Geometric Distribution: Formula and Legacy AP Statistics Guide
Study the geometric model as legacy AP Statistics enrichment while clearly separating it from the revised 2026–27 tested course. This page is legacy/enrichment rather than current-core 2027 exam content.
Direct answer
geometric distribution: The geometric distribution models the trial number of the first success under independent trials with constant success probability p. It is useful enrichment but was removed from the revised AP Statistics course effective 2026–27.
Quick reference: Geometric Distribution: Formula and Legacy AP Statistics Guide
| First success on trial x | P(X=x)=(1−p)ˣ⁻¹p |
|---|---|
| Mean | 1p |
Geometric Distribution: Formula and Legacy AP Statistics Guide: complete lesson
The geometric model describes the trial number of the first success
The geometric distribution models X, the number of independent Bernoulli trials required to obtain the first success, when the probability of success p remains constant from trial to trial. The support is 1, 2, 3, … because at least one trial is required. This page is intentionally labeled legacy/enrichment: College Board removed the geometric distribution from the revised AP Statistics course beginning in 2026–27, so it should not be treated as a current-core 2027 exam topic.
The model remains useful for understanding waiting-time probability, independence, complements, and the contrast between “first success” and “number of successes.” Those ideas transfer to current probability reasoning even though the named geometric topic is no longer in the tested framework.
Check the geometric conditions before using a formula
A geometric setting requires binary trial outcomes that can be classified as success/failure, independence or a defensible approximation to independence, a constant probability p of success, and a variable that counts trials until the first success. If the probability changes as trials proceed, or if the question asks how many successes occur in a fixed number of trials, a geometric model is not appropriate.
Sampling without replacement from a small finite population can violate constant probability and independence. For enrichment exercises, a large population relative to the number of draws can justify an approximation, but the model assumptions should be stated rather than silently ignored.
Probability of the first success on trial k
To have the first success on trial k, the first k−1 trials must be failures and trial k must be a success. Therefore P(X=k)=(1−p)k−1p. The order is built into the event; there is no combination factor because only one sequence qualifies: failure, failure, …, failure, success.
For p=0.30, P(X=4)=0.70³(0.30)=0.1029. A common error multiplies by 4 or uses a binomial coefficient, which would count sequences where the success appears earlier. Those sequences do not satisfy “first success on trial 4.”
Cumulative geometric probabilities are easiest through complements
P(X>k) means no success occurs in the first k trials, so P(X>k)=(1−p)k. Therefore P(X≤k)=1−(1−p)k. This complement form is often cleaner than summing P(X=1)+…+P(X=k), and it highlights the waiting-time meaning of the model.
Be careful with boundaries: “more than 5 trials” means X>5 and requires five failures before trial 6 begins; “at least 5 trials” means X≥5 and requires the first four trials to fail. Because X is discrete, that one-trial boundary changes the exponent.
Mean and standard deviation summarize waiting time
For a geometric random variable counting trials until first success, μ=1/p and σ=√(1−p)/p². If p=0.20, the expected waiting time is 5 trials. That is a long-run average across many independent waiting periods, not a prediction that every first success occurs on the fifth trial.
As p decreases, both the expected wait and its variability increase. Rare successes create long right tails: X can be arbitrarily large even though the probability of very long waits decays geometrically.
The memoryless property is distinctive
The geometric distribution is memoryless: conditional on having already observed failures, the remaining waiting time has the same geometric form. Algebraically, P(X>s+t | X>s)=P(X>t). If a component has a constant 0.1 chance of failure on each independent cycle and it has survived 20 cycles, the model does not make failure on cycle 21 more likely merely because of age. Real physical systems often do age, so the geometric model can be inappropriate despite the elegant property.
Geometric versus binomial is a question about what is fixed
A binomial setting fixes the number of trials n and counts successes. A geometric setting fixes the target of the first success and lets the number of trials vary. “How many defective items appear among the next 20?” is binomial-type counting. “How many items are inspected until the first defective?” is geometric waiting time.
