2027 G3 SEC Mathematics K310 | Learner’s Guide Vol 0178. A probability tree is a map of possible sequences. Every branch asks what could happen next; every complete path is one possible story. When students understand that simple idea, they stop guessing whether to add or multiply and start reasoning from the event being asked about.
This guide concentrates on the structure of tree diagrams: placing probabilities on branches, multiplying down a path, adding paths for alternatives, using complements, and noticing when later probabilities change. The specialised case of drawing objects without replacement was already introduced in Volume 0095. Here it becomes one contrast within a wider method for examining combined events, with new worked examples and interpretation.
1. Verify the current K310 scope before learning an impressive-looking method
The 2027 G3 school-candidate directory, updated 26 October 2026, lists Mathematics as K310, corresponding to earlier reference code 4052. The official K310 2027 syllabus includes probability of single events and simple combined events. It explicitly names possibility diagrams and tree diagrams where appropriate, together with addition and multiplication of probabilities in the contexts of mutually exclusive and independent events.
K310 Mathematics is assessed through two written papers, each lasting 2 hours 15 minutes and carrying 90 marks, with a 50 per cent weighting per paper. Both allow approved calculators. Paper 2’s final question focuses on a real-world context; that does not mean every real-world modelling task will require a tree. Choose a representation because it suits the event structure, not because a diagram looks advanced.
This chapter uses conditional branch reasoning to make sequential events intelligible. It does not present Bayes’ theorem, complex statistical distributions or advanced conditional probability identities as separately mandated K310 syllabus topics. Some examples draw a second item after a first; the changing denominator is an accessible way to see what information a later branch has.
2. Start with the sample space, not a formula
A sample space is the set of possible outcomes under the experiment’s assumptions. When flipping a fair coin once, the outcomes are heads and tails. When flipping it twice, the ordered outcomes are HH, HT, TH and TT. The order matters because HT and TH are different sequences even if both contain one head and one tail.
If the coin is fair and successive flips are independent, each of the four ordered paths has probability 1/4. That result is not obtained by adding 1/2 and 1/2. Each complete path requires both events: one result on the first flip and a specified result on the second. Multiply along the path.
A tree diagram is useful when each stage has named possible outcomes. It does not replace the need to define the experiment. If the second toss happens only when the first is heads, the sample space has a different structure. Read the stopping rule before drawing branches.
3. Build a verbal tree before drawing it
Imagine a first branch that says “Heads, 1/2” and another that says “Tails, 1/2”. From each first-stage outcome, the second flip branches again into Heads, 1/2 and Tails, 1/2. This text description is already a tree model; the visual diagram would only arrange the same information spatially.
Students often rush straight to small fractions written beside lines and forget which stage those fractions describe. Label each stage: first flip, second flip. At the end of each path, write its complete ordered outcome. A branch labelled “heads” is not yet the same as a path labelled “heads then tails”.
An effective tree can be made with pencil or described clearly in a table. In this text-only guide, every example states the branches in words so the method remains accessible without diagrams. During examination practice, draw a clean tree when it helps keep cases separate.
4. The branch-sum check catches impossible models
At a particular node, the outgoing branches must describe all mutually exclusive next-step possibilities for that node. For a fair coin, 1/2 + 1/2 = 1. For a bag containing three red counters and two blue counters, the first-stage probabilities are 3/5 and 2/5, which again total 1.
If the outgoing fractions add to 0.8, ask whether an outcome was omitted or a denominator was copied wrongly. If they total 1.2, the branches may overlap or use inconsistent bases. The check applies at each node, not merely across the first stage.
Sometimes a question gives an incomplete tree and asks for a missing probability. If the displayed outgoing values are 0.35, 0.40 and p for three mutually exclusive exhaustive outcomes, then p = 0.25. But you must first confirm that the three branches indeed cover every allowed next result.
5. Multiply along a path: the word ‘and’ matters
For two fair flips, P(Heads then Tails) = 1/2 × 1/2 = 1/4. Each factor belongs to one stage along one path. The product is a joint outcome: the first result and the second result in that order.
In a bag example with replacement, the first red has probability 3/5. After putting it back and restoring the contents, the second blue has probability 2/5. Thus P(Red then Blue) = (3/5) × (2/5) = 6/25. The phrase “with replacement” explains why the original composition returns before the second stage.
The multiplication is not a magic response to the word “and”. It works because we use the appropriate probability at each step of the particular path. When the earlier result changes the later sample space, the second factor must be the updated branch probability, not the first-stage fraction copied automatically.
