--- title: Probability Distributions and Expected Value subtitle: Conditional reasoning, Bayes' theorem, binomial models, and decisions author: Mathematics Department organization: Your School recipient: High school statistics students date: 2026-09-13 version: "1.0" description: A proof-led high school probability lesson connecting conditional probability, discrete distributions, expected value, binomial models, and decisions. keywords: [conditional probability, Bayes theorem, probability distributions, expected value, binomial distribution, statistics] template: assignment theme: latex-article paper: us-letter titlePage: true toc: true tocDepth: 2 headingNumbers: false equationNumbers: false columns: 1 figurePlacement: none tableCaptionPosition: top header: Probability | From Conditions to Decisions footer: Student lesson and practice pdfProfile: tagged language: en-US --- # Probability Distributions and Expected Value Probability describes uncertainty in a model. It does not promise what will happen next. A fair coin can land heads five times in succession; fairness means that each toss has probability $1/2$ of heads and that long-run relative frequency tends to stabilize near $1/2$ under repeated independent tosses. This lesson develops one chain of ideas: $$ \text{sample space} \longrightarrow \text{conditional probability} \longrightarrow \text{random variable} \longrightarrow \text{distribution} \longrightarrow \text{expected value and decision}. $$ ## Learning goals By the end of this lesson, you should be able to: - describe events as subsets of a sample space; - use addition, multiplication, and complement rules; - distinguish independence from mutual exclusivity; - calculate and interpret conditional probability; - derive and apply Bayes' theorem using tables or trees; - validate a discrete probability distribution; - compute and interpret expectation, variance, and standard deviation; - recognize, derive, and use a binomial distribution; - compare decisions by expected value while stating assumptions and risk. ## 1. Events and probability rules A **sample space** $S$ is the set of possible outcomes in a model. An **event** is a subset of $S$. For events $A$ and $B$: - $A^c$ means that $A$ does not occur; - $A\cap B$ means that both occur; - $A\cup B$ means that at least one occurs. ### Why the addition rule subtracts an overlap Adding $P(A)$ and $P(B)$ counts outcomes in $A\cap B$ twice. Subtracting that overlap once gives $$ P(A\cup B)=P(A)+P(B)-P(A\cap B). $$ The complement rule follows because $A$ and $A^c$ are disjoint and fill the sample space: $$ P(A^c)=1-P(A). $$ ### Worked example 1: a die For a fair six-sided die, let $A$ be “even” and $B$ be “greater than $3$.” Then $$ A=\{2,4,6\},\quad B=\{4,5,6\},\quad A\cap B=\{4,6\}. $$ Therefore $$ P(A\cup B)=\frac36+\frac36-\frac26=\frac23. $$ ## 2. Conditional probability and independence Once $B$ is known to have occurred, the relevant sample space shrinks to $B$. Provided $P(B)>0$, $$ P(A\mid B)=\frac{P(A\cap B)}{P(B)}. $$ Rearranging yields the multiplication rule $$ P(A\cap B)=P(B)P(A\mid B)=P(A)P(B\mid A). $$ Events $A$ and $B$ are **independent** when learning one does not change the probability of the other: $$ P(A\mid B)=P(A). $$ Equivalently, when the relevant probabilities are defined, $$ P(A\cap B)=P(A)P(B). $$ Mutually exclusive events cannot occur together. Nonempty mutually exclusive events are not independent: observing one makes the probability of the other zero. ### Worked example 2: read the denominator named by “given” The following synthetic table summarizes how $100$ students traveled and whether they arrived on time. :::table{#tbl:transport caption="Synthetic transportation and arrival data"} | | Bus | Walk | Total | | --- | ---: | ---: | ---: | | On time | 60 | 25 | 85 | | Late | 12 | 3 | 15 | | Total | 72 | 28 | 100 | ::: Among bus riders, $$ P(\text{on time}\mid\text{bus})=\frac{60}{72}=\frac56. $$ Among students who were on time, $$ P(\text{bus}\mid\text{on