This distinction is more reliable than memorizing formulas. Identify the random variable in words before calculating. If the variable is a trial number, geometric may fit; if it is a number of successes among a predetermined number of trials, binomial is the natural comparison.
Worked geometric-distribution examples
Legacy example 1: First qualified lead
Let X count sales calls until the first qualified lead with constant success probability p=0.180. Under the independence and constant-p assumptions, X has a geometric model. The phrase “until the first” is the structural clue; if the real process changes probability over time, this calculation becomes an approximation rather than a guaranteed model.
P(X=5)=(1−0.180)4(0.180)=0.08138. The cumulative probability of a success within the first 5 trials is 1−(1−0.180)5=0.62926, while the probability of needing more than 5 trials is 0.37074.
The long-run mean waiting time is 1/p=5.556 trials, and the standard deviation is about 5.031 trials. These summaries describe repeated waiting periods; they do not say the next first success will occur near the mean.
Legacy example 2: First defective chip
Let X count chips inspected until the first defective chip with constant success probability p=0.040. Under the independence and constant-p assumptions, X has a geometric model. The phrase “until the first” is the structural clue; if the real process changes probability over time, this calculation becomes an approximation rather than a guaranteed model.
P(X=6)=(1−0.040)5(0.040)=0.03261. The cumulative probability of a success within the first 6 trials is 1−(1−0.040)6=0.21724, while the probability of needing more than 6 trials is 0.78276.
The long-run mean waiting time is 1/p=25.000 trials, and the standard deviation is about 24.495 trials. These summaries describe repeated waiting periods; they do not say the next first success will occur near the mean.
Legacy example 3: First made free throw
Let X count shots until the first made free throw with constant success probability p=0.720. Under the independence and constant-p assumptions, X has a geometric model. The phrase “until the first” is the structural clue; if the real process changes probability over time, this calculation becomes an approximation rather than a guaranteed model.
P(X=4)=(1−0.720)3(0.720)=0.01581. The cumulative probability of a success within the first 4 trials is 1−(1−0.720)4=0.99385, while the probability of needing more than 4 trials is 0.00615.
The long-run mean waiting time is 1/p=1.389 trials, and the standard deviation is about 0.735 trials. These summaries describe repeated waiting periods; they do not say the next first success will occur near the mean.
Legacy example 4: First late train
Let X count days observed until the first late train with constant success probability p=0.120. Under the independence and constant-p assumptions, X has a geometric model. The phrase “until the first” is the structural clue; if the real process changes probability over time, this calculation becomes an approximation rather than a guaranteed model.
P(X=5)=(1−0.120)4(0.120)=0.07196. The cumulative probability of a success within the first 5 trials is 1−(1−0.120)5=0.47227, while the probability of needing more than 5 trials is 0.52773.
The long-run mean waiting time is 1/p=8.333 trials, and the standard deviation is about 7.817 trials. These summaries describe repeated waiting periods; they do not say the next first success will occur near the mean.
Legacy example 5: First survey completion
Let X count contacts until the first completed survey with constant success probability p=0.350. Under the independence and constant-p assumptions, X has a geometric model. The phrase “until the first” is the structural clue; if the real process changes probability over time, this calculation becomes an approximation rather than a guaranteed model.
P(X=6)=(1−0.350)5(0.350)=0.04061. The cumulative probability of a success within the first 6 trials is 1−(1−0.350)6=0.92458, while the probability of needing more than 6 trials is 0.07542.
The long-run mean waiting time is 1/p=2.857 trials, and the standard deviation is about 2.304 trials. These summaries describe repeated waiting periods; they do not say the next first success will occur near the mean.
Legacy example 6: First machine alarm
Let X count cycles until the first alarm with constant success probability p=0.080. Under the independence and constant-p assumptions, X has a geometric model. The phrase “until the first” is the structural clue; if the real process changes probability over time, this calculation becomes an approximation rather than a guaranteed model.
P(X=4)=(1−0.080)3(0.080)=0.06230. The cumulative probability of a success within the first 4 trials is 1−(1−0.080)4=0.28361, while the probability of needing more than 4 trials is 0.71639.
