in a game — the carnival question

ProblemA player pays $2 to roll a die. Roll a 6: win $5. Otherwise: win nothing. Find the expected net gain per roll.
ThinkNet gain = prize minus cost to play. Rolling a 6: $5 − $2 = +$3 net. Not rolling a 6: $0 − $2 = −$2 net.
Represent x | P(X=x) | x · P(X=x) -----+--------+----------- +3 | 1/6 | 0.500 −2 | 5/6 | −1.667 | Sum | −1.167
CalculateE(X) = 3 · (1/6) + (−2) · (5/6) = 3/6 − 10/6 = −7/6 ≈ −$1.17
CheckNegative expected value is consistent with the player facing unfavorable odds — five of six sides produce a loss. ✓
InterpretOn average, the player loses approximately $1.17 per roll. Over many plays, the carnival gains; over many plays, the player loses. This game is not a good financial deal for the player.

Answer: E(X) ≈ −$1.17 per roll. The carnival earns approximately $1.17 per play in the long run.

Data Ethics Note

Many games, lotteries, and gambling opportunities have a negative expected value for the participant. The expected value framework makes these situations transparent. Statistics provides tools to evaluate risk honestly rather than relying on misleading advertising that emphasizes prizes while minimizing the probability of winning.

Section 10.5

Interpreting Expected Value

The expected value is the long-run average across many repetitions — not a prediction for any single trial. If E(X) = 2.90 for a spinner, no single spin will land on 2.90, because 2.90 is not a possible value. If E(X) = −$1.17 for a game, a player might win on any individual roll, but the average across thousands of rolls will be approximately −$1.17 per roll.

Notice also that the expected value formula is a weighted average: values with higher probabilities contribute more. A uniform distribution — where all values have equal probability — produces an expected value equal to the ordinary arithmetic mean of the possible values.

Student Warning

Do not say "I expect to get exactly E(X) on the next trial." Say instead: "Across a large number of trials, the average result will be approximately E(X)." These are very different claims. Expected value is a long-run concept, not a single-trial guarantee.

Practice 10.1–10.5Concept and Skill

Classify each as discrete or continuous: (a) number of siblings, (b) temperature at noon, (c) number of cars in a parking lot, (d) exact time of sunrise.
A discrete random variable X has values 0, 1, 2 with probabilities 0.30, 0.50, 0.20. Verify the distribution is valid, then find E(X).
A game pays $10 with probability 0.05 and $0 otherwise. The game costs $1 to play. Find the expected net gain per play and interpret it.
A student says, "The expected value is 3.2, so I'll get 3.2 on the next trial." Explain the error.

Section 10.6

Variability of a Random Variable

Two random variables can have the same expected value but behave very differently. One might cluster tightly around the average; another might be wildly spread out. In finance, two investments might offer the same expected return, but one is highly volatile and another is stable. To describe the spread of a random variable's distribution, statistics uses variance and standard deviation.

Variance of a Discrete Random Variable

Var(X) = σ²_X = Σ (x − μ_X)² · P(X = x)

Average squared distance from the mean, weighted by probability.

Example 10.6.1

Computing variance from the spinner distribution

ProblemUsing the spinner from 10.4.1 where E(X) = 2.90, find Var(X).
Represent x | P(X=x) | x − 2.90 | (x−2.90)² | (x−2.90)²·P ---+--------+----------+------------+------------- 1 | 0.10 | −1.90 | 3.61 | 0.361 2 | 0.20 | −0.90 | 0.81 | 0.162 3 | 0.40 | +0.10 | 0.01 | 0.004 4 | 0.30 | +1.10 | 1.21 | 0.363 | | | Sum: | 0.890
CalculateVar(X) = 0.361 + 0.162 + 0.004 + 0.363 = 0.890
InterpretThe variance is 0.890 (in squared units). Convert to standard deviation for a directly interpretable measure of spread.

Answer: Var(X) = 0.890

Section 10.7

Standard Deviation of a Random Variable

σ_X = √Var(X) = √[ Σ (x − μ_X)² · P(X = x) ]

In the same units as X — directly interpretable as typical spread from the mean.