6. Add distinct complete paths: the word ‘or’ matters
Suppose the event is “exactly one red in two draws with replacement”. That happens along either Red–Blue or Blue–Red. These complete paths are mutually exclusive: one experiment cannot follow both ordered paths at once. Add their path probabilities.
Under the three-red/two-blue bag model, P(RB) = 3/5 × 2/5 = 6/25 and P(BR) = 2/5 × 3/5 = 6/25. Therefore P(exactly one red) = 6/25 + 6/25 = 12/25. A tree makes it harder to forget the second order.
Do not add individual branch probabilities from different stages to represent a combined event. The two relevant units are complete ordered paths. Once you know what counts as a successful path, addition is a natural final operation.
7. ‘At least one’ often invites a complement
With two draws with replacement, “at least one red” includes RR, RB and BR. You could calculate all three paths and add them. A shorter route is to notice that the only failure is BB. P(BB) = 2/5 × 2/5 = 4/25, so P(at least one red) = 1 − 4/25 = 21/25.
A complement is the event “not A”. Its probability is one minus the probability of A, provided A and not-A exhaust all outcomes. The phrase “at least one” often indicates that the complement “none” may have fewer cases than the direct event.
Other examples include “at least one heads” versus “no heads” or “at least one faulty item” versus “all acceptable items”. Do not automatically use complements if the direct event is easier. Decide by counting paths.
8. Independence means the second probability stays the same
For independent events, knowing the first outcome does not alter the probability distribution of the second. A coin flip followed by a fair die roll is a simple model if the devices do not influence each other. The die still has probability 1/6 for a specified face regardless of the coin.
Likewise, if a counter is replaced and the bag returned to its original composition before the next draw, the red probability remains 3/5 at both stages. That is an independence model under the stated random-draw assumptions. The phrase “with replacement” does not rescue a flawed experiment if, for example, the mixing conditions change unpredictably.
State the assumption, not just the arithmetic. Independence is about the relationship between events. When a question specifies independent trials, multiply probabilities appropriately; when the procedure changes the composition, update the branches.
9. Without replacement: the tree remembers what happened
Start with three red and two blue counters, five total. If the first counter is red and is not returned, the remaining bag contains two red and two blue. Therefore the second-stage branches after Red are 2/4 for Red and 2/4 for Blue. If the first is Blue, the bag contains three red and one blue; the second branches are 3/4 for Red and 1/4 for Blue.
The full path probabilities are RR = (3/5)(2/4) = 3/10; RB = (3/5)(2/4) = 3/10; BR = (2/5)(3/4) = 3/10; BB = (2/5)(1/4) = 1/10. They add to one. Exactly one red now has probability 3/10 + 3/10 = 3/5, not 12/25.
This changes the result without changing the starting bag. That is why a tree must represent the actual process, not just its initial contents. For a deeper practice set focused specifically on update rules, consult Volume 0095.
10. Read branch labels as conditional information
The statement “after Red, the chance of Blue is 2/4” means the next probability is considered under the condition that a red counter has already been removed. The condition is not a decoration: it determines the numbers remaining in the bag.
This is a natural introduction to conditional probability language. For example, if the first draw is Red, the second draw is Blue with probability 1/2 in the five-counter no-replacement example. That is a different question from the probability that the whole experiment yields Red then Blue, which is 3/10.
The first probability refers to a second-stage event given a known first-stage result; the second refers to the joint ordered path before either draw is known. These are connected, but not interchangeable. A common error is to report 1/2 as P(RB) and omit the probability of reaching the Red node.
11. A path probability is not a branch probability
Ask: “What is the probability that a counter is blue on the second draw, given the first was red?” The answer is 2/4 = 1/2. Then ask: “What is the probability that the two draws are red then blue?” That answer is (3/5)(2/4) = 3/10. The difference is the starting point of the question.
A tree records both the chance of arriving at a node and the conditional chances departing from it. If a question begins with “given that”, begin at the appropriate conditioned stage when the information is fixed. If it asks for the original experiment’s path, start at the root.
This distinction supports later reasoning in Science, where a predicted trait conditional on a parental genotype differs from an unconditional population claim. First define which experiment and which information the question assumes.
12. Count favourable outcomes only after checking equal likelihood
For two fair coin flips, HH, HT, TH and TT are four equally likely ordered outcomes. Counting favourable cases over four is legitimate. However, a tree for three red and two blue counters has unequal branch probabilities. Counting the four sequence labels RR, RB, BR and BB as equally likely would give a false probability for each.