The long-run mean waiting time is 1/p=12.500 trials, and the standard deviation is about 11.990 trials. These summaries describe repeated waiting periods; they do not say the next first success will occur near the mean.
Legacy example 7: First six on a die
Let X count die rolls until the first six with constant success probability p=0.167. Under the independence and constant-p assumptions, X has a geometric model. The phrase “until the first” is the structural clue; if the real process changes probability over time, this calculation becomes an approximation rather than a guaranteed model.
P(X=5)=(1−0.167)4(0.167)=0.08038. The cumulative probability of a success within the first 5 trials is 1−(1−0.167)5=0.59812, while the probability of needing more than 5 trials is 0.40188.
The long-run mean waiting time is 1/p=6.000 trials, and the standard deviation is about 5.477 trials. These summaries describe repeated waiting periods; they do not say the next first success will occur near the mean.
Legacy example 8: First successful login
Let X count attempts until the first successful login with constant success probability p=0.650. Under the independence and constant-p assumptions, X has a geometric model. The phrase “until the first” is the structural clue; if the real process changes probability over time, this calculation becomes an approximation rather than a guaranteed model.
P(X=6)=(1−0.650)5(0.650)=0.00341. The cumulative probability of a success within the first 6 trials is 1−(1−0.650)6=0.99816, while the probability of needing more than 6 trials is 0.00184.
The long-run mean waiting time is 1/p=1.538 trials, and the standard deviation is about 0.910 trials. These summaries describe repeated waiting periods; they do not say the next first success will occur near the mean.
Legacy example 9: First rain day
Let X count independent forecast days until the first rain day with constant success probability p=0.220. Under the independence and constant-p assumptions, X has a geometric model. The phrase “until the first” is the structural clue; if the real process changes probability over time, this calculation becomes an approximation rather than a guaranteed model.
P(X=4)=(1−0.220)3(0.220)=0.10440. The cumulative probability of a success within the first 4 trials is 1−(1−0.220)4=0.62985, while the probability of needing more than 4 trials is 0.37015.
The long-run mean waiting time is 1/p=4.545 trials, and the standard deviation is about 4.014 trials. These summaries describe repeated waiting periods; they do not say the next first success will occur near the mean.
Legacy example 10: First accepted application
Let X count reviewed applications until the first acceptance with constant success probability p=0.270. Under the independence and constant-p assumptions, X has a geometric model. The phrase “until the first” is the structural clue; if the real process changes probability over time, this calculation becomes an approximation rather than a guaranteed model.
P(X=5)=(1−0.270)4(0.270)=0.07668. The cumulative probability of a success within the first 5 trials is 1−(1−0.270)5=0.79269, while the probability of needing more than 5 trials is 0.20731.
The long-run mean waiting time is 1/p=3.704 trials, and the standard deviation is about 3.164 trials. These summaries describe repeated waiting periods; they do not say the next first success will occur near the mean.
Legacy example 11: First correct response
Let X count independent quiz attempts until the first correct response with constant success probability p=0.400. Under the independence and constant-p assumptions, X has a geometric model. The phrase “until the first” is the structural clue; if the real process changes probability over time, this calculation becomes an approximation rather than a guaranteed model.
P(X=6)=(1−0.400)5(0.400)=0.03110. The cumulative probability of a success within the first 6 trials is 1−(1−0.400)6=0.95334, while the probability of needing more than 6 trials is 0.04666.
The long-run mean waiting time is 1/p=2.500 trials, and the standard deviation is about 1.936 trials. These summaries describe repeated waiting periods; they do not say the next first success will occur near the mean.
Legacy example 12: First returned package
Let X count shipments observed until the first return with constant success probability p=0.060. Under the independence and constant-p assumptions, X has a geometric model. The phrase “until the first” is the structural clue; if the real process changes probability over time, this calculation becomes an approximation rather than a guaranteed model.
P(X=4)=(1−0.060)3(0.060)=0.04984. The cumulative probability of a success within the first 4 trials is 1−(1−0.060)4=0.21925, while the probability of needing more than 4 trials is 0.78075.
