Addition Rule (general) — formula?
Answer: P(A or B) = P(A) + P(B) − P(A and B)
Explanation
Subtract the overlap so it isn't double-counted. If A and B are mutually exclusive, P(A and B) = 0, so the rule simplifies to P(A or B) = P(A) + P(B).
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AP Statistics: Probability, Random Variables and Sampling Distributions (High School)
Probability RulesRandom VariablesBinomial DistributionGeometric DistributionNormal Distribution & Z-ScoresSampling DistributionsCentral Limit Theorem & Standard ErrorConditions Checklist
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Probability Rules
What does it mean for two events to be mutually exclusive (disjoint)?
Answer: They cannot occur together: P(A and B) = 0.
Explanation
Example: rolling a 2 and rolling a 5 on the same die roll are mutually exclusive.
Complement Rule — formula?
Answer: P(Aᶜ) = 1 − P(A)
Explanation
Very useful for 'at least one' problems: P(at least one success) = 1 − P(no successes).
Conditional probability — formula?
Answer: P(B | A) = P(A and B) / P(A), for P(A) > 0
Explanation
Read as 'the probability of B given A has occurred.' It restricts the sample space to only the outcomes in which A happened.
General Multiplication Rule — formula?
Answer: P(A and B) = P(A) · P(B | A)
Explanation
This is just the conditional probability formula rearranged. It works whether or not A and B are independent.
Definition of independent events
Answer: A and B are independent if P(B | A) = P(B) (equivalently P(A | B) = P(A)) — knowing that one event occurred does not change the probability of the other.
Multiplication Rule for Independent Events — formula?
Answer: P(A and B) = P(A) · P(B)
Explanation
Special case of the general multiplication rule that applies only when A and B are independent.
Can mutually exclusive events also be independent?
Answer: No (unless one event has probability 0). If A and B are mutually exclusive, P(A and B) = 0, but P(A)·P(B) > 0 whenever both events have positive probability, so independence fails.
Explanation
Common AP trap: students assume disjoint means independent. In fact disjoint events are maximally dependent — if A happens, you know for certain B did not.
Random Variables
Expected value (mean) of a discrete random variable X — formula?
Answer: μ_X = Σ xᵢ · P(xᵢ), summed over all possible values xᵢ of X
Explanation
Example: X = 1, 2, 3 with P(X)= .2, .5, .3 → μ_X = 1(.2) + 2(.5) + 3(.3) = 2.1
Variance of a discrete random variable X — formula?
Answer: σ²_X = Σ (xᵢ − μ_X)² · P(xᵢ); σ_X = √σ²_X
Explanation
Continuing the example (μ_X = 2.1): σ²_X = (1−2.1)²(.2) + (2−2.1)²(.5) + (3−2.1)²(.3) = 0.49, so σ_X ≈ 0.70
Effect of a linear transformation Y = a + bX on mean and standard deviation
Answer: μ_Y = a + b·μ_X ; σ_Y = |b|·σ_X (so σ²_Y = b²σ²_X)
Explanation
Adding a constant a shifts the mean but not the spread. Multiplying by b scales both the mean and SD (variance scales by b²).
Mean of a sum or difference of two random variables X and Y
Answer: μ_{X+Y} = μ_X + μ_Y and μ_{X−Y} = μ_X − μ_Y — true always, independence not required
Variance of a sum or difference of two INDEPENDENT random variables
Answer: σ²_{X+Y} = σ²_X + σ²_Y and σ²_{X−Y} = σ²_X + σ²_Y — variances always add, even for a difference
Explanation
Holds only when X and Y are independent. Common error: subtracting variances for X−Y. Combining independent variables always increases variability, whether you add or subtract them.
Binomial Distribution
BINS conditions for a binomial setting
Answer: Binary outcomes (success/failure), Independent trials (check the 10% condition if sampling without replacement), fixed Number of trials n, Same probability of success p on every trial
Binomial probability formula
Answer: P(X = k) = C(n, k) · p^k · (1 − p)^(n − k), where C(n, k) = n! / [k!(n − k)!]
Explanation
Example: n = 5 fair coin flips, P(exactly 3 heads) = C(5,3)(0.5)³(0.5)² = 10(0.125)(0.25) = 0.3125
Mean of a binomial random variable
Answer: μ_X = n·p
Explanation
n = number of trials, p = probability of success on each trial.
Standard deviation of a binomial random variable
Answer: σ_X = √[n·p·(1 − p)]
Explanation
Variance is σ²_X = np(1−p).
Large Counts Condition for approximating a binomial distribution as Normal
Answer: n·p ≥ 10 and n·(1 − p) ≥ 10
Explanation
When satisfied, X is approximately Normal with mean np and SD √(np(1−p)), allowing Normal-based calculations for proportions.
What shape does a binomial distribution have when p = 0.5 (shown here for n = 10)?
Answer: Symmetric, bell-like shape centered at μ = np = 5
Explanation
As p moves away from 0.5 (toward 0 or 1), the distribution becomes skewed unless n is large enough to offset it.
Geometric Distribution
BITS conditions for a geometric setting
Answer: Binary outcomes, Independent trials, Trials continue until the first success (that's what is being counted), Same probability p each trial
Geometric probability formula
Answer: P(X = k) = (1 − p)^(k − 1) · p, for k = 1, 2, 3, ...