time})=\frac{60}{85}=\frac{12}{17}. $$ These conditionals answer different questions and need not be equal. Arrival and transportation are not independent because $5/6\ne85/100$. ## 3. Bayes' theorem The multiplication rule expresses the same intersection in two orders: $$ P(A\cap B)=P(B\mid A)P(A)=P(A\mid B)P(B). $$ Solving for $P(A\mid B)$ gives Bayes' theorem: $$ P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)}. $$ If $A_1,\ldots,A_k$ partition the sample space, then $$ P(B)=\sum_{i=1}^k P(B\mid A_i)P(A_i). $$ Bayes' theorem reverses a condition. It does not allow $P(A\mid B)$ and $P(B\mid A)$ to be interchanged. ### Worked example 3: a base-rate tree Consider a synthetic screening model with $2\%$ prevalence, $90\%$ sensitivity, and $95\%$ specificity. In a modeled population of $10{,}000$, the expected branch counts are shown below. :::figure{#fig:bayes-tree caption="A frequency tree makes the base rate and both conditional error rates visible" alt="Tree diagram for 10000 modeled cases: 200 have the condition, producing 180 positive and 20 negative results; 9800 do not, producing 490 positive and 9310 negative results" width=96 placement=none} ![Frequency tree for a synthetic screening model](bayes-frequency-tree.svg) ::: There are $180+490=670$ positive results, of which $180$ come from the condition-present branch. Therefore $$ P(\text{condition}\mid+)=\frac{180}{670}=\frac{18}{67}\approx0.269. $$ The probability is not $90\%$: sensitivity describes $P(+\mid\text{condition})$, whereas the question asks for the reversed conditional. This synthetic example teaches probability only; it is not a medical recommendation or a claim about a real test. ## 4. Random variables and distributions A **random variable** assigns a numerical value to each outcome. For two fair coin tosses, let $X$ be the number of heads. The outcomes $TT,TH,HT,HH$ map to $0,1,1,2$, so the probability distribution is :::table{#tbl:two-coins caption="Distribution of the number of heads in two fair tosses"} | $x$ | 0 | 1 | 2 | | --- | ---: | ---: | ---: | | $P(X=x)$ | $1/4$ | $1/2$ | $1/4$ | ::: A discrete probability mass function $p(x)$ is valid when $$ p(x)\ge0\quad\text{for every }x, \qquad \sum_x p(x)=1. $$ The possible values of $X$ need not be equally likely, consecutive, or themselves probabilities. ### Worked example 4: theoretical and empirical distributions A theoretical distribution follows from model assumptions, such as equally likely coin-toss sequences. An empirical distribution uses observed relative frequencies. If a machine produced $4$, $12$, and $4$ batches with $0$, $1$, and $2$ defects, respectively, the empirical probabilities are $0.2$, $0.6$, and $0.2$. They summarize these $20$ observations; they do not prove an unchanging physical law. ## 5. Expected value and spread For a discrete random variable $X$, the expected value is the probability-weighted mean $$ \mu=E(X)=\sum_x xP(X=x). $$ It represents a long-run average under repeated trials governed by the same model. It need not be a possible single outcome. The variance and standard deviation are $$ \operatorname{Var}(X)=\sum_x (x-\mu)^2P(X=x), \qquad \sigma=\sqrt{\operatorname{Var}(X)}. $$ An equivalent computational identity is $$ \operatorname{Var}(X)=E(X^2)-[E(X)]^2. $$ ### Worked example 5: mean and standard deviation Suppose $X$ has probabilities $0.2$, $0.5$, and $0.3$ at values $0$, $1$, and $2$. Then $$ E(X)=0(0.2)+1(0.5)+2(0.3)=1.1. $$ Also, $$ E(X^2)=0+1(0.5)+4(0.3)=1.7, $$ so $$ \operatorname{Var}(X)=1.7-1.1^2=0.49, \qquad \sigma=0.7. $$ ### Why linear transformations behave linearly For constants $a$ and $b$, $$ E(aX+b) =\sum_x (ax+b)P(X=x) =a\sum_x xP(X=x)+b\sum_x P(X=x) =aE(X)+b. $$ The final step uses the fact that probabilities sum to $1$. ### Worked example 6: fairness and risk are different A game pays a net profit of $\$3$ with probability $0.4$ and a net loss of $\$2$ with probability $0.6$. Its expected profit is $$ 3(0.4)+(-2)(0.6)=0. $$ The game is fair