The long-run mean waiting time is 1/p=16.667 trials, and the standard deviation is about 16.159 trials. These summaries describe repeated waiting periods; they do not say the next first success will occur near the mean.
Legacy waiting-time laboratory
Legacy waiting-time laboratory 1: school survey
For each waiting-time setting, identify success, p, independence, and the event that ends the process. Contrast P(X=k), P(X≤k), and P(X>k) so that the exponent follows from the number of required failures rather than from memorized syntax. In this school survey, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Legacy waiting-time laboratory 1: school survey,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Legacy waiting-time laboratory 1: school survey,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Legacy waiting-time laboratory 2: public-health study
For each waiting-time setting, identify success, p, independence, and the event that ends the process. Contrast P(X=k), P(X≤k), and P(X>k) so that the exponent follows from the number of required failures rather than from memorized syntax. In this public-health study, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Legacy waiting-time laboratory 2: public-health study,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Legacy waiting-time laboratory 2: public-health study,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Legacy waiting-time laboratory 3: manufacturing process
For each waiting-time setting, identify success, p, independence, and the event that ends the process. Contrast P(X=k), P(X≤k), and P(X>k) so that the exponent follows from the number of required failures rather than from memorized syntax. In this manufacturing process, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Legacy waiting-time laboratory 3: manufacturing process,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Legacy waiting-time laboratory 3: manufacturing process,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Legacy waiting-time laboratory 4: transportation system
For each waiting-time setting, identify success, p, independence, and the event that ends the process. Contrast P(X=k), P(X≤k), and P(X>k) so that the exponent follows from the number of required failures rather than from memorized syntax. In this transportation system, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Legacy waiting-time laboratory 4: transportation system,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Legacy waiting-time laboratory 4: transportation system,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Legacy waiting-time laboratory 5: consumer study
For each waiting-time setting, identify success, p, independence, and the event that ends the process. Contrast P(X=k), P(X≤k), and P(X>k) so that the exponent follows from the number of required failures rather than from memorized syntax. In this consumer study, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Legacy waiting-time laboratory 5: consumer study,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Legacy waiting-time laboratory 5: consumer study,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Legacy waiting-time laboratory 6: environmental monitoring
For each waiting-time setting, identify success, p, independence, and the event that ends the process. Contrast P(X=k), P(X≤k), and P(X>k) so that the exponent follows from the number of required failures rather than from memorized syntax. In this environmental monitoring, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Legacy waiting-time laboratory 6: environmental monitoring,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Legacy waiting-time laboratory 6: environmental monitoring,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Legacy waiting-time laboratory 7: sports analysis
For each waiting-time setting, identify success, p, independence, and the event that ends the process. Contrast P(X=k), P(X≤k), and P(X>k) so that the exponent follows from the number of required failures rather than from memorized syntax. In this sports analysis, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Legacy waiting-time laboratory 7: sports analysis,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Legacy waiting-time laboratory 7: sports analysis,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Legacy waiting-time laboratory 8: education program
For each waiting-time setting, identify success, p, independence, and the event that ends the process. Contrast P(X=k), P(X≤k), and P(X>k) so that the exponent follows from the number of required failures rather than from memorized syntax. In this education program, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Legacy waiting-time laboratory 8: education program,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Legacy waiting-time laboratory 8: education program,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Legacy waiting-time laboratory 9: service operation
For each waiting-time setting, identify success, p, independence, and the event that ends the process. Contrast P(X=k), P(X≤k), and P(X>k) so that the exponent follows from the number of required failures rather than from memorized syntax. In this service operation, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Legacy waiting-time laboratory 9: service operation,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Legacy waiting-time laboratory 9: service operation,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Legacy waiting-time laboratory 10: technology experiment
For each waiting-time setting, identify success, p, independence, and the event that ends the process. Contrast P(X=k), P(X≤k), and P(X>k) so that the exponent follows from the number of required failures rather than from memorized syntax. In this technology experiment, write the target in words before using notation so the calculation remains tied to the variable, population, or model actually being studied.
