Probability and Random Variables — AP Statistics Stats Study Guide

For: AP Statistics candidates sitting AP Statistics.

Covers: Basic probability rules, conditional probability and independence, discrete and continuous random variables, mean and standard deviation of random variables, and binomial/geometric probability distributions.

You should already know: Algebra 2, basic probability intuition.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Statistics style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Probability and Random Variables?

Probability is the mathematical framework for quantifying uncertainty in random events, while random variables assign numerical values to outcomes of these random processes, forming the core of all statistical inference in the AP Statistics curriculum. Standard notation uses $P(A)$ for the probability of event $A$, and capital letters ($X$, $Y$) to represent random variables, with lowercase letters ($x$, $y$) for observed values of those variables. This unit makes up 10-20% of your total AP Statistics exam score, per the official CED, and underpins all later topics including sampling distributions, confidence intervals, and hypothesis testing.

2. Basic probability rules

All probability calculations follow a set of universal rules, built on the definition of a sample space $S$, the set of all possible outcomes of a random process, and events, which are subsets of that sample space:

1. Probability range: $0\le P(A)\le 1$ for any event $A$, where 0 represents an impossible event and 1 represents a certain event.
2. Complement rule: The probability of an event not occurring is $P(A^c)=1-P(A)$, where $A^c$ is the complement of event $A$.
3. General addition rule: For any two events $A$ and $B$, the probability of either event occurring is $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, where $A\cap B$ is the intersection of $A$ and $B$ (both events occurring).
4. Mutually exclusive addition rule: If $A$ and $B$ cannot occur at the same time (no overlapping outcomes), $P(A\cap B)=0$, so $P(A\cup B)=P(A)+P(B)$.
5. Total probability rule: The sum of probabilities for all outcomes in the sample space equals 1.

Worked example

You roll a fair 6-sided die. Event $A$ = rolling an even number, Event $B$ = rolling a number greater than or equal to 4. Calculate $P(A\cup B)$.

1. Sample space $S=\{1,2,3,4,5,6\}$, each outcome has probability $\frac{1}{6}$
2. $P(A)=P(2,4,6)=\frac{3}{6}=0.5$, $P(B)=P(4,5,6)=\frac{3}{6}=0.5$
3. Overlap $P(A\cap B)=P(4,6)=\frac{2}{6}\approx 0.333$
4. Apply general addition rule: $0.5+0.5-0.333=0.667$, or $\frac{2}{3}$

Examiners frequently test the general addition rule, so never assume events are mutually exclusive unless explicitly stated.

3. Conditional probability and independence

Conditional probability is the probability of an event occurring given that another event has already happened, written $P(A|B)$ (read "probability of A given B"). The formula is derived by restricting the sample space to only outcomes where $B$ is true: $$P(A|B)=\frac{P(A\cap B)}{P(B)}\text{for }P(B)>0$$

Two events are independent if the occurrence of one does not change the probability of the other occurring. All three of the following conditions are equivalent, and any can be used to verify independence on the exam:

1. $P(A|B)=P(A)$
2. $P(B|A)=P(B)$
3. $P(A\cap B)=P(A)\times P(B)$

Worked example

A survey of 100 high school students finds 60 play soccer, 35 play basketball, and 20 play both sports. a) What is the probability a student plays basketball given they play soccer? b) Are playing soccer and playing basketball independent events? Justify your answer.

1. Part a: $P(B|S)=\frac{P(B\cap S)}{P(S)}=\frac{0.2}{0.6}=\frac{1}{3}\approx 0.333$
2. Part b: Test the independence condition $P(S)\times P(B)=0.6\times 0.35=0.21$, which is not equal to $P(S\cap B)=0.2$. The events are not independent.

Always show the full calculation when verifying independence on free response questions; stating the result without work will lose points.

4. Random variables — discrete and continuous

A random variable (RV) is a variable that takes numerical values based on the outcome of a random process. There are two core categories of random variables tested on the AP exam:

1. Discrete random variables: Take a countable set of distinct values (e.g., number of heads in 10 coin flips, number of customers entering a store in an hour). They are described by a probability mass function (PMF) $p(x)=P(X=x)$, where the sum of all $p(x)$ values equals 1, and each $p(x)$ is between 0 and 1.
2. Continuous random variables: Take all values in an interval of real numbers (e.g., height of a randomly selected student, time to finish an exam). They are described by a probability density function (PDF), where the area under the PDF over an interval gives the probability of the RV falling in that interval. For any single value $a$, $P(X=a)=0$, since there is no area over a single point. The total area under the entire PDF equals 1.

