4.12 The Geometric Distribution

If X is the trial number of the first success and each trial is independent with success probability p, the geometric probability mass function is P(X = x) = (1 − p)^(x−1) · p, for x = 1, 2, 3, … That means you get x−1 failures (each with probability 1−p) then a success (probability p). The geometric distribution has mean μ = 1/p and standard deviation σ = sqrt(1 − p)/p, and it’s memoryless (past failures don’t change future chances). This matches the AP CED UNC-3.E and UNC-3.F learning objectives. For more review and AP-style practice on Topic 4.12, check the Geometric Distribution study guide (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC) and the Unit 4 overview (https://library.fiveable.me/ap-statistics/unit-4). Want extra practice? There are 1000+ AP Stats problems at (https://library.fiveable.me/practice/ap-statistics).

Use a binomial when you have a fixed number of independent Bernoulli trials (n known), and you ask how many successes occur (X = 0,1,...,n). Use a geometric when you repeat independent Bernoulli trials until you get the first success and the random variable is the trial number of that first success (X = 1,2,3,...). Key checks from the CED: trials are independent, two outcomes (success/failure), same success probability p each trial. Binomial PMF: P(X = k) = C(n,k) p^k (1−p)^{n−k}. Geometric PMF: P(X = x) = (1−p)^{x−1} p. Means differ: binomial μ = np, geometric μ = 1/p. Also remember geometric is memoryless (the “clock” restarts). On the AP exam, verify which quantity the question asks for (fixed n vs “first” or “until”), then pick the matching model. For a quick review of geometric problems and AP-aligned examples, see the Topic 4.12 study guide (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC); for unit review and lots of practice, check Unit 4 (https://library.fiveable.me/ap-statistics/unit-4) and practice questions (https://library.fiveable.me/practice/ap-statistics).

Short answer: X counts the trial number of the first success—not the total number of successes. More detail: For a geometric random variable (CED UNC-3.E), X = the trial on which the first success occurs (support x = 1,2,3,...). The PMF is P(X = x) = (1 − p)^{x−1} p. So X = 1 means you got a success on the first trial; X = 4 means the first three were failures and the fourth was the first success. The mean is μ = 1/p and σ = sqrt(1−p)/p (CED UNC-3.F). This is different from a binomial which counts the number of successes in a fixed number of trials—geometric waits for the first success. If you want practice problems or a quick review of these formulas and interpretations, check the topic study guide (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC) and more unit resources (https://library.fiveable.me/ap-statistics/unit-4).

Think of a geometric random variable X as “which trial gives the first success” when you do independent Bernoulli trials with success probability p (CED UNC-3.E). To get P(X = x): - You need the first x−1 trials to be failures. Each failure has probability 1−p, and independence means multiply them: (1−p)^(x−1). - Then trial x must be a success, probability p. - Multiply those: P(X = x) = (1−p)^(x−1) p. The support is x = 1,2,3,... So the formula counts the exact sequence: fail, fail, …, fail (x−1 times), then success. That’s also why the geometric mean is 1/p (UNC-3.F.1): on average you wait 1/p trials for the first success. For more AP-aligned examples and quick practice, see the geometric distribution study guide (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC) and try related practice problems (https://library.fiveable.me/practice/ap-statistics).

Short answer: geometric models the trial number of the first success; negative binomial models the trial number (or total failures) when you get the r-th success. Details you should know for AP Stats (CED-aligned): - Geometric X: number of the trial of the first success. PMF: P(X = x) = (1−p)^{x−1} p for x = 1,2,3,...; mean = 1/p; memoryless property. (CED UNC-3.E, UNC-3.F, UNC-3.G) - Negative binomial: extends that idea to the r-th success. One common form gives the probability the r-th success occurs on trial x (x = r, r+1, ...). Geometric is just the negative binomial with r = 1. - Be careful: some books define the negative binomial as “number of failures before the r-th success” (support and formula shift accordingly). For more review and AP-style examples, see the Topic 4.12 study guide (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC) and try practice problems at (https://library.fiveable.me/practice/ap-statistics).

