Tree diagrams make the sample space of a two-step experiment visible

What to assume they know

• Listing every outcome: the sample space — sample space, elements, n(S), and reading a size off the experiment
• An event is a selection from the sample space — an event as a subset, and probability by counting
• Theoretical probability: counting favourable outcomes when all are equally likely — the equally likely assumption
• Fair, unbiased, and memoryless: the gambler's fallacy — independence as the chapter uses the word, and the working definition the explanation supplies
• Multiplying two fractions, and adding fractions with a common denominator

What they should be able to do

• State what a multi-step experiment is, using the chapter's own definition
• Read a tree diagram: identify a branch, a path, a step and the outcome column
• Explain why the number of complete paths is the product of the branchings
• State what the number written on a branch is the probability of
• Compute the probability of an outcome by counting paths, and say when that is valid
• Compute the probability of a path by multiplying along it, and show the two agree for a fair coin tossed twice
• Build a tree for an experiment whose branches carry different numbers, and explain why counting paths now gives the wrong answer
• Build a tree for a draw made without replacement, and explain what changes on the second layer
• Compute the size of a multi-step sample space without drawing the tree
• Choose between a tree and a table for a given two-step experiment

Where it usually goes wrong

• "The tree is just a neat way of writing the list." It is the multiplication rule drawn. A list of four items tells you nothing about where four came from.
• "The branch numbers add." At a single split they add to 1. Along a path they multiply. Confusing the two is the standard error, and the fair-coin tree hides it because 1/2 + 1/2 and 1/2 × 1/2 are both easy numbers to write down.
• "Every path is equally likely." Only when every branch at every split is. Exercise Set 7.4 Q1 kills this on the page after the figure.
• "Count the entries in the last column." For the fruit baskets that gives 1/4 in place of 1/6, and for the pens 1/3 in place of 29/81.
• "Without replacement is a different topic." Same tree, different numbers on the second layer. Nothing about the drawing changes.
• "Two steps means multiply the answer by 2." Two steps means multiply the number of branches, which is not the same and is often not 2.
• "You always have to draw the tree." For a size, multiply the branchings. For six outfits, the chapter itself asks for a table.
• "A tree needs independent steps." The chapter's definition says so and its own starred Q10 does not obey it. Trees handle dependent steps perfectly well; that is what the second layer's changed numbers are for.

Questions to check understanding

• Draw a tree diagram for a described two-step experiment and read the sample space off it
• State the size of a multi-step sample space without drawing the tree
• Compute the probability of a complete outcome by multiplying along its path
• Compute the probability of an event made of several paths, by multiplying and then adding
• Draw a tree for a draw made without replacement and label both layers correctly
• Present a two-step sample space as a table — the form End-of-Chapter Q6 asks for
• Compare the sizes of the with-replacement and without-replacement sample spaces for the same set-up
• Reasoning question: a student draws a correct tree for the two fruit baskets, then counts four outcomes and answers 1/4. What has gone wrong, and what is the right answer?

Examples worth working on the board

Inputs, not answers, except where the chapter prints the result itself. Values marked Verified are worked out here or an added reading of a printed page; the chapter prints no answers and this volume has no appended answer key.