In the no-replacement example, those paths have probabilities 3/10, 3/10, 3/10 and 1/10. Four labels do not imply four equal chances. To calculate a composite event, add actual path probabilities.
The general counting rule “favourable outcomes divided by total equally likely outcomes” contains a crucial condition: equally likely. When the simplified labels disguise unequal numbers of elementary arrangements, a tree with explicit probabilities is safer.
13. When events are mutually exclusive and when they are independent
Mutually exclusive events cannot happen at the same time within the stated experiment. On one die roll, “show 1” and “show 6” are mutually exclusive. Their probabilities add when the question asks for one or the other. Independent events are different: the result of one does not change the chance of the other.
Two possible faces on a single roll are not independent merely because both are possible. If the die has shown 1, it cannot simultaneously show 6 on that roll. Conversely, heads on the first flip and heads on the second independent flip are not mutually exclusive; they can occur together.
Students often memorise “or means add; and means multiply” without checking event relations. The safer principle is: for mutually exclusive alternatives, add their probabilities. Along a properly conditioned sequence, multiply branch probabilities. If the events overlap, plain addition can double-count the intersection.
14. A union that needs inclusion–exclusion rather than blind addition
Roll a fair six-sided die once. Let A mean an even number, {2,4,6}, and B mean a number greater than 4, {5,6}. P(A) = 3/6, P(B) = 2/6, but outcome 6 belongs to both. Adding them gives 5/6 and counts the 6 twice.
The correct probability of A or B is P(A) + P(B) − P(A and B) = 3/6 + 2/6 − 1/6 = 4/6 = 2/3. The favourable distinct faces are {2,4,5,6}. This is not necessarily a tree problem; a simple listed sample space is clearer here.
Use a tree when order or branching conditions matter. Use a possibility diagram, list or Venn diagram when the structure is more naturally about overlapping categories. Good mathematical communication includes choosing the simplest correct representation.
15. Missing branches in real-world wording
Consider: “A volunteer arrives on time with probability 0.8. If late, the session is cancelled. Otherwise, the session proceeds to a second check that passes with probability 0.9.” A tree should branch first into on-time 0.8 and late 0.2. Only the on-time branch continues to the second check. Do not draw a second check after the late branch when the scenario says the session ends.
The probability of proceeding and passing is 0.8 × 0.9 = 0.72. The probability of ending without a pass is 1 − 0.72 = 0.28, covering late arrival and an on-time arrival followed by failure. These two failure routes have probabilities 0.2 and 0.8 × 0.1 = 0.08.
This illustrates a stopping rule. A diagram that forces every branch to an identical number of stages may be visually neat but conceptually wrong.
16. Branch sums can reveal a copying mistake before calculation
Suppose the first stage has Red 0.4 and Blue 0.6. After Red, a student writes Good 0.7 and Not good 0.2. The two second-stage fractions total 0.9. Unless the wording specifies a third possibility, the branch set is incomplete or incorrect. A missing 0.1 is not a rounding nuisance if the probabilities are otherwise exact as given.
After fixing the branches, inspect the event definition. “At least one good result” could include paths from different first-stage branches. Before combining them, label each complete outcome and its probability. This avoids a correct-looking sum of incomplete cases.
An exam candidate should use three checks: outgoing branch sums, total final path sum and a reasonableness check against the event’s possible range. All calculated probabilities should lie between 0 and 1 inclusive.
17. Work backwards when a final probability is known
A two-stage independent experiment has P(A) = 0.6 on the first stage and P(B) = p on the second. The problem states P(A then B) = 0.18. The path equation gives 0.6p = 0.18, so p = 0.30. Verify that the other second-stage outcome has probability 0.70 if B and not-B exhaust the branch.
Now remove the independence assumption: if B’s probability depends on A, the same equation identifies the branch probability after A but does not automatically determine B after not-A. The available data are insufficient for that other branch.
Working backwards is especially useful for missing values on a tree diagram. It also trains evidence discipline: a given joint probability does not reveal every conditional probability in a larger experiment.
18. A tree with three stages
Flip a fair coin three times independently. Eight ordered paths exist: HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. Each has probability (1/2)³ = 1/8. Exactly two heads occurs along HHT, HTH and THH, so its probability is 3/8.
At least one heads can be found by complement: only TTT fails, so the result is 1 − 1/8 = 7/8. Exactly one tails means the same three paths as exactly two heads and therefore also has probability 3/8. Naming the event precisely helps detect such equivalences.
