How do you count outcomes without assuming too much?
You will be able to: List a sample space and calculate event and complement probabilities.
How do you count outcomes without assuming too much?
A fair six-sided die can land on 1 through 6. “At least 5” includes two outcomes; “not at least 5” includes the other four.
A useful starting point: How can repeated trials estimate a probability? →
Words and symbols before equations
- Sample space
- All possible distinct outcomes of a random process.
- Event
- A collection of outcomes of interest.
- Complement
- All outcomes outside an event, written Aᶜ.
- Equally likely
- Each elementary outcome has the same probability.
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
Probabilities lie between 0 and 1, and the complete sample space has probability 1. If elementary outcomes are equally likely, divide favorable outcomes by all outcomes.
For the die, P(at least 5)=2/6=1/3. The complement has probability 1−1/3=2/3 because the two events partition every possible outcome.
Counting labels alone fails when outcomes are not equally likely. Three coin tosses produce counts 0,1,2,3 heads, but these four counts have probabilities 1/8,3/8,3/8,1/8. List equally likely sequences before combining them.
A worked example, step by step
A uniformly selected card is numbered 1 through 10. Find the probability of a multiple of 3 and its complement.
- The sample space contains ten equally likely card labels.
- Multiples of 3 are 3,6,9: three favorable outcomes.
- P(multiple of 3)=3/10=0.30.
- The complement probability is 1−0.30=0.70, covering the other seven labels.
Favorable-count divided by total-count requires equally likely elementary outcomes.
Are the possible totals from two fair dice equally likely?
Compare with an explanation
No. Different totals arise from different numbers of ordered pairs.
Predict. Change one thing. Explain.
Use the eight equally likely three-toss sequences. Select at least two heads, then identify its complement directly from the unselected sequences.
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 questionOne of 12 numbered tokens is chosen uniformly. Let E be an even number greater than 6. Find E and its complement probability.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: All 12 labels are equally likely.
- 1 point: E={8,10,12}, giving 3 favorable outcomes.
- 1 point: P(E)=3/12=0.25.
- 1 point: P(Eᶜ)=0.75; its outcomes include all labels except 8,10,12.
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 1Why list a sample space?
To identify possible outcomes without omissions or duplicates.
RECALL 2What is an event’s complement?
Every outcome not in the event.
RECALL 3When is simple counting valid?
When the counted elementary outcomes are equally likely.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do you count outcomes without assuming too much?
- 0≤P(A)≤1.
- P(sample space)=1.
- P(Aᶜ)=1−P(A).
Remember: Favorable-count divided by total-count requires equally likely elementary outcomes.
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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