Worked example

Classify each random variable as discrete or continuous: a) Number of correct answers on a 50-question multiple choice test b) Weight of a randomly chosen bag of chips c) Number of rain days in a month Solution: a) Discrete (countable values 0 to 50), b) Continuous (can take any value in a range, measured to infinite precision), c) Discrete (countable whole number values)

A common exam trap asks for the probability of a continuous RV taking a single value; the correct answer is always 0.

5. Mean and SD of a random variable

The mean of a random variable, also called the expected value $E(X)$ or ${\mu }_X$, is the long-run average value of the RV if the random process is repeated infinitely many times. For discrete RVs, it is calculated by weighting each possible value by its probability: $${\mu }_X=E(X)=\sum x_{i}p(x_i)$$

Variance ${\sigma }_X^2$ measures the spread of the RV around its mean, calculated as the average squared deviation from the mean: $${\sigma }_X^2=Var(X)=\sum (x_i-{\mu }_X{)}^{2}p(x_i)$$ A faster alternative formula for variance, recommended for exam use, is $Var(X)=E(X^2)-[E(X){]}^2$, where $E(X^2)=\sum x_i^{2}p(x_i)$.

Standard deviation ${\sigma }_X$ is the square root of variance, and has the same units as the original random variable, making it easier to interpret than variance: $${\sigma }_X=\sqrt{Var(X)}$$

Worked example

A discrete RV $X$ has the following PMF: $x=1,p=0.2$; $x=2,p=0.3$; $x=3,p=0.5$. Calculate $E(X)$, $Var(X)$, and ${\sigma }_X$.

1. $E(X)=(1\times 0.2)+(2\times 0.3)+(3\times 0.5)=0.2+0.6+1.5=2.3$
2. $E(X^2)=(1^2\times 0.2)+(2^2\times 0.3)+(3^2\times 0.5)=0.2+1.2+4.5=5.9$
3. $Var(X)=5.9-(2.3{)}^2=5.9-5.29=0.61$
4. ${\sigma }_X=\sqrt{0.61}\approx 0.781$

6. Binomial and geometric distributions

The binomial and geometric distributions are the two most commonly tested discrete probability distributions on the AP Statistics exam, each applying to specific random process scenarios:

Binomial distribution

A random variable follows a binomial distribution $X\sim \text{Binomial}(n,p)$ if the process meets the BIN conditions:

• B: Binary outcomes (each trial is either a success or failure)
• I: Independent trials (outcome of one trial does not affect others)
• N: Fixed number of trials $n$
• P: Constant probability of success $p$ on each trial

For a binomial RV, the probability of exactly $k$ successes in $n$ trials is: $$P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k(1-p{)}^{n-k}$$ where $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$ is the combination of $n$ items taken $k$ at a time. The mean and variance of a binomial RV are: $${\mu }_X=np{\sigma }_X^2=np(1-p)$$

Geometric distribution

A random variable follows a geometric distribution $Y\sim \text{Geometric}(p)$ if the process counts the number of trials until the first success, meeting the same binary, independent, and constant $p$ conditions as the binomial, but with no fixed number of trials. The probability the first success occurs on the $k$th trial is: $$P(Y=k)=(1-p{)}^{k-1}p$$ The mean and variance of a geometric RV are: $${\mu }_Y=\frac{1}{p}{\sigma }_Y^2=\frac{1-p}{p^2}$$

Worked example

You flip a fair coin ($p=0.5$ for heads) repeatedly. a) What is the probability you get exactly 3 heads in 5 flips? b) What is the probability the first head appears on the 4th flip?

1. Part a: $X\sim \text{Binomial}(n=5,p=0.5)$: $P(X=3)=\left(\genfrac{}{}{0ex}{}{5}{3}\right)(0.5{)}^3(0.5{)}^2=10\times 0.125\times 0.25=0.3125$
2. Part b: $Y\sim \text{Geometric}(p=0.5)$: $P(Y=4)=(0.5{)}^3\times 0.5=0.0625$

Always explicitly state the conditions for your chosen distribution when answering free response questions to earn full credit.