Longer trees create more opportunities to miss a branch. Before multiplying, write the count and labels of all complete paths, or use the structure to group them. Do not mistake a visually larger tree for a conceptually harder probability problem.
19. A useful comparison: independent coins versus a drawn sequence
A fair coin flipped twice has P(HH) = 1/4, P(HT) = 1/4, P(TH) = 1/4 and P(TT) = 1/4. The four ordered outcomes are equally likely. A coloured-counter bag with unequal contents does not usually share that uniform four-path property.
Even in the same bag, replacement changes path probabilities. With three red and two blue counters, the with-replacement RR chance is 9/25; without replacement it is 3/10. In the first model, a red result leaves the next red chance at 3/5. In the second, it reduces it to 2/4.
Both methods are legitimate within their own assumptions. An answer that forgets whether an item was replaced is not slightly imprecise; it is a solution to a different experiment.
20. Probability describes a model; frequency describes observed counts
If a fair coin is flipped one hundred times, the probability of heads on each independent flip is 1/2. That does not guarantee exactly fifty heads in one hundred flips. The observed relative frequency may be above or below 0.5 in a particular sample.
Suppose an original classroom example records 56 heads in 100 tosses. The observed proportion is 0.56, while the model predicts probability 0.5 on each trial. These quantities can be compared without pretending one must equal the other. Observed variation alone does not prove the coin unfair.
The same distinction will be important in the following Biology chapter on simple inheritance ratios. A predicted 3:1 phenotype ratio is a probabilistic expectation under model assumptions, not a guarantee that each group of four offspring has exactly three of one type.
21. The difference between a model and a forecast
A tree built from stipulated probabilities is a mathematical model. It can answer the question exactly within those assumptions. A forecast for a real school event is a further claim: it assumes that actual participants, processes or equipment behave as the model says.
For example, if 80 per cent of applications are approved in a fictional historical sample, that number is not automatically an independent 0.8 probability for every future person. Circumstances and selection criteria may vary. A tree that multiplies 0.8 by itself several times needs justification for treating the events as independent.
Real-world problem solving rewards clear assumptions. Present the tree calculation, then separate the mathematical result from the empirical claim about future frequencies.
22. Visual alternatives to a tree
For two dice, a 6 × 6 possibility table may be clearer than a tree with six first-stage branches and thirty-six final paths. For simultaneous membership in two groups, a Venn diagram may better reveal overlap. For drawing counters in a fixed order, a tree makes stage-dependent denominators immediately visible.
Students should recognise the structural question: Does the second stage depend on what happened in the first? Do we need ordered outcomes? Are there many equally likely combinations? Is the event an intersection or a union? The answer suggests the representation.
The 2027 K310 syllabus explicitly says possibility diagrams and tree diagrams are used where appropriate. It does not demand a tree for every probability task. A neat correct sample-space table can earn mathematical clarity without wasting space.
23. One complete examination-style model
A fictional club offers two independent trials. First, a student spins a fair spinner with four equal sectors numbered 1–4. Then the student flips a fair coin. A prize requires an even spinner number and Heads. The even probability is 2/4 = 1/2, and the Heads probability is 1/2. Their independence gives a prize probability of 1/4.
If the event were “an even spinner number or Heads”, the outcomes overlap. By inclusion–exclusion, P(even or Heads) = 1/2 + 1/2 − 1/4 = 3/4. A student who simply adds the two half-probabilities would incorrectly report certainty.
Now suppose the game is repeated independently twice and the question asks for two prizes. The probability is (1/4)² = 1/16. If it asks for at least one prize, the complement is (3/4)² = 9/16, so the answer is 1 − 9/16 = 7/16. This shows how one clear event definition can support several different probability questions.
24. Thirty original practice questions with worked answers
Do these in order: branch checks, two-stage path calculations, complements and dependency, followed by interpretation. All contexts are fictional practice. Unless specified otherwise, coins and dice are fair, draws are random and statements of independence are part of the problem.
01 · Sample space
Question. Write the outcomes of two coin flips.
Worked answer. HH, HT, TH and TT. Reason. The order of flips matters.
02 · Root branch
Question. A bag has 3 red and 2 blue counters. State the probability of red first.
Worked answer. 3/5. Reason. There are three red counters among five.
03 · Branch sum
Question. At a node, one outcome is 0.35. What is its only alternative’s probability?
Worked answer. 0.65. Reason. Exhaustive mutually exclusive branches sum to one.
