What does a third independent choice add?
You will be able to: Enumerate ordered outcomes and use a spatial product model without confusing it with probability density.
What does a third independent choice add?
Two coin tosses give HH, HT, TH and TT. A third toss splits each possibility into two, producing eight distinct sequences.
A useful starting point: How do you count outcomes without assuming too much? →
Words and symbols before equations
- Ordered outcome
- A result that preserves which event happened first, second and third.
- Product sample space
- All combinations of one outcome from each component process.
- Independent tosses
- Earlier results do not change later head probabilities.
- Condition
- Information that restricts the eligible outcomes.
What this picture assumes
Three fair, independent coin tosses have eight equally likely sequences. H=1 and T=0 locate the corners of a schematic cube; coordinates are outcomes, not physical distances. Conditioning on the first toss leaves four equally likely sequences.
Read the picture in three steps
- Read the axes and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- 3 matching sequences among 8 eligible sequences. Probability = 3/8 = 0.375. All eight sequences are equally likely.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
The cube assigns one axis to each toss: T=0 and H=1. Each corner represents one sequence such as HTH. Its position records choices, not a physical distance or frequency.
Fair independent tosses give each corner probability (1/2)³=1/8. Exactly two heads occurs at HHT, HTH and THH, so its probability is 3/8. At least two also includes HHH, giving 4/8.
Given that the first toss is H, restrict to four eligible corners: HHH,HHT,HTH,HTT. Three of these have at least two heads, so the conditional probability is 3/4. Camera rotation changes neither the event nor its probability. Unequal or dependent tosses would require different weights.
A worked example, step by step
Find P(exactly one head) for three fair independent tosses, then find it given the first toss is H.
- The complete space has eight equally likely sequences.
- Exactly one head occurs at HTT,THT,TTH, so P=3/8.
- Given first H, the eligible sequences are HHH,HHT,HTH,HTT.
- Only HTT qualifies, so the conditional probability is 1/4.
The cube has eight outcome points, not a filled volume of continuous probability. Conditioning changes the reference set.
Does HHT represent the same ordered result as HTH?
Compare with an explanation
No. The tail occurs on a different toss, so the outcomes are distinct.
Predict. Change one thing. Explain.
Select exactly one head and turn on the first-H condition. Rotate the cube to inspect the four eligible corners; confirm the fraction using the complete 2D outcome list.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
3 matching sequences among 8 eligible sequences. Probability = 3/8 = 0.375. All eight sequences are equally likely.
| Sequence | Eligible? | Meets event? | Conditional weight |
|---|---|---|---|
| TTT | Yes | No | 1/8 |
| TTH | Yes | Yes | 1/8 |
| THT | Yes | Yes | 1/8 |
| THH | Yes | No | 1/8 |
| HTT | Yes | Yes | 1/8 |
| HTH | Yes | No | 1/8 |
| HHT | Yes | No | 1/8 |
| HHH | Yes | No | 1/8 |
Three fair, independent coin tosses have eight equally likely sequences. H=1 and T=0 locate the corners of a schematic cube; coordinates are outcomes, not physical distances. Conditioning on the first toss leaves four equally likely sequences.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using the probability values, reference groups, graph scales or model assumptions. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionFor three fair independent tosses, find P(no heads), P(at least one head), and P(no heads given first H).
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: No heads is TTT: probability 1/8.
- 1 point: At least one is its complement: 7/8.
- 1 point: Given first H, all eligible sequences already contain a head.
- 1 point: Thus the conditional probability of no heads is 0.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What do cube axes represent?
The outcomes of tosses 1, 2 and 3.
RECALL 2What changes when the camera rotates?
Only the viewing angle.
RECALL 3Why are the eight corners equally weighted?
The tosses are fair and independent.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
What does a third independent choice add?
- Three binary choices give 2³=8 ordered outcomes.
- Each has probability 1/8 only for fair independent tosses.
- Condition first, then count within the eligible set.
Remember: The cube has eight outcome points, not a filled volume of continuous probability. Conditioning changes the reference set.
Conditions: Three fair, independent coin tosses have eight equally likely sequences. H=1 and T=0 locate the corners of a schematic cube; coordinates are outcomes, not physical distances. Conditioning on the first toss leaves four equally likely sequences.
Refresh Kid · AP Statistics Unit 2 · Objectives 2.4.A · Review edition
Framework, scope and review status
Mapped to College Board, AP Statistics CED, Topic 2.4, objectives 2.4.A. Framework effective Fall 2026, checked September 17, 2026. Unit 2 includes probability, random variables, probability models and introductory sampling distributions; it is part of the revised five-unit course.
Examples and datasets are synthetic, independently authored teaching material. Assumptions about independence, replacement and equal likelihood are stated before calculations. Simulation estimates fluctuate. Discrete probability is summed; continuous probability is area. Sampling distributions and randomization distributions use different repetition mechanisms. Formal inference comes in later units.
The Organic Chemistry Tutor companion title and destination were checked; the full video was not reviewed. Khan Academy’s destination was checked, but its lesson content was not fully readable by the research tool. OpenStax provides optional reference reading. No provider scripts, questions or graphics were copied. Refresh Kid is not affiliated with these providers.
GitHub’s 3D website collection informed optional spatial inspection. Our original three-toss outcome cube uses self-hosted Three.js with its MIT license. Eight corners represent eight equally likely sequences of three fair independent tosses. Conditioning removes ineligible sequences; camera rotation never changes probabilities. Quantitative graphs remain 2D to avoid perspective distortion. Complete labeled diagrams, outcome lists and explanations remain available without 3D.
Independent teacher review and observation of students remain pending. Technical checks do not certify statistical accuracy, accessibility or learning effectiveness. This is a review edition.
Released AP Statistics questions and scoring guides are optional. Older exams use the earlier framework, so check alignment before selecting parts. All practice on this page is original, not official AP material.
Learn → Explore → Practice → Review is informed by the IES learning guide. This implementation has not yet been evaluated with learners.
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