A high-quality solution should also include a self-check tailored to this topic: compare the sign, units, boundary, probability range, or design logic with what the scenario makes plausible. If the numerical output contradicts that check, revisit the setup before changing the conclusion. In “Legacy waiting-time laboratory 10: technology experiment,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Finally, distinguish what the statistical method establishes from what the study design does not establish. The strongest response gives the numerical or graphical evidence and then limits the claim to the population, process, association, or legacy-enrichment scope justified by the data-generating mechanism. In “Legacy waiting-time laboratory 10: technology experiment,” apply this check to the named variables and numerical direction rather than treating it as a reusable sentence from another exercise. Count the failures implied by the boundary before choosing the exponent, and keep this waiting-time calculation labeled as legacy enrichment for the revised AP course.
Geometric Distribution: Formula and Legacy AP Statistics Guide: 22 multiple-choice questions
These legacy/enrichment items practice waiting-time formulas and model conditions; they are not presented as current 2027 tested content.
Question 1. Geometric Distribution (Legacy/Enrichment)
Independent trials at a public high school in Metro East during a two-month observation window have success probability 0.309. Let X be the trial of the first success. Find P(X=7) and E(X), noting the 2027 status of this topic.
Answer: C
P(X=7)=(1−p)6p=(0.691)6(0.309)=0.0336. E(X)=1p=3.236 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 2. Geometric Distribution (Legacy/Enrichment)
Independent trials at a wildlife clinic in Capital Region during a multiweek validation study have success probability 0.35. Let X be the trial of the first success. Find P(X=8) and E(X), noting the 2027 status of this topic.
Answer: D
P(X=8)=(1−p)7p=(0.65)7(0.35)=0.0172. E(X)=1p=2.857 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 3. Geometric Distribution (Legacy/Enrichment)
Independent trials at a grocery cooperative in North Valley during a six-week field trial have success probability 0.467. Let X be the trial of the first success. Find P(X=6) and E(X), noting the 2027 status of this topic.
Answer: A
P(X=6)=(1−p)5p=(0.533)5(0.467)=0.0201. E(X)=1p=2.141 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 4. Geometric Distribution (Legacy/Enrichment)
Independent trials at a community college in Coastal Plains during a monthly quality review have success probability 0.424. Let X be the trial of the first success. Find P(X=2) and E(X), noting the 2027 status of this topic.
Answer: A
P(X=2)=(1−p)1p=(0.576)1(0.424)=0.2442. E(X)=1p=2.358 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 5. Geometric Distribution (Legacy/Enrichment)
Independent trials at a community bank in Midwest consortium during a spring 2027 pilot have success probability 0.311. Let X be the trial of the first success. Find P(X=2) and E(X), noting the 2027 status of this topic.
Answer: D
P(X=2)=(1−p)1p=(0.689)1(0.311)=0.2143. E(X)=1p=3.215 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 6. Geometric Distribution (Legacy/Enrichment)
Independent trials at a recycling program in Mountain Region during a summer implementation review have success probability 0.48. Let X be the trial of the first success. Find P(X=6) and E(X), noting the 2027 status of this topic.
Answer: B
P(X=6)=(1−p)5p=(0.52)5(0.48)=0.0182. E(X)=1p=2.083 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 7. Geometric Distribution (Legacy/Enrichment)
Independent trials at a school district in South Harbor during a baseline measurement week have success probability 0.193. Let X be the trial of the first success. Find P(X=2) and E(X), noting the 2027 status of this topic.
Answer: B
P(X=2)=(1−p)1p=(0.807)1(0.193)=0.1558. E(X)=1p=5.181 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 8. Geometric Distribution (Legacy/Enrichment)
Independent trials at a municipal water office in Desert County during a semester-long cohort study have success probability 0.236. Let X be the trial of the first success. Find P(X=8) and E(X), noting the 2027 status of this topic.
Answer: C
P(X=8)=(1−p)7p=(0.764)7(0.236)=0.0359. E(X)=1p=4.237 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 9. Geometric Distribution (Legacy/Enrichment)
Independent trials at a city recreation department in Cedar Grove during a yearly program evaluation have success probability 0.518. Let X be the trial of the first success. Find P(X=6) and E(X), noting the 2027 status of this topic.