Example 10.7.1

A complete distribution analysis

ProblemX has distribution: P(X=0) = 0.25, P(X=1) = 0.50, P(X=2) = 0.25. Find E(X), Var(X), and σ_X.
Calculate E(X)E(X) = 0(0.25) + 1(0.50) + 2(0.25) = 0 + 0.50 + 0.50 = 1.00
Calculate Var(X) x | P(X=x) | (x−1)² | (x−1)²·P ---+--------+--------+---------- 0 | 0.25 | 1 | 0.25 1 | 0.50 | 0 | 0.00 2 | 0.25 | 1 | 0.25 Var(X) = 0.50
Calculate σ_Xσ_X = √0.50 ≈ 0.707
CheckThis symmetric distribution has its center at x = 1, which matches E(X) = 1. ✓
InterpretThe distribution is centered at 1.00. Outcomes typically differ from the mean by about 0.71 units. The moderate spread reflects that outcomes of 0 and 2 are each one unit away from the center and together have total probability 0.50.

Answer: E(X) = 1.00, Var(X) = 0.50, σ_X ≈ 0.707

Key Distinction — Random Variable vs. Data Set

The variance and standard deviation of a random variable use the probability distribution — not raw data. These describe the theoretical spread of the distribution. The sample standard deviation s from Chapter 5 describes spread in an observed data set. Both measure spread, but they come from different sources and are computed differently.

Section 10.8

Binomial Settings

Many random processes share a common structure: a procedure is repeated a fixed number of times, each repetition has exactly two possible outcomes, and each repetition is independent of the others with the same probability of success. This structure is called a binomial setting. When all four conditions hold, the count of successes follows a binomial distribution.

Binary

Each trial has exactly two outcomes: success or failure. The labels "success" and "failure" are just names — neither implies a value judgment.

Independent

The result of one trial does not affect the probability on any other trial. This is the most commonly overlooked condition.

Fixed n

The total number of trials n is fixed in advance before the experiment begins. It does not depend on the outcomes.

Same p

The probability of success p is the same for every trial. It cannot change from trial to trial.

Examples and Non-Examples

Situation	Binomial?	Condition Check
Flip a fair coin 10 times; X = number of heads	Yes	Binary ✓ Independent ✓ n=10 fixed ✓ p=0.5 constant ✓
Guess all 5 questions on a 4-choice quiz; X = correct	Yes	Binary ✓ Independent ✓ n=5 fixed ✓ p=0.25 constant ✓
Draw 5 cards without replacement; X = number of hearts	No	Independence fails — probability of heart changes after each draw
Flip a coin until first head; X = number of flips	No	Fixed n fails — number of flips is not set in advance; depends on outcome

Student Warning

Do not assume that any situation involving "success and failure" is automatically binomial. Always check all four BINS conditions before using the binomial formula. The independence condition and the fixed-n condition are the most commonly overlooked.

Section 10.9

Binomial Probabilities

When all four BINS conditions are satisfied, the probability of exactly k successes in n trials — with probability of success p — is given by the binomial probability formula.

Binomial Probability Formula

P(X = k) = C(n,k) · p^k · (1−p)^(n−k)

where n = number of trials, k = number of successes (0 through n), p = probability of success per trial, and C(n,k) = n! / [k!(n−k)!] is the binomial coefficient ("n choose k").

The binomial coefficient C(n,k) counts the number of different arrangements of k successes among n trials. It is the same as the combination nCr from Chapter 9. The formula has two parts: C(n,k) counts the arrangements; p^k(1−p)^n−k gives the probability of any single such arrangement. Multiplying them gives the total probability of exactly k successes in any order.

Example 10.9.1

Binomial probability — guessing on a quiz

ProblemA five-question multiple choice quiz has four answer choices per question. A student randomly guesses every answer. Find P(X = 2), the probability of getting exactly 2 correct.
ThinkCheck BINS: Binary (correct/incorrect) ✓, Independent ✓, n = 5 fixed ✓, p = 0.25 per question ✓. Binomial with n = 5, p = 0.25, k = 2.
Calculate C(5,2) = 5! / (2! · 3!) = 120 / (2 · 6) = 10 P(X=2) = 10 · (0.25)² · (0.75)³ = 10 · 0.0625 · 0.4219 = 10 · 0.02637 ≈ 0.2637
CheckThe expected number correct is 5 × 0.25 = 1.25. Getting exactly 2 is close to but above the expected value, so a probability around 0.26 is plausible. ✓
InterpretThere is approximately a 26.4% chance a randomly guessing student gets exactly 2 correct on this quiz.

