Why multiply along a probability tree?
You will be able to: Use conditional branches for sequential draws with and without replacement.
Why multiply along a probability tree?
A bag contains 3 orange and 2 blue tokens. Drawing one token changes what remains unless it is returned before the second draw.
A useful starting point: What changes when you learn an event occurred? →
Words and symbols before equations
- Branch probability
- A probability conditional on the path already followed.
- Replacement
- Returning the first item before drawing again.
- Path probability
- The probability of a complete ordered sequence.
- General multiplication rule
- P(A∩B)=P(A)P(B∣A).
What this picture assumes
The bag initially contains 3 orange and 2 blue objects. Each draw is random. Without replacement, the first object stays out; with replacement it is returned and the bag is remixed.
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.
- Without replacement: P(OO) = 0.3. Mixed colors = P(OB)+P(BO) = 0.6. All four path probabilities sum to 1. O means orange; B means blue.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the mathematics
Without replacement, an orange first draw has probability 3/5. Then only 2 orange tokens remain among 4, so P(second orange given first orange)=2/4.
Multiply along the orange–orange path: (3/5)(2/4)=3/10. To find one token of each color in either order, add the disjoint orange–blue and blue–orange paths.
With replacement and thorough remixing, the second color probabilities remain 3/5 and 2/5. Returning the token restores the composition; the model also assumes independent random draws.
A worked example, step by step
Find the probability of exactly one orange in two draws from 3 orange and 2 blue tokens, without replacement.
- Orange then blue has probability (3/5)(2/4)=3/10.
- Blue then orange has probability (2/5)(3/4)=3/10.
- These ordered paths cannot both describe the same two-draw result, so add them.
- The probability of exactly one orange is 3/10+3/10=3/5.
Do not multiply unchanged marginal probabilities when an earlier draw changes the composition.
Why is the second denominator 4 without replacement?
Compare with an explanation
Only four tokens remain after the first token is removed.
Predict. Change one thing. Explain.
Toggle replacement. Compare the orange–orange path with the two mixed-color paths. Explain each second-draw denominator.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Without replacement: P(OO) = 0.3. Mixed colors = P(OB)+P(BO) = 0.6. All four path probabilities sum to 1. O means orange; B means blue.
| Path | Probability |
|---|---|
| OO | 0.3 |
| OB | 0.3 |
| BO | 0.3 |
| BB | 0.1 |
The bag initially contains 3 orange and 2 blue objects. Each draw is random. Without replacement, the first object stays out; with replacement it is returned and the bag is remixed.
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 questionA bag has 4 green and 2 white tokens. Find P(two white) and P(one of each) without replacement.
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Compare with the answer and four-point rubric
- 1 point: Two white: (2/6)(1/5)=1/15.
- 1 point: Green then white: (4/6)(2/5)=4/15.
- 1 point: White then green: (2/6)(4/5)=4/15.
- 1 point: One of each: 8/15, adding the disjoint ordered paths.
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 labels a second-level branch?
A conditional probability given the first result.
RECALL 2Why add path probabilities?
Distinct complete paths are disjoint outcomes.
RECALL 3What does replacement change?
It restores the original composition before the next draw.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why multiply along a probability tree?
- Multiply conditional probabilities along a path.
- Add disjoint paths making the event.
- At every node, outgoing branch probabilities sum to 1.
Remember: Do not multiply unchanged marginal probabilities when an earlier draw changes the composition.
Conditions: The bag initially contains 3 orange and 2 blue objects. Each draw is random. Without replacement, the first object stays out; with replacement it is returned and the bag is remixed.
Refresh Kid · AP Statistics Unit 2 · Objectives 2.6.A · Review edition
Framework, scope and review status
Mapped to College Board, AP Statistics CED, Topic 2.6, objectives 2.6.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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