Probability, Random Variables, and Probability Distributions

What you'll learn

This revision guide covers probability foundations, random variables, and their distributions as tested in AP Statistics. You'll learn to distinguish between discrete and continuous random variables, calculate and interpret expected values and standard deviations, and apply the binomial and geometric distributions to real-world scenarios. These concepts form the foundation for statistical inference.

Key terms and definitions

Random variable — a numerical outcome of a random process, typically denoted by a capital letter (e.g., X)

Probability distribution — a function that assigns probabilities to all possible values of a random variable; probabilities must sum to 1 for discrete variables

Expected value (mean) — the long-run average value of a random variable, calculated as μ = E(X) = Σx·P(x) for discrete variables

Variance — a measure of spread for a random variable, calculated as σ² = Σ(x - μ)²·P(x) or σ² = E(X²) - [E(X)]²

Discrete random variable — a random variable that takes on a countable number of distinct values (often whole numbers)

Continuous random variable — a random variable that can take any value within an interval, with probabilities represented by areas under a density curve

Binomial distribution — the probability distribution for the number of successes in a fixed number of independent trials with constant probability of success

Geometric distribution — the probability distribution for the number of trials needed to get the first success in a sequence of independent trials

Core concepts

Discrete random variables and probability distributions

A discrete random variable takes specific, countable values. Its probability distribution must satisfy two conditions:

• Each probability must be between 0 and 1: 0 ≤ P(X = x) ≤ 1
• All probabilities must sum to 1: ΣP(X = x) = 1

Representing discrete distributions:

You can display probability distributions using tables, histograms (with bars touching and height representing probability), or formulas. For example, if X represents the number of heads in three coin flips:

The expected value represents the balance point or center of the distribution. Calculate it by multiplying each value by its probability and summing:

E(X) = μ = Σx·P(x)

The variance measures spread around the mean:

Var(X) = σ² = Σ(x - μ)²·P(x)

Or equivalently: σ² = E(X²) - [E(X)]² = Σx²·P(x) - μ²

The standard deviation is σ = √σ²

Linear transformations of random variables

When you transform a random variable using Y = a + bX:

Effect on mean:

• E(a + bX) = a + bE(X)
• μᵧ = a + bμₓ

Effect on variance and standard deviation:

• Var(a + bX) = b²Var(X)
• σᵧ² = b²σₓ²
• Adding a constant shifts the distribution but doesn't change spread
• Multiplying by a constant scales both center and spread

Example: If X has mean 10 and standard deviation 2, and Y = 3X + 5:

• E(Y) = 5 + 3(10) = 35
• SD(Y) = |3|(2) = 6

Combining independent random variables

When X and Y are independent random variables, their combined behaviour follows specific rules.

For the sum or difference:

E(X ± Y) = E(X) ± E(Y)

This addition/subtraction rule for means works whether or not X and Y are independent.

Var(X ± Y) = Var(X) + Var(Y)

Note that variances always add, even for differences. This rule requires independence.

SD(X ± Y) = √[Var(X) + Var(Y)]

Example: A café sells coffee (C) and pastries (P). Daily coffee sales have mean £240 and SD £18. Pastry sales have mean £85 and SD £12. For total daily revenue T = C + P:

• E(T) = 240 + 85 = £325
• SD(T) = √(18² + 12²) = √468 ≈ £21.63

Binomial distributions

The binomial distribution models the number of successes in a fixed number of independent trials. It requires four conditions (BINS):

• Binary outcomes: each trial has exactly two outcomes (success/failure)
• Independent trials: outcome of one trial doesn't affect others
• Number of trials: fixed in advance
• Same probability: probability of success (p) constant for each trial

Notation: X ~ Binomial(n, p) or X ~ B(n, p)

where n = number of trials and p = probability of success

Calculating probabilities:

P(X = k) = C(n,k) · p^k · (1-p)^(n-k)

where C(n,k) = n!/(k!(n-k)!) is the number of combinations

Parameters:

• Mean: μ = np
• Standard deviation: σ = √[np(1-p)]

When to use: Situations involving counting successes in a fixed number of independent trials with constant probability, such as:

• Number of correct answers when guessing on multiple-choice questions
• Number of defective items in a fixed sample size
• Number of free throws made out of 10 attempts

Geometric distributions

The geometric distribution models the number of trials needed to get the first success in a sequence of independent trials. It satisfies the same conditions as binomial (binary, independent, same probability) but doesn't have a fixed number of trials.

Notation: X ~ Geometric(p)

where p = probability of success on each trial

Calculating probabilities:

P(X = k) = (1-p)^(k-1) · p

This represents (k-1) failures followed by a success on the kth trial.

Parameters:

• Mean: μ = 1/p
• Standard deviation: σ = √[(1-p)/p²]

When to use: Situations asking "how many trials until the first success":

• Number of items inspected until finding the first defect
• Number of free throws attempted until making the first basket
• Number of customers served until finding someone who purchases

Continuous random variables and density curves

Unlike discrete variables, continuous random variables can take any value in an interval. Their probability distributions are represented by density curves (also called probability density functions).