Answer: P(X = 2) ≈ 0.2637

Example 10.9.2

Full binomial distribution — all values of k

ProblemConstruct the complete probability distribution for the number of correct guesses on the five-question quiz (n = 5, p = 0.25).
Calculate P(X=0) = C(5,0)·(0.25)⁰·(0.75)⁵ = 1·1·0.2373 ≈ 0.2373 P(X=1) = C(5,1)·(0.25)¹·(0.75)⁴ = 5·0.25·0.3164 ≈ 0.3955 P(X=2) = C(5,2)·(0.25)²·(0.75)³ = 10·0.0625·0.4219 ≈ 0.2637 P(X=3) = C(5,3)·(0.25)³·(0.75)² = 10·0.0156·0.5625 ≈ 0.0879 P(X=4) = C(5,4)·(0.25)⁴·(0.75)¹ = 5·0.0039·0.75 ≈ 0.0146 P(X=5) = C(5,5)·(0.25)⁵·(0.75)⁰ = 1·0.00098·1 ≈ 0.0010
CheckSum = 0.2373 + 0.3955 + 0.2637 + 0.0879 + 0.0146 + 0.0010 = 1.0000 ✓
InterpretThe distribution is right-skewed. A randomly guessing student is most likely to get 1 correct (≈39.6%). Getting 4 or 5 correct is very unlikely (≈1.6% total).

Answer: See table. Distribution is right-skewed; P(X=1) is highest at ≈ 0.3955.

x (correct answers)	P(X = x)
0	0.2373
1	0.3955
2	0.2637
3	0.0879
4	0.0146
5	0.0010
Sum	1.0000

Technology Note

On most graphing calculators: binompdf(n, p, k) returns P(X = k); binomcdf(n, p, k) returns P(X ≤ k). In spreadsheets: =BINOM.DIST(k, n, p, FALSE) for exact probability; =BINOM.DIST(k, n, p, TRUE) for cumulative. Always verify with a rough manual calculation to confirm input order and values are correct.

Section 10.10

Mean and Standard Deviation of a Binomial Random Variable

When X is binomial with n trials and probability of success p, there are simple shortcut formulas for the mean and standard deviation that do not require constructing the full distribution table.

Binomial Mean and Standard Deviation

μ_X = np σ_X = √( np(1−p) )

These formulas follow from the general expected value and variance formulas applied to the binomial distribution. At the high school level, knowing, applying, and interpreting them is sufficient.

Example 10.10.1

Mean and standard deviation — the quiz

ProblemFor the five-question quiz with n = 5, p = 0.25, find µ_X and σ_X.
Calculate μ_X = np = 5 × 0.25 = 1.25 σ_X = √(np(1−p)) = √(5 × 0.25 × 0.75) = √0.9375 ≈ 0.968
InterpretA randomly guessing student would average 1.25 correct answers across many repetitions of this quiz. The standard deviation of ≈ 0.97 means the number correct typically differs from the mean by about 1 question.

Answer: μ_X = 1.25, σ_X ≈ 0.968

Example 10.10.2

Binomial in a public health context

ProblemA disease affects 8% of the population. A random sample of 200 people is tested. (a) Check BINS conditions. (b) Find µ_X and σ_X. (c) Interpret both.
Check BINSBinary: has disease or not ✓ · Independent: large population, each person approximately independent ✓ · Fixed n = 200 ✓ · Same p = 0.08 per person ✓. This is approximately a binomial setting.
Calculate μ_X = 200 × 0.08 = 16 σ_X = √(200 × 0.08 × 0.92) = √14.72 ≈ 3.84
InterpretIn a random sample of 200 people, we expect an average of 16 people to have the disease. The standard deviation of ≈ 3.84 means the actual count in samples of 200 typically falls within about 4 people of 16. Samples with 12–20 cases would all be within one standard deviation of the mean.