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Forgetting to subtract the intersection in the general addition rule, leading to probabilities greater than 1. Why it happens: Students incorrectly assume all events are mutually exclusive. Correct move: Always use the general addition rule unless you are explicitly told events cannot overlap.
• Pitfall 2: Calculating $P(A|B)$ as $\frac{P(A)}{P(B)}$ instead of $\frac{P(A\cap B)}{P(B)}$. Why it happens: Students mix up the order of conditional probability and forget the restriction to the sample space of the given event. Correct move: Write the conditional probability formula explicitly before plugging in any numbers, and label which event is the "given" first.
• Pitfall 3: Assigning a non-zero probability to a single value of a continuous random variable. Why it happens: Students apply discrete RV rules to continuous RVs. Correct move: For continuous RVs, only calculate probability over intervals, and state $P(X=a)=0$ if asked for a single point probability.
• Pitfall 4: Using the binomial distribution for problems counting trials until the first success. Why it happens: Students focus only on binary outcomes and independent trials, missing the fixed number of trials requirement for binomial distributions. Correct move: Ask: "Am I counting successes in a fixed number of trials (binomial) or trials until the first success (geometric)?"
• Pitfall 5: Using the sample mean formula $\frac{\sum x}{n}$ for expected value of a random variable. Why it happens: Students confuse descriptive statistics means with probability expected values. Correct move: Weight each RV value by its probability, only use the sample mean formula for observed data sets with equal weight for each observation.

8. Practice Questions (AP Statistics Style)

Question 1

A local coffee shop finds that 30% of customers buy a pastry, 40% buy a coffee, and 20% buy both. a) What is the probability a randomly selected customer buys a pastry or a coffee? b) What is the probability a customer buys coffee given they bought a pastry? c) Are buying a pastry and buying coffee independent events? Justify your answer.

Solution 1

a) Use the general addition rule: $P(P\cup C)=0.3+0.4-0.2=0.5$, or 50% b) Conditional probability: $P(C|P)=\frac{0.2}{0.3}\approx 0.667$, or $\frac{2}{3}$ c) Test independence: $P(P)\times P(C)=0.3\times 0.4=0.12$, which is not equal to $P(P\cap C)=0.2$. The events are not independent.

Question 2

A discrete random variable $X$ represents the number of hours a student spends studying per week, with PMF: $x=2,p=0.1$; $x=4,p=0.3$; $x=6,p=0.4$; $x=8,p=0.2$. a) Calculate the expected number of study hours per week. b) Calculate the standard deviation of study hours.

Solution 2

a) $E(X)=(2\times 0.1)+(4\times 0.3)+(6\times 0.4)+(8\times 0.2)=0.2+1.2+2.4+1.6=5.4$ hours b) $E(X^2)=(4\times 0.1)+(16\times 0.3)+(36\times 0.4)+(64\times 0.2)=0.4+4.8+14.4+12.8=32.4$ $Var(X)=32.4-(5.4{)}^2=32.4-29.16=3.24$ ${\sigma }_X=\sqrt{3.24}=1.8$ hours

Question 3

A student guesses randomly on a 10-question multiple choice test, where each question has 4 options, only one correct. a) What is the probability the student gets exactly 3 questions correct? b) What is the probability the student gets at least 1 question correct? c) What is the expected number of correct answers?

Solution 3

This is a binomial distribution: $n=10$ trials, $p=0.25$ probability of a correct guess, independent trials, binary outcomes. a) $P(X=3)=\left(\genfrac{}{}{0ex}{}{10}{3}\right)(0.25{)}^3(0.75{)}^7=120\times 0.015625\times 0.1335\approx 0.250$, or 25% b) Use the complement rule: $P(X\ge 1)=1-P(X=0)=1-(0.75{)}^{10}\approx 1-0.0563=0.9437$, or 94.4% c) Expected value: $\mu =np=10\times 0.25=2.5$ correct answers

9. Quick Reference Cheatsheet

10. What's Next

This unit is the foundational building block for all remaining content in the AP Statistics curriculum. Next, you will apply these probability rules to sampling distributions, which describe how sample statistics (like sample mean or sample proportion) vary across repeated samples, and form the basis for confidence intervals and hypothesis testing, which make up 30-40% of your total exam score. Understanding random variables and their distributions will also help you interpret the results of statistical studies, a key skill for the investigative task in the free response section of the AP Stats exam.

If you have any questions about probability rules, distributions, or how to apply these concepts to exam questions, you can ask Ollie at any time for personalized explanations, worked examples, or extra practice problems. Head to the homepage, to access more AP Statistics study guides, practice tests, and AI tutoring support tailored to your learning needs.