Read sampling without replacement with Book Isle: update the remaining balls at every level

A probability tree often goes wrong when its second level retains original counts or a conditional branch is mistaken for a complete-path probability. Read a problem in Book Isle and work through an original two-red/one-blue, two-draw, without-replacement example in your notes. Its conditions differ from replacement sampling, so answers cannot be interchanged. The app does not calculate probabilities: draw a tree on paper and record results in words, retaining the problem, remaining states and reasons for multiplication or addition.

Record randomness and no replacement in the problem

Locate a relevant problem in a lawfully obtained text, recording title, edition, chapter and problem number with an available return bookmark. Our invented bag contains equal-sized red balls R1 and R2 and blue ball B. Each draw selects uniformly from remaining balls, without replacement, for two ordered draws. R1 and R2 share a colour but retain identities for counting. Write uniform selection, order and no replacement at the start, not merely two red/one blue. Keep replacement or simultaneous-draw problems separate rather than changing them to fit this tree; their sample spaces may differ.

Branch red and blue from the three-ball state

Write the first-red probability explicitly as 2/3, meaning two thirds; first blue is 1/3, summing to one. Two counts red balls and three all balls, not the number of draws. Record root state two red/one blue and label both branches. If assigning 1/2 to each colour, return to uniformly selected individual balls: colour categories have unequal counts. Two colours do not imply equal probabilities, and two first-level branches do not force every later split in half. This example includes certain and impossible second-level outcomes.

After red, retain one red and one blue

On the first-red branch, one red and one blue remain, total two. Second red and second blue each have probability 1/2. Write that remaining state at the node: the denominator changes from three to two because one ball was removed without replacement. A second-red 2/3 accidentally restores the removed ball; 1/3 changes only the numerator. Repair the model from the problem rather than overwriting a final answer. This 1/2 is conditional on first red, not the probability of red on both draws.

After blue, the next draw must be red

On the first-blue branch, two red and zero blue remain, total two. Second red is 2/2=1 and second blue 0/2=0. Keep a zero-probability blue branch labelled impossible in a diagnostic tree or omit it visually, but explain why blue-blue cannot occur. A blue first-level branch does not require a nonzero blue second-level branch. If second red is written 1/2, reread R1 and R2: two individual outcomes belong to one colour event. Conditions changed, so the second draw is not an independent copy of the first. This is the crucial difference from replacement sampling.

Multiply along paths for ordered colour results

Multiply full paths: red-red is 2/3×1/2=1/3; red-blue is also 1/3; blue-red is 1/3×1=1/3; blue-blue is 1/3×0=0. Retain both factors to identify the second-level condition. Do not add segments along one path. Addition combines mutually exclusive departures at a node or distinct paths belonging to an event. A red-red result exceeding first-red probability 2/3 signals a multiplication/addition error. All four ordered colour paths must sum to one. The zero path contributes nothing but identifies where an impossible result is excluded.

State whether the event distinguishes order

Exactly one red and one blue includes red-blue and blue-red, giving 1/3+1/3=2/3. First red then blue includes only red-blue, probability 1/3. At least one blue is also 2/3 here because the single blue ball makes two blues impossible; this equality is not universal. Contrast second-red given first-blue, probability one, with the complete blue-red path, probability 1/3, keeping conditional probability distinct from a joint event. Retain the question’s event wording beside every result so later review does not leave unidentified fractions.

Recount six labelled elementary paths

List six ordered individual paths: R1 then R2, R1 then B, R2 then R1, R2 then B, B then R1, B then R2. Each is 1/3×1/2=1/6. Red-red, red-blue and blue-red each contain two, hence each totals 1/3. B then B is absent because the blue ball left the bag. Merely merging labels into three colour classes does not establish equal probability; count their equally likely constituent paths first. Equal class sizes happen here and are not universal. Extra weights or size differences in another problem require checking rather than assuming every path is 1/6.

Reread three conditions after returning

Save problem location, three assumptions, remaining node states, path products and event combinations. Exit reading and return to the same problem number. Without the answer, explain why the second denominator is two, why second blue after first blue is zero and why exactly one of each colour combines two paths. Revisit the relevant node if unclear rather than merely recalculating a final fraction. Manually record short conditions and location if selection is unavailable; do not claim automatic formula capture. Keep replacement practice in a clearly labelled separate record instead of overwriting this one. Their value lies in differing conditions, not matching answers.

Related tutorials and sources

Read probability examples in Book Isle: separate mutually exclusive and independent events with four outcomes; Follow worked examples in Book Isle: track symbols, units and substitutions.

Sources checked on 8 November 2026: the official Book Isle App Store listing and Apple Lookup show version 1.1.4, requiring iOS / iPadOS 15.6 or later. This workflow uses local reading, search, bookmarks and notes for manual checking. Readers record the samples and conclusions; automatic chart recognition, statistical testing or proof verification is not assumed. Confirm available controls against the installed version and file type. The exercise does not claim step-by-step device testing in this batch.

Method reference: OpenStax on tree and Venn diagrams distinguishes replacement from no replacement and explains multiplication along paths. The two-red/one-blue example and six labelled paths are original practice material.