Key properties of density curves:

• The curve lies entirely on or above the horizontal axis
• The total area under the curve equals 1
• Probability = area under the curve over an interval
• P(X = any exact value) = 0 (area of a line is zero)
• Therefore P(X < a) = P(X ≤ a) for continuous variables

Important point: You cannot find probabilities by multiplying values by heights for continuous distributions. You must use areas.

For the uniform distribution on interval [a, b]:

• The density curve is a rectangle
• Height = 1/(b - a)
• Mean: μ = (a + b)/2
• Standard deviation: σ = (b - a)/√12

Worked examples

Example 1: Discrete probability distribution

The number of goals (G) scored by a football team in a match has this probability distribution:

(a) Verify this is a valid probability distribution. [1 mark]

(b) Calculate the expected number of goals per match. [2 marks]

(c) Calculate the standard deviation of goals per match. [3 marks]

Solution:

(a) Sum of probabilities = 0.15 + 0.35 + 0.30 + 0.15 + 0.05 = 1.00 ✓ All probabilities are between 0 and 1 ✓ Therefore valid probability distribution.

(b) E(G) = Σg·P(g) = 0(0.15) + 1(0.35) + 2(0.30) + 3(0.15) + 4(0.05) = 0 + 0.35 + 0.60 + 0.45 + 0.20 = 1.60 goals

(c) First calculate E(G²): E(G²) = 0²(0.15) + 1²(0.35) + 2²(0.30) + 3²(0.15) + 4²(0.05) = 0 + 0.35 + 1.20 + 1.35 + 0.80 = 3.70

Var(G) = E(G²) - [E(G)]² = 3.70 - (1.60)² = 3.70 - 2.56 = 1.14

SD(G) = √1.14 = 1.07 goals (to 2 d.p.)

Example 2: Binomial distribution

A quality control inspector knows that 8% of components from a production line are defective. She randomly selects 15 components for inspection.

(a) Explain why a binomial distribution is appropriate for modelling X, the number of defective components. [2 marks]

(b) Find the probability that exactly 2 components are defective. [2 marks]

(c) Calculate the mean and standard deviation of X. [2 marks]

Solution:

(a) Binary outcomes: each component is defective or not defective ✓ Independent: random selection means one component doesn't affect another ✓ Number of trials fixed at n = 15 Same probability: p = 0.08 for each component Therefore X ~ B(15, 0.08)

(b) P(X = 2) = C(15,2) × (0.08)² × (0.92)¹³ = 105 × 0.0064 × 0.3383 = 0.227 (to 3 d.p.)

(c) Mean: μ = np = 15 × 0.08 = 1.2 components Standard deviation: σ = √[np(1-p)] = √[15 × 0.08 × 0.92] = √1.104 = 1.05 components (to 2 d.p.)

Example 3: Geometric distribution

A basketball player has a 65% success rate on free throws. Let X represent the number of attempts needed to make her first successful free throw.

(a) What is the probability she makes her first basket on the third attempt? [2 marks]

(b) Calculate the expected number of attempts needed. [1 mark]

Solution:

(a) X ~ Geometric(0.65) P(X = 3) = (0.35)² × (0.65) = 0.1225 × 0.65 = 0.0796 or 7.96%

(b) E(X) = 1/p = 1/0.65 = 1.54 attempts

Common mistakes and how to avoid them

• Confusing binomial and geometric: Binomial counts successes in n trials; geometric counts trials until first success. Read the question carefully: "How many successes?" suggests binomial; "How many attempts until?" suggests geometric.

• Forgetting variances add (not standard deviations): When combining independent random variables, always add variances first, then take the square root. Don't add standard deviations directly.

• Adding variances when finding differences: Remember Var(X - Y) = Var(X) + Var(Y), not minus. Variances always add for independent variables.

• Using P(X = x) for continuous variables: For continuous random variables, probabilities of exact values are always zero. Use intervals and areas instead.

• Incorrect transformation of variance: When finding Var(a + bX), remember only b affects variance: Var(a + bX) = b²Var(X). The constant a shifts the distribution but doesn't change spread.

• Not checking independence: The variance addition rule Var(X + Y) = Var(X) + Var(Y) only works when X and Y are independent. Always verify or state this assumption.

Exam technique for "Probability, Random Variables, and Probability Distributions"

• Show all working clearly: For probability calculations, write out the formula first, substitute values, then calculate. Examiners award method marks even if the final answer is incorrect.

• State distributions explicitly: Write "X ~ B(n, p)" or "X ~ Geometric(p)" to show you've identified the correct model. This often earns a mark and helps organize your solution.

• Check conditions when required: If asked to justify using binomial or geometric distributions, explicitly verify BINS (for binomial) or the geometric requirements. Use bullet points for clarity.

• Use calculator efficiently: Learn your calculator's binomial and geometric probability functions. However, show the formula you're using in your working, even if calculating on the calculator.

Quick revision summary

Discrete random variables have countable values with probabilities summing to 1. Calculate expected value as Σx·P(x) and variance as Σ(x - μ)²·P(x) or E(X²) - [E(X)]². For linear transformations, E(a + bX) = a + bE(X) and Var(a + bX) = b²Var(X). When combining independent variables, means add/subtract but variances always add. Binomial distribution models number of successes in n independent trials with constant probability p; geometric models trials until first success. Continuous variables have probabilities represented by areas under density curves, with P(X = exact value) = 0.