in expected-value terms, but an individual play still ends at $3$ or $-2$, never at $0$. Equal expected values can also have different spreads, so expectation alone does not encode risk tolerance. ## 6. The binomial distribution A count $X$ is binomial when: 1. there is a fixed number $n$ of trials; 2. each trial has two labeled outcomes; 3. trials are independent; 4. the success probability $p$ is constant. For exactly $k$ successes, any particular success-failure order has probability $p^k(1-p)^{n-k}$. There are $\binom nk$ orders, so $$ P(X=k)=\binom nk p^k(1-p)^{n-k}, \qquad k=0,1,\ldots,n. $$ :::figure{#fig:binomial caption="The binomial distribution for four fair independent trials" alt="Bar chart for zero through four successes with probabilities one sixteenth, four sixteenths, six sixteenths, four sixteenths, and one sixteenth" width=88 placement=none} ![Binomial probability bars for n equals 4 and p equals one half](binomial-four-fair.svg) ::: ### Why the binomial mean is $np$ Let $I_j=1$ when trial $j$ succeeds and $0$ otherwise. Then $X=I_1+\cdots+I_n$ and $E(I_j)=p$. Linearity of expectation gives $$ E(X)=E(I_1)+\cdots+E(I_n)=np. $$ For independent trials, $$ \operatorname{Var}(X)=np(1-p). $$ ### Worked example 7: exact binomial probability If $X\sim\operatorname{Bin}(5,0.4)$, then $$ P(X=3)=\binom53(0.4)^3(0.6)^2=0.2304. $$ ### Worked example 8: “at least one” by complement For ten independent trials with success probability $0.1$, $$ P(X\ge1)=1-P(X=0)=1-(0.9)^{10}\approx0.6513. $$ ## 7. Expected value in decisions Expected value compares long-run consequences under stated probabilities and payoffs. It does not establish that the probabilities are accurate, include nonfinancial consequences, or choose a risk preference for the decision maker. ### Worked example 9: two strategies A safe strategy yields $\$4$. A risky strategy yields $\$10$ with probability $0.5$ and loses $\$2$ with probability $0.5$. Both have expected value $\$4$: $$ E(\text{risky})=10(0.5)-2(0.5)=4. $$ The strategies are equal by expected value but not by possible outcomes or spread. ### Worked example 10: a decision under a weather model Suppose a planning model assigns probability $0.3$ to rain. An outdoor event yields payoff $120$ without rain and $-80$ with rain; an indoor option yields $50$ either way. Then $$ E(\text{outdoor})=0.7(120)+0.3(-80)=60, $$ compared with $E(\text{indoor})=50$. The result supports the outdoor option only if the probability and payoff assumptions are defensible and the decision maker accepts its downside. ## 8. Common reasoning errors > [!WARNING] > A probability statement is incomplete without a model and conditioning information. A numerical expected value is incomplete without units, time horizon, and an account of risk. Check for these errors: - reversing $P(A\mid B)$ and $P(B\mid A)$; - calling mutually exclusive events independent; - using the binomial formula when $p$ changes or trials depend on one another; - accepting a distribution with a negative probability or total other than $1$; - interpreting expectation as the most likely or guaranteed outcome; - comparing decisions while omitting relevant outcomes or using unsupported probabilities. # Practice set Give exact fractions where practical and round requested decimals to four places. ## A. Events and conditional probability 1. List the sample space for two fair coin tosses and find the probability of exactly one head. 2. For a fair die, let $A$ be “even” and $B$ be “greater than $3$.” Find $P(A\cap B)$ and $P(A\cup B)$. 3. Using Table 1, find $P(\text{bus})$. 4. Using Table 1, find $P(\text{on time}\mid\text{bus})$. 5. Using Table 1, find $P(\text{bus}\mid\text{on time})$. 6. Are “bus” and “on time” independent in Table 1? Justify numerically. 7. Using Table 1, find $P(\text{late}\cup\text{walk})$. 8. Suppose $P(A)=0.4$, $P(B)=0.5$, and $P(A\cap B)=0.2$. Determine whether $A$ and $B$ are independent and find $P(A\cup B)$. ## B. Random variables and distributions 9. Is $P(X=0)=0.15$, $P(X=1)=0.50$, $P(X=2)=0.35$ a valid distribution? Explain. 