04 · Two flips
Question. Find P(H then T) for fair independent flips.
Worked answer. 1/4. Reason. Multiply 1/2 by 1/2 along the path.
05 · Exactly one head
Question. Find the chance of exactly one head in two flips.
Worked answer. 1/2. Reason. HT and TH contribute 1/4 each.
06 · At least one head
Question. Find the chance of at least one head in two flips.
Worked answer. 3/4. Reason. Subtract the TT probability 1/4 from one.
07 · With-replacement RR
Question. Three red and two blue counters. Draw twice with replacement. Find P(RR).
Worked answer. 9/25. Reason. Multiply 3/5 × 3/5.
08 · With-replacement RB
Question. Find P(RB) in that model.
Worked answer. 6/25. Reason. Multiply 3/5 × 2/5.
09 · With-replacement one red
Question. Find P(exactly one red) in that model.
Worked answer. 12/25. Reason. RB or BR: 6/25 + 6/25.
10 · With-replacement any red
Question. Find P(at least one red) in that model.
Worked answer. 21/25. Reason. Complement BB: 1 − 4/25.
11 · Without replacement RR
Question. Use the same starting bag; do not replace the first. Find P(RR).
Worked answer. 3/10. Reason. Multiply 3/5 × 2/4.
12 · Without replacement RB
Question. Find P(RB).
Worked answer. 3/10. Reason. After first red, two blue remain out of four.
13 · Without replacement BR
Question. Find P(BR).
Worked answer. 3/10. Reason. Multiply 2/5 × 3/4.
14 · Without replacement BB
Question. Find P(BB).
Worked answer. 1/10. Reason. Multiply 2/5 × 1/4.
15 · Sum all four
Question. Check the total for RR, RB, BR and BB without replacement.
Worked answer. 1. Reason. 3/10 + 3/10 + 3/10 + 1/10 equals one.
16 · Given the first red
Question. With no replacement, what is the chance the second is blue, given the first was red?
Worked answer. 1/2. Reason. The remaining bag has two blue among four; this is a branch probability.
17 · Joint RB instead
Question. Before drawing, find the chance of red then blue without replacement.
Worked answer. 3/10. Reason. Include the chance of reaching first red: 3/5 × 2/4.
18 · Two red among four?
Question. A bag has 4 red and 3 blue. Without replacement, find P(two red).
Worked answer. 2/7. Reason. Multiply 4/7 × 3/6 = 12/42.
19 · Blue both times
Question. In the same bag, find P(two blue) without replacement.
Worked answer. 1/7. Reason. Multiply 3/7 × 2/6 = 6/42.
20 · At least one red
Question. In that bag, without replacement, find P(at least one red).
Worked answer. 6/7. Reason. Subtract the probability of two blue, 1/7.
21 · With-replacement complement
Question. In that seven-counter bag, replace after the first draw. Find P(at least one red).
Worked answer. 40/49. Reason. 1 − (3/7)² = 40/49.
22 · Coin and spinner
Question. Fair coin plus four-sector fair spinner. Find P(Heads and even).
Worked answer. 1/4. Reason. Multiply 1/2 × 2/4.
23 · Union instead
Question. In that game, find P(Heads or even).
Worked answer. 3/4. Reason. Add 1/2 and 1/2, then subtract intersection 1/4.
24 · Two game wins
Question. Each independent game is won with probability 1/4. Find two wins.
Worked answer. 1/16. Reason. Multiply 1/4 by 1/4.
25 · At least one win
Question. For those two games, find P(at least one win).
Worked answer. 7/16. Reason. 1 − (3/4)².
26 · Three tosses
Question. Find P(exactly two heads in three fair flips).
Worked answer. 3/8. Reason. HHT, HTH and THH each have probability 1/8.
27 · Missing branch
Question. A node has three outcomes of probabilities 0.20, 0.55 and p. Find p.
Worked answer. 0.25. Reason. 1 − 0.20 − 0.55.
28 · Reverse path
Question. P(A) = 0.6 and P(A then B) = 0.18. Find P(B after A).
Worked answer. 0.3. Reason. Solve 0.6p = 0.18; no independence assumption is needed for this path equation.
29 · Mutually exclusive
Question. One fair die roll: find P(1 or 6).
Worked answer. 1/3. Reason. The events are disjoint, so add 1/6 + 1/6.
30 · Model versus count
Question. A fair coin gave 56 heads in 100 independent tosses. Does that prove its chance of heads is 0.56?