Explanation
k = trial number on which the first success occurs. Example: p = 0.25, P(first success on 3rd trial) = (0.75)²(0.25) = 0.1406
Mean of a geometric random variable
Answer: μ_X = 1/p
Explanation
The expected number of trials until the first success.
Standard deviation of a geometric random variable
Answer: σ_X = √[(1 − p)/p²]
Explanation
Variance is σ²_X = (1−p)/p².
Identify the distribution shape shown here (p = 0.2)
Answer: Geometric distribution — it is always right-skewed, with the highest probability at k = 1 and decreasing probabilities afterward
Explanation
Unlike the binomial distribution, the geometric distribution is never symmetric; there is no upper bound on k.
Normal Distribution & Z-Scores
Empirical Rule (68–95–99.7 Rule) — statement
Answer: For approximately Normal data: about 68% of values fall within 1 SD of the mean, about 95% within 2 SD, and about 99.7% within 3 SD
Explanation
Example: if μ=100, σ=15 (like IQ scores), about 68% of scores fall between 85 and 115, and about 95% fall between 70 and 130.
Z-score formula
Answer: z = (x − μ) / σ
Explanation
Converts a value x into the number of standard deviations it lies from the mean, so probabilities can be found using the standard Normal distribution. Example: x=120, μ=100, σ=15 → z = (120−100)/15 ≈ 1.33
What is the standard Normal distribution?
Answer: The Normal distribution with mean μ = 0 and standard deviation σ = 1, denoted N(0, 1)
Explanation
Any Normal variable can be standardized via z = (x−μ)/σ, letting a single table or calculator function find probabilities for any Normal distribution.
Sampling Distributions
Definition: sampling distribution
Answer: The distribution of values taken by a statistic (e.g., p̂ or x̄) across all possible samples of the same size n from the same population
Explanation
It describes sample-to-sample variability of a statistic, not the variability of individual observations within one sample.
Distinguish: parameter vs. statistic
Answer: Parameter: a number describing the population (e.g., p, μ) — fixed, usually unknown. Statistic: a number calculated from a sample (e.g., p̂, x̄) — used to estimate the parameter, and it varies from sample to sample.
Sampling distribution of a sample proportion p̂ — mean and standard deviation
Answer: μ_p̂ = p ; σ_p̂ = √[p(1 − p)/n]
Explanation
p is the true population proportion and n is the sample size. As n increases, σ_p̂ decreases, giving more precise estimates.
Conditions for the sampling distribution of p̂ to be approximately Normal
Answer: Random: data come from a random sample/random assignment. 10% Condition: n ≤ 0.10N, so sampled observations are roughly independent. Large Counts: n·p ≥ 10 and n·(1 − p) ≥ 10
Sampling distribution of a sample mean x̄ — mean and standard deviation
Answer: μ_x̄ = μ ; σ_x̄ = σ / √n
Explanation
μ and σ are the population mean and standard deviation. σ_x̄ is the standard deviation of the sampling distribution; when σ is unknown and estimated by s, this quantity is called the standard error.
Central Limit Theorem & Standard Error
Central Limit Theorem (CLT) — statement
Answer: As the sample size n increases, the sampling distribution of the sample mean x̄ becomes approximately Normal, regardless of the shape of the population distribution, provided n is sufficiently large (commonly n ≥ 30).
Explanation
If the population itself is already Normal, x̄ is exactly Normal for any sample size n.
Conditions for using the CLT / Normal approximation for x̄
Answer: Random: data from a random sample. 10% Condition: n ≤ 0.10N for independence. Large Sample: n ≥ 30, OR the population distribution is already known to be approximately Normal.
Standard error — definition and formulas for p̂ and x̄
Answer: Standard error (SE) is the estimated standard deviation of a sampling distribution, using a sample statistic in place of the unknown population parameter: SE_p̂ = √[p̂(1 − p̂)/n] ; SE_x̄ = s / √n
Explanation
Used whenever the true parameter (p or σ) is unknown, which is the usual case in inference (confidence intervals, hypothesis tests).
Distinguish standard deviation of a statistic vs. standard error
Answer: Standard deviation of the sampling distribution uses the true population parameter (e.g., σ_x̄ = σ/√n); standard error substitutes a sample estimate for the unknown parameter (e.g., SE_x̄ = s/√n)
How does the standard error of x̄ change as sample size n increases (σ = 10)?
Answer: SE decreases as n increases, but at a decreasing rate — SE is proportional to 1/√n, so to cut SE in half you must quadruple the sample size
Explanation
At n=10, SE≈3.16; at n=100, SE≈1.0 (a 10-fold increase in n only reduces SE by about a factor of 3.16).
Conditions Checklist
What three types of conditions should you check before any probability/inference procedure in AP Statistics?
Answer: Random (how the data were collected), Independence/10% Condition (individual observations are roughly independent; sample ≤ 10% of population if sampling without replacement), Normal/Large Counts (a shape condition appropriate to the procedure, e.g., Large Counts np≥10 & n(1−p)≥10 for proportions, or n≥30 / population known Normal for means)
Explanation
Often remembered as the 'Random, Independent, Normal' framework underlying nearly every AP Statistics inference procedure — always state and check all three before proceeding.
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