10. Explain why probabilities $0.4$, $0.7$, and $-0.1$ do not define a distribution even though they sum to $1$. 11. Construct the distribution of the number of heads in two fair tosses. 12. Let $Y$ be the sum of two fair dice. State its possible values and find $P(Y=7)$. 13. In $20$ observed batches, $4$ had no defects, $12$ had one, and $4$ had two. Construct the empirical distribution of defect count $X$. 14. For the distribution in Exercise 13, find $P(X\ge1)$. 15. A random variable has values $-1$ and $2$ with probabilities $0.6$ and $0.4$. Find $P(X>0)$ and $P(X\le-1)$. 16. Using the distribution in Exercise 9, find $P(X\le1)$ and $P(X>1)$. ## C. Expected value and spread 17. For $P(X=0)=0.2$, $P(X=1)=0.5$, and $P(X=2)=0.3$, find $E(X)$. 18. For the same distribution, find $\operatorname{Var}(X)$ and $\sigma$. 19. A game pays net $\$3$ with probability $0.4$ and net $-\$2$ with probability $0.6$. Find its expected net payoff and interpret it. 20. A spinner's gross prizes are $\$0$, $\$5$, and $\$20$ with probabilities $0.6$, $0.3$, and $0.1$. Find the fair entry price by expected value. 21. A warranty model assigns probability $0.03$ to a $\$100$ claim and otherwise pays $0$. Find the expected claim per contract. 22. If $E(X)=4.2$, find $E(2X+5)$ and justify the rule used. 23. Compare a guaranteed payoff of $\$4$ with a payoff of $\$10$ with probability $0.5$ and $-\$2$ with probability $0.5$. Compare expected values and risk. 24. In $40$ observations, $X=0$ occurred $5$ times, $X=1$ occurred $20$ times, and $X=2$ occurred $15$ times. Find the empirical expected value. ## D. Binomial models 25. Eight independent components each pass a test with probability $0.6$. Is the number passing binomial? Give $n$ and $p$. 26. Three cards are drawn without replacement from a deck containing five red and five blue cards. Explain why the number of red cards is not binomial under the standard definition. 27. If $X\sim\operatorname{Bin}(5,0.4)$, find $P(X=3)$. 28. If $X\sim\operatorname{Bin}(6,0.2)$, find $P(X=0)$. 29. If $X\sim\operatorname{Bin}(10,0.1)$, find $P(X\ge1)$. 30. If $X\sim\operatorname{Bin}(20,0.35)$, find the mean, variance, and standard deviation. 31. Construct the full distribution for $X\sim\operatorname{Bin}(4,0.5)$. 32. A student guesses on five independent four-choice questions. Find the probability of exactly two correct answers and the expected number correct. ## E. Bayes' theorem and decisions 33. In the synthetic screening model, find $P(+)$. 34. In the same model, find $P(\text{condition}\mid+)$. 35. In the same model, find $P(\text{condition}\mid-)$. 36. Factory A makes $60\%$ of a product and has a $1\%$ defect rate. Factory B makes $40\%$ and has a $3\%$ defect rate. Find $P(\text{B}\mid\text{defective})$. 37. A safe strategy returns $\$40$. A risky strategy returns $\$100$ with probability $0.6$ and $-\$80$ with probability $0.4$. Compare expected values. 38. A raffle has $500$ tickets, one $\$800$ prize, and a $\$5$ ticket price. Find the expected net payoff for one ticket. 39. Option A costs $\$40$ with certainty. Option B costs $\$150$ with probability $0.2$ and $\$0$ otherwise. Compare expected costs and explain why expected value may not settle the choice. 40. Under a model with rain probability $0.3$, an outdoor plan yields $120$ without rain and $-80$ with rain; an indoor plan yields $50$. Compare expected values and state one assumption that should be challenged before deciding. ## References - Common Core State Standards for Mathematics, HSS-CP and HSS-MD. - OpenStax, *Introductory Statistics 2e*, Chapters 3-4, CC BY 4.0. - NIST/SEMATECH, *e-Handbook of Statistical Methods*, binomial distribution reference. All data and decisions in this packet are synthetic teaching models. They are not medical, financial, insurance, engineering, or operational advice. Independent mathematics-teacher review and classroom trial remain outstanding.