Answer: μ_X = 16, σ_X ≈ 3.84. The expected number of cases is 16; typical samples range roughly 12–20.

MGU Library Connection

This public health example connects to Chapter 12: Sampling Distributions, where behavior of sample proportions across many samples is studied.
MGU Biology and Public Health pages extend this reasoning to disease screening, prevalence estimation, and population-level decisions.
The Finance and Economics pages apply expected value and binomial distributions to actuarial science and risk modeling.

Section 10.11

Geometric Settings as an Introduction

A geometric setting resembles a binomial setting in important ways: trials are independent, each is binary, and the probability of success p is constant. The key difference is the question being asked. Instead of fixing n trials and counting successes, the geometric setting asks: how many trials are needed to get the first success? The number of trials until and including the first success is a geometric random variable.

Feature	Binomial Setting	Geometric Setting
Number of trials	Fixed at n	Not fixed — determined by outcome
What X counts	Number of successes in n trials	Number of trials until first success
Possible values of X	0, 1, 2, …, n	1, 2, 3, … (no upper limit)
Mean	µ = np	µ = 1/p

Geometric Probability Formula

P(X = k) = (1−p)^(k−1) · p

k is the trial number on which the first success occurs (k = 1, 2, 3, …)

Example 10.11.1

Geometric probability — rolling a 6

ProblemA player rolls a fair die and wants to get a 6. What is the probability the first 6 appears on exactly the 4th roll?
ThinkGeometric setting: binary (6 or not), independent rolls, counting until first success, p = 1/6 per roll.
Calculate P(X = 4) = (5/6)³ · (1/6) = (125/216) · (1/6) = 125/1296 ≈ 0.0965
InterpretThere is approximately a 9.65% chance that the first 6 appears on exactly the fourth roll. The first three rolls must all be non-6 (probability 5/6 each), and the fourth must be a 6 (probability 1/6).

Answer: P(X = 4) ≈ 0.0965. Expected rolls until first 6: 1/(1/6) = 6.

Expected Value of a Geometric Random Variable

The mean of a geometric random variable is µ_X = 1/p. If p = 1/6, then on average it takes 1/(1/6) = 6 trials to get the first success. This makes intuitive sense: if each trial has a 1-in-6 chance, about 6 trials are needed on average. Geometric distributions are studied in greater depth in AP Statistics and college probability courses.

Practice 10.6–10.11Skill Practice

X has distribution: P(X=5)=0.20, P(X=10)=0.35, P(X=15)=0.30, P(X=20)=0.15. Find E(X), Var(X), and σ_X.
A fair coin is flipped 8 times. Let X = number of heads. (a) Check BINS. (b) Find P(X=5). (c) Find P(X≥7). (d) Find µ_X and σ_X.
A manufacturer finds 3% of items are defective. A batch of 50 items is selected. (a) Verify BINS. (b) Find µ_X and σ_X. (c) Find P(X=0).
A basketball player makes free throws with probability 0.80. She takes 10 shots. Find P(X=8) and P(X≥9).
A geometric setting: probability of success is 0.20 per trial. Find P(X=3) and the expected number of trials until first success.

Section 10.12

Random Variables in Games, Finance, and Risk

Expected value and standard deviation of random variables appear throughout everyday decision-making. Every game of chance assigns a probability distribution to the player's winnings. The expected value tells the player what to expect on average over many plays. In virtually every lottery, casino game, or carnival game, the expected value for the participant is negative — by design, because the operator must cover costs and earn a margin over many plays.

Understanding expected value does not mean a person should never play a game. People sometimes pay for entertainment, social experience, or the excitement of uncertainty. But knowing the expected value makes the financial cost of that choice transparent. The same reasoning applies to insurance: a policyholder typically pays more in premiums over time than they receive in claims, but they purchase protection against catastrophic loss — trading a known small cost for elimination of an uncertain large one.