Answer: A
P(X=6)=(1−p)5p=(0.482)5(0.518)=0.0135. E(X)=1p=1.931 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 10. Geometric Distribution (Legacy/Enrichment)
Independent trials at a recycling program in Coastal Plains during a follow-up evaluation period have success probability 0.391. Let X be the trial of the first success. Find P(X=3) and E(X), noting the 2027 status of this topic.
Answer: C
P(X=3)=(1−p)2p=(0.609)2(0.391)=0.145. E(X)=1p=2.558 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 11. Geometric Distribution (Legacy/Enrichment)
Independent trials at a regional hospital in Lakeside district during a winter readiness review have success probability 0.22. Let X be the trial of the first success. Find P(X=8) and E(X), noting the 2027 status of this topic.
Answer: D
P(X=8)=(1−p)7p=(0.78)7(0.22)=0.0386. E(X)=1p=4.545 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 12. Geometric Distribution (Legacy/Enrichment)
Independent trials at a housing authority in Capital Region during a multiweek validation study have success probability 0.33. Let X be the trial of the first success. Find P(X=6) and E(X), noting the 2027 status of this topic.
Answer: A
P(X=6)=(1−p)5p=(0.67)5(0.33)=0.0446. E(X)=1p=3.03 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 13. Geometric Distribution (Legacy/Enrichment)
Independent trials at a university advising center in Capital Region during a yearly program evaluation have success probability 0.235. Let X be the trial of the first success. Find P(X=7) and E(X), noting the 2027 status of this topic.
Answer: A
P(X=7)=(1−p)6p=(0.765)6(0.235)=0.0471. E(X)=1p=4.255 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 14. Geometric Distribution (Legacy/Enrichment)
Independent trials at a farm cooperative in Cedar Grove during a fall 2026 audit have success probability 0.205. Let X be the trial of the first success. Find P(X=7) and E(X), noting the 2027 status of this topic.
Answer: D
P(X=7)=(1−p)6p=(0.795)6(0.205)=0.0518. E(X)=1p=4.878 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 15. Geometric Distribution (Legacy/Enrichment)
Independent trials at a wildlife clinic in Atlantic Corridor during a community outreach cycle have success probability 0.524. Let X be the trial of the first success. Find P(X=6) and E(X), noting the 2027 status of this topic.
Answer: B
P(X=6)=(1−p)5p=(0.476)5(0.524)=0.0128. E(X)=1p=1.908 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 16. Geometric Distribution (Legacy/Enrichment)
Independent trials at a city recreation department in Pine Ridge during a weekday operations study have success probability 0.47. Let X be the trial of the first success. Find P(X=4) and E(X), noting the 2027 status of this topic.
Answer: C
P(X=4)=(1−p)3p=(0.53)3(0.47)=0.07. E(X)=1p=2.128 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 17. Geometric Distribution (Legacy/Enrichment)
Independent trials at a solar installer in Westview during a two-month observation window have success probability 0.328. Let X be the trial of the first success. Find P(X=7) and E(X), noting the 2027 status of this topic.
Answer: B
P(X=7)=(1−p)6p=(0.672)6(0.328)=0.0302. E(X)=1p=3.049 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 18. Geometric Distribution (Legacy/Enrichment)
Independent trials at a city transit agency in Atlantic Corridor during a multiweek validation study have success probability 0.355. Let X be the trial of the first success. Find P(X=6) and E(X), noting the 2027 status of this topic.
Answer: C
P(X=6)=(1−p)5p=(0.645)5(0.355)=0.0396. E(X)=1p=2.817 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 19. Geometric Distribution (Legacy/Enrichment)
Independent trials at a municipal water office in North Valley during a spring 2027 pilot have success probability 0.47. Let X be the trial of the first success. Find P(X=3) and E(X), noting the 2027 status of this topic.