Worked answer. No. Reason. The result is an observed relative frequency, not proof of the theoretical event probability.
25. Diagnose the five errors that cost understanding
Error one: forgetting a second order. Exactly one red in two draws includes red-blue and blue-red. Error two: forgetting to update. Removing a red changes the bag before the second draw. Error three: forgetting the root. A second-stage branch probability is not the same as the original joint path probability.
Error four: treating outcomes as equally likely when they are not. Four labelled paths can have four different probabilities. Error five: multiplying or adding before identifying the event. State whether the question asks for an intersection, alternative, complement or conditional stage.
Most incorrect trees can be repaired by tracing one complete possible story in words: “First blue, then red; the second probability is based on what remains.” If that story contradicts the diagram, fix the branch before touching the arithmetic.
26. Choose accuracy only after completing the probability
Where possible, keep fractions exact through the calculation. If P = 3/7 × 2/6, simplify the fraction rather than rounding 3/7 before multiplying. A precise rational probability is especially useful if the next question asks for a complement or a repeated experiment.
The K310 syllabus notes require non-exact numerical answers generally to three significant figures unless otherwise instructed. Many probability answers here are exact fractions; leave them as fractions when a question permits. If asked for a percentage or decimal, carry out the conversion at the end.
Numerical precision does not rescue the wrong tree. A beautifully rounded answer to the wrong event is still wrong. For the wider exact-versus-approximate discipline, see Volume 0149.
27. Advance to biological genetic crosses without confusing models
The next Science chapter explores one-gene inheritance in an intentionally simple fictional plant. A Punnett diagram lists possible allele combinations from two parents; probability reasoning explains their expected frequencies. Students who understand path multiplication will recognise why, under a stipulated model, P(A from one parent and a from the other) is a product of the respective gamete probabilities.
A Punnett diagram is not simply a probability tree with biological labels, however. It depends on assumptions about alleles, segregation, dominance and fertilisation. A tree’s arithmetic does not establish those biological facts. Those need to be supplied or justified separately.
This is the progression from mathematical structure to scientific interpretation: first calculate what a model predicts, then explain what an observed sample can and cannot confirm. That separation is a powerful general examination habit.
28. A practical week of deliberate improvement
On day one, describe trees in words and check branch sums. On day two, compute two-stage path products. On day three, recognise all paths that satisfy exactly-one and at-least-one events. On day four, compare with-replacement and without-replacement models. On day five, solve missing probabilities and stopping-rule cases. On day six, choose between trees, lists and possibility diagrams. On day seven, complete a mixed timed set and explain each event definition before calculating.
Keep a mistake log based on the step where reasoning failed. If you always forget BR as an alternative to RB, practise ordered labels. If you update denominators correctly but misread “given that”, practise conditional language. If you calculate accurately but confuse models with real frequencies, practise interpretive conclusions.
Improvement is not a matter of drawing more branches. It means being able to justify each branch, each multiplication and each final addition in a sentence.
29. Student questions answered in plain language
Must I draw a tree for every probability question? No. The K310 syllabus names tree and possibility diagrams where appropriate. Use the representation that fits the outcome structure.
Is without replacement always dependent? In the ordinary finite bag model, earlier draws change later composition, so the probabilities generally depend on prior results. Edge cases can behave differently, so follow the actual model.
Can I add the probabilities of Red and Blue? If you mean a single draw that must be either red or blue and those are exhaustive, the result is one. If you mean a two-stage event, specify which complete paths count.
Does a predicted probability guarantee an observed ratio? No. A predicted chance is a property of the stipulated process. Finite samples vary.
Why is HT different from TH? They are different ordered sequences, even though each contains one head and one tail. The event “exactly one head” includes both.
30. Continue through the mathematics and EMS routes
Use the Mathematics Hub for probability and modelling foundations, PSLE Learning Guide for earlier fractions and counting ideas, and the G3 SEC learner guide for the wider examination route. Related owners include Volume 0095 on sequential no-replacement draws and Volume 0160 on misleading aggregate rates.
The Science Hub supports the interpretation of repeated events in experiments; the English Hub helps students state exactly which condition and outcome a conclusion concerns. Always use the official SEAB K310 2027 syllabus for the scope and examination format.
Good tree diagrams are less about drawing and more about responsibility for every fraction. Each branch must belong to a defined stage; each path must describe a complete possible outcome; each addition must combine distinct successful paths. If you can explain those three relationships, you are doing probability rather than guessing at formulae.