Example 10.12.1

Insurance and expected value

ProblemA homeowner pays $900/year for flood insurance. The company estimates: 3% chance of a flood with average $25,000 claim; 97% chance of no flood and no claim. (a) Find the expected annual payout. (b) Is the premium profitable? (c) Why might the homeowner buy insurance even with a negative expected value?
Calculate (a) E(claim) = 25,000 × 0.03 + 0 × 0.97 = 750 + 0 = $750
Interpret (b)The company expects to pay $750 per policyholder per year. The premium is $900 > $750, so the company covers expected claims, operating costs, and earns a margin. Profitable. ✓
Interpret (c)The homeowner faces a small probability (3%) of a devastating $25,000 loss. By paying $900 per year, they replace that uncertain catastrophe with a known small annual cost. Even if the expected value of buying insurance is slightly negative for the homeowner, the dramatic reduction in financial variability — and exposure to catastrophic loss — makes the purchase rational and often necessary.

Answer: E(claim) = $750. Premium of $900 is profitable. Insurance transfers variability, not just expected value.

Finance and Investment

In finance, investors study both expected return and standard deviation of returns. The standard deviation is often called risk. Two investments with equal expected returns but different standard deviations present different risk profiles. A risk-averse investor prefers lower variability for the same expected return. This trade-off between expected value and variability is central to financial decision-making and portfolio theory.

Practice 10.12Applications

A raffle sells 200 tickets at $3 each. Prizes: one prize of $200, two prizes of $50, five prizes of $10. (a) Find the net gain distribution. (b) Find E(X). (c) Interpret the expected value.
A die game: roll a fair die, win $6 if it shows 1, win $2 if it shows 2 or 3, win nothing otherwise. The game costs $2 to play. (a) Define net gain X. (b) Find E(X). (c) Is this game in the player's favor?
Two investments each have expected return $500. Investment A has σ = $80; Investment B has σ = $300. (a) Which is riskier? (b) Which would a risk-averse investor likely prefer? (c) What type of investor might prefer B?

Section 10.13

Common Mistakes with Variables and Distributions

Mistake 1 — Confusing Expected Value with Most Likely Outcome

"E(X) = 1.25, so I expect to get 1.25 correct on the next quiz."

Correct: E(X) is the long-run average. It is not a prediction for a single trial. 1.25 is not even a possible value on the quiz.

Mistake 2 — Applying Binomial Without Checking BINS

Using the binomial formula for 5 cards drawn without replacement from a deck.

Correct: Independence fails — the probability of success changes after each draw. The binomial formula requires independence between trials.

Mistake 3 — Computing "Exactly k" When "At Least k" Is Asked

P(X ≥ 3) = P(X = 3) = C(5,3)(0.4)³(0.6)²

Correct: P(X ≥ 3) = P(X=3) + P(X=4) + P(X=5). "At least," "at most," "more than," and "fewer than" are different from "exactly." Read carefully.

Mistake 4 — Omitting the Binomial Coefficient

P(X=2) = (0.25)²(0.75)³ ≈ 0.0264 (missing the C(5,2) = 10 factor)

Correct: P(X=2) = C(5,2) · (0.25)² · (0.75)³ ≈ 0.2637. The coefficient accounts for the 10 different arrangements of 2 successes in 5 trials. Without it, the answer is ten times too small.

Mistake 5 — Averaging Values Instead of Using Weighted Average

E(X) = (0 + 1 + 2) / 3 = 1.00 for a distribution where P(X=0)=0.10, P(X=1)=0.50, P(X=2)=0.40 Correct: E(X) = 0(0.10) + 1(0.50) + 2(0.40) = 1.30. Use Σ x·P(X=x), not a simple average of values.

Mistake 6 — Using σ_X = np Instead of √(np(1−p))

For n=20, p=0.6: σ_X = 20 × 0.6 = 12

Correct: σ_X = √(20 × 0.6 × 0.4) = √4.80 ≈ 2.19. The mean formula np gives 12, but that is µ_X, not σ_X. Standard deviation uses √(np(1−p)).

Mistake 7 — Confusing Binomial and Geometric Settings

"I need to find P(X=4) for a binomial when the question asks how many trials until the first success."

Correct: If the number of trials is fixed and you count successes, use binomial. If you count trials until the first success, use geometric. Ask: Is n fixed in advance?

