Does knowing one event change the other’s chance?
You will be able to: Check independence with a conditional or product calculation.
Does knowing one event change the other’s chance?
A fair coin toss and an independent die roll can both meet your event definitions. Independence is about unchanged probabilities, not about events being unable to coincide.
A useful starting point: Why multiply along a probability tree? →
Words and symbols before equations
- Independent events
- Learning one event occurred does not change the other’s probability.
- Product criterion
- Independence requires P(A∩B)=P(A)P(B).
- Dependent events
- The independence criterion fails.
- Marginal probability
- An event’s probability without conditioning on the other.
What this picture assumes
P(A)=0.40 and P(B)=0.30 are fixed. The intersection ranges from 0 to 0.30. Rectangular region areas encode probabilities in this synthetic model; it is not a measured population.
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.
- Union 0.58; intersection 0.12. Events are not disjoint and independent in this model. Independence needs intersection 0.12, not zero.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
If P(A)=0.40 and P(B)=0.30, independence would give intersection 0.40×0.30=0.12. Compare that value with the actual intersection before multiplying marginals.
Equivalently, for P(B)>0, compare P(A given B) with P(A). If both are 0.40, the information B does not change the probability of A.
Independent positive-probability events can overlap. Disjoint positive-probability events cannot be independent. Data-table conditional proportions can suggest an association, but an observed sample difference alone is not proof of a population relationship.
| Feature | Disjoint | Independent |
|---|---|---|
| Can both happen? | No | Yes, if both probabilities are positive |
| Intersection | Zero | Product of marginal probabilities |
| Effect of learning B | Excludes A | Leaves probability of A unchanged |
A worked example, step by step
Let P(A)=0.5, P(B)=0.4 and P(A∩B)=0.2. Check independence in two ways.
- The product of marginals is 0.5×0.4=0.2.
- It equals the stated intersection, satisfying the product criterion.
- Also P(A given B)=0.2/0.4=0.5.
- That equals P(A), so the events are independent under this probability model.
Separate descriptions or separate times do not by themselves prove independence.
If an independent fair coin has shown five heads, is a tail now more likely?
Compare with an explanation
No. The next tail probability remains 1/2 under the independent fair-coin model.
Predict. Change one thing. Explain.
Set the overlap to 0.12 and then to 0.20. Compare the product criterion and conditional probability at each setting.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Union 0.58; intersection 0.12. Events are not disjoint and independent in this model. Independence needs intersection 0.12, not zero.
| Quantity | Calculation | Value |
|---|---|---|
| P(A or B) | 0.40 + 0.30 − 0.12 | 0.58 |
| P(A given B) | 0.12 / 0.30 | 0.4 |
| P(B given A) | 0.12 / 0.40 | 0.3 |
| Independence target | 0.40 × 0.30 | 0.12 |
P(A)=0.40 and P(B)=0.30 are fixed. The intersection ranges from 0 to 0.30. Rectangular region areas encode probabilities in this synthetic model; it is not a measured population.
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 questionP(A)=0.6, P(B)=0.5, P(A∩B)=0.2. Determine independence and calculate P(A given B).
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Compare with the answer and four-point rubric
- 1 point: Independent overlap would be 0.6×0.5=0.3.
- 1 point: Actual overlap 0.2 differs, so the events are dependent.
- 1 point: P(A given B)=0.2/0.5=0.4.
- 1 point: This differs from the marginal P(A)=0.6 and confirms the conclusion.
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 1How do you test independence numerically?
Compare the actual intersection with the product of marginals.
RECALL 2Can independent events happen together?
Yes.
RECALL 3Does a streak change the next independent probability?
No, if the model probability stays fixed.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Does knowing one event change the other’s chance?
- Independent iff P(A∩B)=P(A)P(B).
- If P(B)>0, compare P(A∣B) to P(A).
- Disjoint is not independent.
Remember: Separate descriptions or separate times do not by themselves prove independence.
Conditions: P(A)=0.40 and P(B)=0.30 are fixed. The intersection ranges from 0 to 0.30. Rectangular region areas encode probabilities in this synthetic model; it is not a measured population.
Refresh Kid · AP Statistics Unit 2 · Objectives 2.7.A · Review edition
Framework, scope and review status
Mapped to College Board, AP Statistics CED, Topic 2.7, objectives 2.7.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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