• §7.4's opening (p. 168). A tree diagram lays out every possible outcome of a multi-step experiment, which the chapter defines as an experiment made of a series of independent trials — tossing a coin twice, or rolling a die three times. Each branch stands for a possible result, and branches split to show the routes for the later steps. The chapter names two uses: seeing a multi-step experiment whole, where each complete route is one outcome, and listing every outcome of a sample space. It then notes that everything before §7.4 has been single-step.
• Example 7 and Fig. 7.6 (p. 168), read on the printed page. The experiment is a fair coin tossed twice. The figure has three column headings — First Toss, Second Toss, Outcome. From an unlabelled point on the left, two branches rise and fall to a coin marked H and a coin marked T, each branch labelled with a half. Each of those two coins splits again into two branches, again each labelled with a half, ending at coins marked H and T. The right-hand column shows four rows of two coins each: HH, HT, TH, TT. There are six branch labels in the artwork and every one of them is a half. No quarter appears anywhere in the figure. The chapter's accompanying instructions describe drawing a line to each result of the first toss and then two lines from each of those, and ask whether you can see that the diagram shows four possible outcomes; it concludes S = {HH, HT, TH, TT}.
• p. 169's first sentence, which is the one to be careful about. It says the theoretical probability of each outcome is written on the branch representing that outcome, and then reprints the favourable-over-possible formula and computes the probability of HH as 1/4 = 0.25 or 25%. What is actually written on the branches is a half — the probability of one step — while the probability of the complete outcome HH is a quarter. So the sentence describes per-step numbers as though they were per-outcome numbers. Read on the printed page and confirmed on the printed page.
• Section 3's counting, made general. Verified: two results at the first step and two at the second give 2 × 2 = 4 complete paths. Three rolls of a die would give 6 × 6 × 6 = 216. The chapter counts to four on the page and never states the rule; stating it is what makes the tree a tool rather than a picture.
• Section 6, the multiplication the figure draws. Verified: along the top path the labels are 1/2 and 1/2, and 1/2 × 1/2 = 1/4, which is exactly the chapter's own answer for HH, obtained a different way. Put the two derivations side by side. The agreement is what licenses the multiplication, and the multiplication is what will survive when the counting stops working.
• Think and Reflect (p. 169). Can you compute the probability of one head and one tail? Verified: the qualifying outcomes are HT and TH, so 2/4 = 1/2. By multiplication: each path is 1/2 × 1/2 = 1/4 and the two add to 1/2. Note that the two paths must be added — this is the chapter's only place where a probability is assembled from more than one path, and it is asked as an open question.
• Exercise Set 7.4 Q1 (p. 169) — the break, and the most important item in this brief. Basket A holds one apple and two oranges; Basket B holds one banana and one mango; one fruit is taken at random from each. Asked for a tree of all the pairs, the sample space, and the probability of getting an apple and a banana. Verified: the physical picks number 3 × 2 = 6 and they are equally likely; the distinct type-pairs number 4 — apple-banana, apple-mango, orange-banana, orange-mango. Multiplying along the apple-banana path gives 1/3 × 1/2 = 1/6. Counting the four type-pairs gives 1/4, and it is wrong, because the orange branch is twice as heavy as the apple branch. So this question, one page after Fig. 7.6, is where the chapter's own method fails and its drawing survives. The tree is the right answer to draw either way: put 1/3 on the apple branch and 2/3 on the orange branch and the arithmetic comes out.
• Exercise Set 7.4 Q2 (p. 169) — the break, worse. A box holds 3 red, 4 black and 2 green pens; you take one without looking, put it back, and your friend does the same. Asked for the possible colour outcomes and a tree, and then to guess the probability that both pens are the same colour. Verified: nine pens in all; the three branch weights are 3/9, 4/9 and 2/9; the nine colour-pairs are not equally likely; and P(same colour) = (3/9)² + (4/9)² + (2/9)² = (9 + 16 + 4)/81 = 29/81 ≈ 0.358. Counting three of the nine colour-pairs would give 1/3 ≈ 0.333, which is wrong. The chapter's own verb — guess — is a fair admission that it has not handed over the tool.
• End-of-Chapter Q10 (pp. 171–172) — the starred question that changes the second layer. A basket holds 4 red and 5 blue balls; one is drawn and set aside, then a second is drawn. Asked for a tree with outcomes and probabilities, then the probability of a red followed by a blue, and the probability of two blues. Verified: the first layer carries 4/9 and 5/9; the second layer depends on what was removed, so the red-first branch splits 3/8 and 5/8 while the blue-first branch splits 4/8 and 4/8. Then P(red then blue) = (4/9)(5/8) = 20/72 = 5/18 ≈ 0.278, and P(two blues) = (5/9)(4/8) = 20/72 = 5/18 as well. The two answers are the same number, and the reason is visible on the tree: both paths multiply 5 and 4 over 9 and 8, only in the other order. That coincidence is worth showing — it is a small piece of structure the question does not point at.
• End-of-Chapter Q13 (p. 172). A box of 4 balls numbered 1 to 4. (i) draw, record, replace, draw again; (ii) draw, record, and draw again without replacing; (iii) the sizes of the two sample spaces. Verified: 4 × 4 = 16 and 4 × 3 = 12. This is the cleanest pair in the chapter for section 12, because only one number in the experiment has changed.
• End-of-Chapter Q12 (iii) (p. 172). Three coins are tossed; a head is wanted on the first one, with the total number of heads coming to two. Verified: HHT and HTH qualify out of eight, so 2/8 = 1/4. A three-layer tree with eight paths, which is the largest tree the chapter's exercises demand.
• End-of-Chapter Q6 (p. 171), for section 11. Two shirts, one red and one blue, and three kinds of trousers — jeans, khakis and shorts. List every one-shirt-one-trouser outfit, and display the answer as a table. Verified: 2 × 3 = 6 outfits. The chapter itself asks for a table here rather than a tree, which makes it the natural place to say when each representation is better: a table for two steps with many options each, a tree when the branch numbers differ or when there are three or more steps.
• End-of-Chapter Q14 (p. 172). A coin tossed and one of 6 numbered cards drawn at the same time. Verified: 12 elements. A two-step sample space where the two steps are of different kinds, so the tree has two branches at the first level and six at the second.

Figures to have open

• Fig. 7.6 (p. 168) redrawn and fully able to be shown moving: three column headings, a start point, two layers of branches with a half on each of the six, and the four two-coin outcomes at the right. All six branch labels and the H and T lettering are inside the artwork — read on the printed page — so rebuild it, do not lift it. The redraw must be able to show the quarter appearing at the end of a path, which the printed figure never does.
• The fruit-basket tree with unequal first-layer weights, 1/3 and 2/3, drawn beside the four-type-pair list with its wrong 1/4. Not in the book, and the topic's single most important addition.
• The pen tree with three first-layer branches at 3/9, 4/9 and 2/9, and the three same-colour paths highlighted for addition. Not in the book.
• The without-replacement ball tree for End-of-Chapter Q10, with the second layer's numbers visibly recomputed after the first ball is removed. Not in the book; the question asks the student to draw it and the chapter supplies no model.
• A 2 × 3 outfit table beside the same information as a tree, for section 11. Standard schematic.
• No photograph is needed.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics (NCF-SE 2023), Chapter 7, §7.4 "Tree diagrams" (pp. 168–169), comprising the definition and two uses on p. 168, Example 7 with Fig. 7.6 on p. 168, the probability of HH and the reprinted formula on p. 169, and the Think and Reflect on p. 169.
• Fig. 7.6 (p. 168), with its three column headings and six branch labels, read on the printed page.
• Exercise Set 7.4, Q1 and Q2 (p. 169) — the whole of the set, and both questions break the count-the-paths method.
• End-of-Chapter Q6 (p. 171), and the starred Q10 (pp. 171–172), Q12 (iii) and Q13 (p. 172), Q14 (p. 172).
• The word independent is first used in Example 6 (p. 164), handled in Fair, unbiased, and memoryless: the gambler's fallacy.
• The Chapter Summary (p. 173) says tree diagrams help list and picture all the outcomes of a random experiment and help compute probabilities of events. It states no multiplication rule. Checked on the printed page.