Section 10.14

Preparing for Normal Distributions

This chapter studied discrete random variables: those that take a countable number of values. The binomial distribution is the most important example. In Chapter 11, attention shifts to continuous random variables, which can take any value in an interval. The most important continuous distribution in all of statistics is the normal distribution.

Instead of a table of probabilities for specific values, the normal distribution is described by a smooth bell-shaped curve. Probability is measured as area under that curve rather than as a sum of bar heights. The central ideas of this chapter carry forward directly: a distribution still has a mean and standard deviation; the mean describes the center; the standard deviation describes spread; probabilities are still interpreted as long-run proportions. Only the method of computing probabilities changes.

The standardization technique using z-scores, introduced in Chapter 5, will take on new meaning in Chapter 11: transforming any normal distribution into a standard normal distribution so that probabilities can be found from a single table or technology tool. The work done here with expected value, variability, and the logic of probability distributions is the direct foundation for everything that follows.

Bridge to Chapter 11

Discrete distributions use bar heights to show probability. Continuous distributions use area under a curve. In Chapter 11, you will learn that the area under a normal curve between two values equals the probability that the variable falls in that interval — extending the probability histogram idea to a smooth, continuous model. The normal distribution's mean and standard deviation function exactly as they do here: center and spread. z-scores let you convert any normal variable to the standard scale to find probabilities.

Chapter 10 — Back Matter

Chapter Summary

Sec 10.1–10.2A random variable assigns numerical values to outcomes of a random experiment. Discrete random variables take a countable number of values; continuous random variables take values from an interval. Possible values come from listing the sample space and assigning numbers to each outcome.
Sec 10.3A probability distribution lists all possible values and their probabilities. It is valid when all probabilities are between 0 and 1 and they sum to exactly 1. A probability histogram gives a geometric picture of the distribution.
Sec 10.4–10.5The expected value E(X) = Σ x·P(X=x) is the long-run average of the random variable, not a prediction for a single trial. It is a weighted average where each value is weighted by its probability.
Sec 10.6–10.7Variance measures average squared distance from the mean, weighted by probability. Standard deviation is the square root of variance and is in the same units as X. These describe theoretical spread of the distribution, distinct from sample statistics s and s².
Sec 10.8A binomial setting requires four BINS conditions: Binary outcomes, Independence, fixed Number of trials n, Same probability p. Check every condition before using the binomial formula.
Sec 10.9The binomial formula P(X=k) = C(n,k)·p^k·(1−p)^(n−k) gives the probability of exactly k successes. The binomial coefficient C(n,k) counts the arrangements. Technology provides exact probabilities for large n.
Sec 10.10For a binomial random variable: µ_X = np and σ_X = √(np(1−p)). These shortcuts apply only when all four BINS conditions are satisfied.
Sec 10.11A geometric setting counts the number of trials until the first success. P(X=k) = (1−p)^(k−1)·p. The mean is 1/p. Unlike binomial, the number of trials is not fixed in advance.
Sec 10.12Random variable ideas apply to games, lotteries, insurance, and finance. Expected value measures long-run average gain or loss. Standard deviation measures risk. Insurance trades expected-value loss for elimination of catastrophic variability.
Sec 10.13Seven common mistakes: confusing E(X) with the most likely outcome; skipping BINS; computing "exactly" instead of "at least"; omitting C(n,k); simple-averaging values in a distribution; using np for standard deviation; confusing binomial and geometric settings.
Sec 10.14Continuous random variables require a density curve rather than a probability table. Chapter 11 extends distribution ideas to the normal curve, where probability equals area. The mean, standard deviation, and probability interpretation from Chapter 10 carry forward unchanged.