If X is geometric (number of the trial of the first success) with success probability p (iid Bernoulli trials, support x = 1,2,3,...), the parameters are: - Mean (expected value): μX = 1/p. Interpretation: on average the first success happens on trial 1/p (in units of “trials”). - Standard deviation: σX = sqrt(1 − p) / p. Interpretation: typical spread (in trials) around that average. Reminder: the PMF is P(X = x) = (1 − p)^(x−1) p. These formulas are on the AP formula sheet and in the CED (UNC-3.F.1). Use them only when the trials are independent with constant p. For more explanation and examples, see the Topic 4.12 study guide (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC). Practice problems are helpful—try the AP practice bank (https://library.fiveable.me/practice/ap-statistics).

Use the geometric distribution when a problem describes repeated independent Bernoulli trials (only success/failure), each trial has the same success probability p, and the random variable X is the trial number of the FIRST success. That’s the UNC-3.E situation on the AP CED. Key signs: “What’s the probability the first success occurs on the 3rd trial?” or “How many trials until the first success?” Use P(X = x) = (1−p)^{x−1} p for x = 1,2,3,… (Topic 4.12). Mean = 1/p and SD = sqrt(1−p)/p (UNC-3.F). Don’t use geometric when you’re counting how many successes in a fixed number of trials—that’s binomial. The geometric also has the memoryless property: past failures don’t change future chances. For practice and more examples, check the geometric study guide (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC) and Unit 4 overview (https://library.fiveable.me/ap-statistics/unit-4). You’ll see these exact ideas tested under Skill 3.A and 3.B on the exam.

Think of X as "the trial number of the first success." If the first success happens on trial x, that means the first x−1 trials were failures and trial x was a success. Each failure has probability 1−p and the final success has probability p, so you multiply (1−p) for each of the x−1 failures and then p for the success: P(X = x) = (1−p)^(x−1) p. Using (1−p)^x would mean you had x failures before the success—that can’t happen if the success is on trial x. Quick example: p = 0.2, first success on trial 3 → P = (0.8)^2·0.2 = 0.128. This matches the CED formula (UNC-3.E.2). For more on interpreting and practicing geometric problems, check the Fiveable geometric study guide (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC) and more practice problems (https://library.fiveable.me/practice/ap-statistics).

If “first success occurs after k trials” you want P(X > k) for a geometric X (first-success trial). Use the geometric PMF P(X = x) = (1−p)^{x−1} p (CED UNC-3.E.2) and sum the tail or use the shortcut: P(X > k) = (1−p)^k. Why that works: “first success after k trials” means the first k trials are all failures, each with probability (1−p) and trials are iid (memoryless). Example: p = 0.2, k = 3 → P(first success after 3 trials) = (1−0.2)^3 = 0.8^3 = 0.512. Always check geometric conditions on the exam: independent Bernoulli trials, constant p, support x = 1,2,…. For more practice and a short guide, see the Topic 4.12 study guide (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC). For broader Unit 4 review and lots of practice problems, use (https://library.fiveable.me/ap-statistics/unit-4) and (https://library.fiveable.me/practice/ap-statistics).

Memoryless means that for a geometric random variable (first-success trial X with success prob p), the chance of needing k more trials doesn’t depend on how many failures you’ve already seen. Formally, P(X > n + k | X > n) = P(X > k). Practically: if you’ve already seen 4 failures, the probability the first success happens 3 trials from now is the same as if you were starting fresh and asking for success on trial 3: P(X = 3) = (1−p)^{2} p. How to use it: when a problem conditions on “no success so far,” drop the past and treat the remaining trials like a new geometric process with the same p. That simplifies conditional probability and expectation calculations (mean = 1/p still applies to future waiting time). This is an AP-style tool you’ll see in Topic 4.12 (UNC-3.E, UNC-3.G). For extra practice and examples, check the Topic 4.12 study guide (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC) and more unit practice (https://library.fiveable.me/ap-statistics/unit-4) or practice problems (https://library.fiveable.me/practice/ap-statistics).