Answer: B
P(X=3)=(1−p)2p=(0.53)2(0.47)=0.132. E(X)=1p=2.128 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 20. Geometric Distribution (Legacy/Enrichment)
Independent trials at a municipal water office in Riverbend during a fall 2026 audit have success probability 0.399. Let X be the trial of the first success. Find P(X=2) and E(X), noting the 2027 status of this topic.
Answer: C
P(X=2)=(1−p)1p=(0.601)1(0.399)=0.2398. E(X)=1p=2.506 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 21. Geometric Distribution (Legacy/Enrichment)
Independent trials at a solar installer in Capital Region during a service-improvement study have success probability 0.477. Let X be the trial of the first success. Find P(X=5) and E(X), noting the 2027 status of this topic.
Answer: A
P(X=5)=(1−p)4p=(0.523)4(0.477)=0.0357. E(X)=1p=2.096 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Question 22. Geometric Distribution (Legacy/Enrichment)
Independent trials at a regional manufacturer in Midwest consortium during a semester-long cohort study have success probability 0.268. Let X be the trial of the first success. Find P(X=6) and E(X), noting the 2027 status of this topic.
Answer: D
P(X=6)=(1−p)5p=(0.732)5(0.268)=0.0563. E(X)=1p=3.731 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Geometric Distribution: Formula and Legacy AP Statistics Guide: 6 free-response questions
For each legacy free-response prompt, state the success event and constant p assumption, translate the waiting-time boundary carefully, and label the result as enrichment rather than current 2027 tested content.
FRQ set 1: Geometric Distribution (Legacy/Enrichment)
Scenario. Independent trials at a regional airport authority in Pine Ridge during a summer implementation review have success probability 0.16. Let X be the trial of the first success. Find P(X=5) and E(X), noting the 2027 status of this topic.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(X=5)=(1−p)4p=(0.84)4(0.16)=0.0797. E(X)=1p=6.25 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
FRQ set 2: Geometric Distribution (Legacy/Enrichment)
Scenario. Independent trials at a regional airport authority in Midwest consortium during a summer implementation review have success probability 0.473. Let X be the trial of the first success. Find P(X=2) and E(X), noting the 2027 status of this topic.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(X=2)=(1−p)1p=(0.527)1(0.473)=0.2493. E(X)=1p=2.114 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
FRQ set 3: Geometric Distribution (Legacy/Enrichment)
Scenario. Independent trials at a municipal water office in Midwest consortium during a school-year data collection have success probability 0.327. Let X be the trial of the first success. Find P(X=6) and E(X), noting the 2027 status of this topic.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(X=6)=(1−p)5p=(0.673)5(0.327)=0.0451. E(X)=1p=3.058 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
FRQ set 4: Geometric Distribution (Legacy/Enrichment)
Scenario. Independent trials at a municipal water office in South Harbor during a winter readiness review have success probability 0.288. Let X be the trial of the first success. Find P(X=7) and E(X), noting the 2027 status of this topic.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
Geometric Distribution: Formula and Legacy AP Statistics Guide — FRQ set 4: Geometric Distribution (Legacy/Enrichment): P(X=7)=(1−p)6p=(0.712)6(0.288)=0.0375. E(X)=1p=3.472 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
FRQ set 5: Geometric Distribution (Legacy/Enrichment)
Scenario. Independent trials at a municipal water office in Desert County during a school-year data collection have success probability 0.506. Let X be the trial of the first success. Find P(X=2) and E(X), noting the 2027 status of this topic.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(X=2)=(1−p)1p=(0.494)1(0.506)=0.25. E(X)=1p=1.976 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
FRQ set 6: Geometric Distribution (Legacy/Enrichment)
Scenario. Independent trials at a city transit agency in Lakeside district during a summer implementation review have success probability 0.509. Let X be the trial of the first success. Find P(X=6) and E(X), noting the 2027 status of this topic.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(X=6)=(1−p)5p=(0.491)5(0.509)=0.0145. E(X)=1p=1.965 trials. Geometric distribution is legacy/enrichment, not required core content for the revised May 2027 exam.
Continue with the next connected AP Statistics skill
The most useful next step is to connect this topic to a neighboring method rather than repeating the same question type indefinitely.