Key Terms Review

random variable

A variable whose numerical value is determined by the outcome of a random experiment. Denoted by a capital letter (X, Y, Z).

discrete random variable

A random variable that takes a countable number of possible values — usually whole numbers representing counts.

continuous random variable

A random variable that can take any value in an interval, including fractions and decimals. Modeled by density curves in Chapter 11.

probability distribution

A complete description of all possible values of a random variable and the probability of each value.

expected value

The long-run average of a random variable: E(X) = Σ x·P(X=x). Also called the mean of the distribution (µ_X).

variance

Average squared distance from the mean, weighted by probability: σ²_X = Σ(x−µ_X)²·P(X=x).

standard deviation

Square root of the variance: σ_X = √Var(X). In the same units as X; measures typical spread from the mean.

binomial setting

A random experiment satisfying all four BINS conditions: Binary outcomes, Independence, fixed N, Same probability p.

binomial coefficient

C(n,k) = n!/[k!(n−k)!]. Counts the number of ways to arrange k successes among n trials.

binomial formula

P(X=k) = C(n,k)·p^k·(1−p)^(n−k). Gives the probability of exactly k successes in a binomial setting.

geometric setting

Independent binary trials with same p, where the variable of interest is the number of trials until the first success.

geometric formula

P(X=k) = (1−p)^(k−1)·p. Gives the probability that the first success occurs on trial k. Mean = 1/p.

long-run average

What the expected value represents: the average outcome across a very large number of repetitions of the experiment.

probability histogram

A bar graph for a discrete distribution with possible values on the horizontal axis and probabilities as bar heights. Total area equals 1.

BINS conditions

The four requirements for a binomial setting: Binary, Independent, fixed N, Same p. All four must hold.

risk (in finance)

In financial contexts, the standard deviation of investment returns. Higher standard deviation indicates greater variability and financial risk.

Concept Review Questions

Part AConcept Check

Explain what a random variable is. How does it differ from a fixed variable or a constant?
What is the difference between a discrete and a continuous random variable? Give one example of each not from this chapter.
What two conditions must every valid probability distribution satisfy?
Explain what the expected value represents. Why should it not be interpreted as a prediction for a single trial?
State the four BINS conditions for a binomial setting. Explain why each is necessary.
How does a geometric setting differ from a binomial setting?
A student says: "Expected value is −$3, so I'm guaranteed to lose $3 every time I play." Explain the error.
Explain in words why the binomial coefficient C(n,k) is required in the binomial probability formula.

Part BSkill Practice — Distributions and Expected Value

Determine whether each table is a valid probability distribution. Explain your answer.
- (a) Values: 0, 1, 2. Probabilities: 0.30, 0.45, 0.25.
- (b) Values: 1, 2, 3. Probabilities: 0.20, 0.35, 0.55.
- (c) Values: 1, 2, 3. Probabilities: 0.10, −0.05, 0.95.
A random variable X has distribution: P(X=2)=0.15, P(X=4)=0.30, P(X=6)=0.40, P(X=8)=0.15. (a) Verify validity. (b) Find E(X). (c) Find Var(X). (d) Find σ_X. (e) Interpret E(X) in context.
Compute each binomial coefficient without a calculator: (a) C(6,2), (b) C(5,0), (c) C(7,7), (d) C(8,3).

Part CSkill Practice — Binomial Distributions

A fair coin is flipped 6 times. Let X = number of heads. (a) Verify BINS. (b) Find P(X=3). (c) Find P(X=0). (d) Find P(X≥5). (e) Find µ_X and σ_X.
A quiz has 10 true-false questions; a student randomly guesses every answer. (a) Identify n and p. (b) Find P(X=7). (c) Find E(X) and σ_X. (d) Find P(X≥8) using complementary counting.
A manufacturing plant has a 5% defect rate. An inspector randomly selects 20 items. (a) Check BINS. (b) Find P(X=2). (c) Find P(X=0). (d) Find µ_X and σ_X.
A basketball player makes free throws at 72%. She takes 8 shots. (a) Verify BINS. (b) Find P(X=6). (c) Find P(X≥7). (d) Find µ_X and σ_X and interpret each.

Part DApplications

A player draws one card from a standard deck. Win $10 if it is an ace, win $5 if it is a face card (J, Q, K), lose $2 for any other card. (a) Define X and list all possible values. (b) Build the probability distribution table. (c) Calculate E(X). (d) Is this game favorable for the player?
A school raffle sells 500 tickets for $2 each. Prizes: one $300 prize, two $100 prizes, five $20 prizes. (a) Write the net gain distribution from the ticket-buyer's perspective. (b) Find E(X). (c) Interpret the expected value in context.
Historical data show 15% of patients experience a mild side effect from a medication. A trial enrolls 25 patients. (a) Find P(X=3). (b) Find P(X>5). (c) Find µ_X and σ_X. (d) Would 10 cases be unusually high? Use the mean and standard deviation to explain.
A rare sports card appears in 4% of randomly selected packs. Packs are independent. (a) What is the probability the first copy appears in exactly the 5th pack? (b) What is the expected number of packs to open? (c) What is the probability of finding the card within the first 3 packs?