Use the geometric PMF and the memoryless idea: P(X > k) means the first success hasn't happened in the first k trials—i.e., the first k trials are all failures. Since trials are independent and failure prob = 1 − p, P(X > k) = (1 − p)^k. Why this works: P(X = x) = (1 − p)^{x−1} p, so summing P(X = k+1) + P(X = k+2) + … equals the probability that the first k trials were failures, which is (1 − p)^k. Example: if p = 0.2, P(X > 3) = (0.8)^3 = 0.512. This is exactly the UNC-3.E memoryless/PMF idea in the CED. For more examples and practice on Topic 4.12, see the Fiveable geometric distribution study guide (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC) and try problems at (https://library.fiveable.me/practice/ap-statistics).

Use x = 1, 2, 3, ... for the AP Stats geometric. The CED defines a geometric random variable X as “the number of the trial on which the first success occurs,” with PMF P(X = x) = (1 − p)^{x−1} p and support x = 1, 2, 3, … (so the mean is 1/p and σ = sqrt(1−p)/p). That’s what the AP exam expects (Topic 4.12, UNC-3.E and UNC-3.F). Why the confusion? Some books define a different version that counts the number of failures before the first success—that one starts at 0 and has PMF P(Y = k) = (1−p)^k p for k = 0,1,2,... But that version isn’t the AP convention, so use the “trial number of first success” form on the exam. For a quick refresher, see the Topic 4.12 study guide (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC). For more practice, check the AP practice problems (https://library.fiveable.me/practice/ap-statistics).

Think of μ = 1/p as the long-run average trial on which the first success happens. If X is geometric (X = trial number of first success, UNC-3.E), then 1/p tells you where the “center” of that distribution sits in the problem’s units—i.e., in number of trials. Example: if p = 0.2 (20% chance each trial), μ = 1/0.2 = 5, so if you repeated the independent Bernoulli process many times you'd expect the first success to occur about on the 5th trial on average. Always state units and context (trials, people, attempts) when you interpret μ, per UNC-3.G.1. For extra practice translating contexts to formulas, check the Geometric Distribution study guide on Fiveable (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC) and try problems from the Unit 4 practice set (https://library.fiveable.me/practice/ap-statistics).

1) Check the experiment: are you repeating independent trials with exactly two outcomes each (success vs failure)? (CED: Bernoulli trials, iid). 2) Ask what the random variable measures: is X the trial number when the FIRST success occurs? If yes, it’s geometric (UNC-3.E.1). 3) Verify constant success probability p for every trial. If p changes, it’s not geometric. 4) Use the PMF: P(X = x) = (1 − p)^(x−1) p for x = 1, 2, 3, … If your probability matches that form, you’ve got a geometric model (UNC-3.E.2). 5) Quick parameter check: mean = 1/p and SD = sqrt(1−p)/p (UNC-3.F.1). If these make sense in context, interpret units accordingly (UNC-3.G.1). 6) If you need more practice or examples, review the Topic 4.12 study guide (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC) and try related practice problems (https://library.fiveable.me/practice/ap-statistics).

The formula comes from the variance of a geometric random variable (first-success version). For a geometric X with P(success)=p, the variance is Var(X) = (1−p)/p^2. The standard deviation is just the square root of variance: σ = sqrt((1−p)/p^2) = sqrt(1−p)/p. Why that variance? Two quick ways to see it: - Use the negative binomial fact: a geometric is a special case of Negative Binomial with r = 1. For NB(r,p), Var = r(1−p)/p^2, so with r = 1 you get (1−p)/p^2. - Or compute directly from the PMF P(X = x) = (1−p)^{x−1}p, find E(X)=1/p and E(X^2) = (2−p)/p^2, and then Var(X)=E(X^2)−[E(X)]^2 = (1−p)/p^2. So intuitively: as p gets smaller, you expect more trials (mean 1/p) and much more variability (variance scales like 1/p^2), and the (1−p) factor adjusts for the chance of failures. For AP review, that standard-deviation formula is in the CED (Topic 4.12); see the Geometric study guide (https://library.fiveable.me/ap-statistics/unit-4/geometric-distribution/study-guide/wcfJTWpCUGP5PNhthkEC). For extra practice, check the AP practice problems (https://library.fiveable.me/practice/ap-statistics).