VectorPaint two-draw probability trees: connect replacement assumptions to four paths

English translation of the Chinese original, translated on 2026-11-08. Original publication dates, attribution, code and version notes are retained.

Every branch weight depends on the drawing rules. Omitting replacement can change interpretation of the second level. State a model with two red objects and one blue, sampled twice independently and uniformly with replacement, then construct four colour paths manually. This is an invented mathematical exercise, not observed sampling or an app-run random experiment or probability calculation.

State object identities and the rule for each draw

List three distinguishable objects externally: red A, red B and blue A. Assume each object is equally likely on every draw, record its colour and replace it immediately before the second draw. Identical independent mechanisms are model assumptions, not something proved by a few physical trials. Display colours while preserving object identities, so red and blue are not mistaken for one object each with half probability. Title the graphic “two draws with replacement: colour outcomes” before adding percentages or conclusions. Different real rules require a separate model, not renamed colours.

Branch the first draw from one root

Create a landscape canvas with one root on the left and first-red/first-blue nodes centrally. Label branches 2/3 and 1/3, reading them explicitly as two thirds and one third to avoid reversing numerator/denominator in prose. Both branches leave the root; blue must not sprout from the red node. Branch length serves layout rather than probability encoding; fractions provide the weights. Check that they sum to one before extending the tree. Include words as well as colours so monochrome viewing remains readable.

Use the same second-level weights after replacement

From first red and first blue, branch to second red and second blue. Replacement means both second-level pairs remain red 2/3 and blue 1/3, without silently using halves below first red. Label leaves RR,RB,BR,BB with order explicitly meaning first then second. RB and BR have the same colour composition but are different ordered paths in this tree. Check outgoing weights sum to one separately at both internal nodes, not only at the root.

Multiply along each path and verify the total

Write each path calculation beside its leaf: RR is 2/3 times 2/3 = 4/9; RB is 2/3 times 1/3 = 2/9; BR is 1/3 times 2/3 = 2/9; BB is 1/3 times 1/3 = 1/9. Multiply along one path rather than adding its branch weights. The four leaves sum to 9/9 = 1; inspect missing or duplicated leaves. These are externally calculated expressions entered as text. Connecting arrows in the application neither assigns conditional probabilities nor performs multiplication.

Identify leaves before aggregating an event

For “exactly one blue”, add an external annotation identifying RB and BR, summing to 4/9. “At least one blue” also includes BB and totals 5/9. These are different conditions and must not share a label. Highlight leaves with boxes or pale bands without hiding connections, retaining root-to-result traceability. Write included path names rather than relying solely on coloured selection. Aggregate disjoint complete paths, without adding an intermediate red node again and double-counting outcomes.

Use a no-replacement control to expose missing assumptions

Save a small separate control explicitly without replacement. After first red, one red and one blue remain, giving second probabilities 1/2 each. After first blue, two reds remain, so second red is one and blue zero. RR,RB,BR therefore each have probability 1/3 and BB zero, unlike the original. Preserve the replacement diagram and the same colour order. Changing the title alone cannot retain original second-level fractions. If omitting a zero-weight branch visually, state that BB is impossible instead of silently dropping an outcome.

Check fractions and path ownership in output

Save the editable source and inspect each numerator/denominator in a viewing-size PNG. With slash fractions, avoid wrapping 2/3 into unrelated lines; stacked fractions also need clear separation and hierarchy. Trace all four paths, then match event annotations to leaves. Object colours are not a data source, so original colour marks suffice without real-draw photographs. For an invisible connection, inspect stacking and text occlusion before declaring zero probability. Export format does not verify independence assumptions.

Troubleshoot rules, branches, products and sums

If totals fail, inspect missing leaves before recomputing products. If one branching pair does not sum to one, revisit object identities and replacement. If exactly one blue is labelled 5/9, repair event scope rather than changing a leaf to force the total. Unequal colour counts in two trials do not refute these model fractions, and this task contains no observed random-experiment data. Deliver assumptions, four path calculations, both event scopes and the no-replacement control so readers distinguish assumptions, derivations and observations. A completed diagram implies neither app-generated random numbers nor probability guarantees for real decisions.