Error Analysis

Each problem contains an error. Identify it and write the correct approach.

Problem 1

E(X) = (0 + 1 + 2) / 3 = 1.00 for distribution P(X=0)=0.20, P(X=1)=0.50, P(X=2)=0.30

What is the error? Write the correct calculation and result.

Problem 2

Cards drawn without replacement from a deck — student applies binomial formula for P(X=2 hearts in 5 draws).

Which BINS condition fails? What type of distribution would be more appropriate?

Problem 3

P(X ≥ 3) for Binomial(n=5, p=0.4) written as P(X=3) = C(5,3)(0.4)³(0.6)²

Identify the error. Write the correct approach for finding P(X ≥ 3).

Problem 4

For Binomial(n=20, p=0.6): σ_X = 20 × 0.6 = 12

What formula was incorrectly applied? Write the correct formula and calculation.

Problem 5

P(pass) = P(X ≥ 6) for Binomial(n=10, p=0.25) written as C(10,6)(0.25)⁶(0.75)⁴ alone.

Identify and correct the error. Describe the full calculation required.

Chapter 10 Checkpoint

Complete without assistance — assesses all key skills and concepts from Chapter 10

Section 1 — Probability Distribution and Expected Value

A game at a school fair charges $3 to play. A player rolls a fair six-sided die: roll a 1 → win $6; roll a 2 or 3 → win $1; roll a 4, 5, or 6 → win nothing. Let X = net gain (prize minus cost to play).

List all possible values of X.
Write the probability distribution table for X.
Verify that the distribution is valid.
Compute E(X).
Interpret E(X) in the context of this game.
Is this game favorable for the player or for the fair? Explain.

Section 2 — Binomial Setting and Probabilities

A medical screening test correctly identifies a condition 90% of the time. A clinic tests 8 patients who all have the condition.

Check all four BINS conditions. State n and p.
Let X = number of patients correctly identified. Find P(X = 8).
Find P(X = 6).
Find P(X ≥ 7).
Find µ_X and σ_X and interpret each in context.
Suppose only 4 of the 8 patients were correctly identified. Is this an unusually low result? Use the mean and standard deviation to explain.

Section 3 — Distribution Analysis

A discrete random variable Y has: P(Y=10)=0.20, P(Y=20)=0.35, P(Y=30)=0.30, P(Y=40)=0.15.

Verify this is a valid distribution.
Compute E(Y).
Compute Var(Y) and σ_Y.
Interpret E(Y): suppose Y represents the score on a randomly selected exam problem.
Find P(Y > 20).

Section 4 — Short Answer

A lottery ticket has E(net gain) = −$1.50. What does this mean for a person who buys one ticket?
A binomial random variable has n = 30 and p = 0.7. Without computing, state the mean. Explain how you know.
Why is C(n,k) necessary in the binomial probability formula? What does it count?
Explain in one sentence how a geometric setting differs from a binomial setting.
A student says the standard deviation of a random variable is the same as the standard deviation of a data set. Is this true? Explain.

Looking Ahead

Chapter 11: Normal Distributions and Standardization

Chapter 10 studied discrete random variables — variables that take a countable list of values. Chapter 11 moves to continuous random variables, where probability is measured as area under a smooth curve rather than as a sum of bar heights. The normal distribution is the most important model in all of statistics: it describes measurement variation in nature, test score distributions, manufacturing tolerances, financial returns, sampling distributions, and much more. The mean and standard deviation you computed in Chapter 10 take on the same roles in the normal model. The z-score formula from Chapter 5 becomes the key to finding normal probabilities — translating any normal variable to a standard scale so that a single table or technology tool can answer any probability question.