Secondary 4 G2 Mathematics Tutor Bukit Timah | Probability Trees and Combined Events. Did you know? If a bag contains three red counters and two blue counters, choosing a red counter changes the next draw—unless that first counter is replaced. A single sentence can change the probability of the whole event.
Probability becomes easier when Secondary 4 G2 Mathematics students can describe the experiment, draw every possible branch and decide whether an earlier event changes a later one. This guide offers quick answers, original worked examples and a step-by-step probability-tree method for families around Bukit Timah, Beauty World, Sixth Avenue and King Albert Park.
For candidates taking the first 2027 SEC G2 Mathematics K210, the official syllabus identifies S2 Probability: simple events, combined events, possibility diagrams and tree diagrams. The combined-events and addition/multiplication rules are underlined in the 2027 K210 specification, meaning those portions are tested only in Paper 2 Section B. That section offers a choice between a Geometry and Measurement question and a Statistics and Probability question. Candidates graduating in 2026 should follow their school-confirmed N(A)-Level Mathematics 4045 syllabus instead; see the 2026 and the 2027 SEAB listings.
Quick answer: read the event before drawing a tree
- List possible outcomes. For two coin tosses, HH, HT, TH and TT are distinct outcomes; the order matters.
- Draw branches for each stage. Each stage must include all possible outcomes, and its probabilities should sum to 1.
- Decide what changes. With replacement, the bag returns to its original composition. Without replacement, one item has gone.
- Multiply along a pathway. For a particular ordered sequence, multiply the probability at each stage given what came before.
- Add disjoint successful pathways. For “one red and one blue in either order”, add the RB and BR pathway results.
- Check the probability range. Any final probability must be between 0 and 1.
A tree is not just a picture. It records the sequence of decisions and conditions in the experiment. Drawing the wrong denominator on the second branch is often the first wrong step, even when all later multiplication is correct.
Independent, dependent and mutually exclusive are different ideas
| Term | Meaning | Useful example |
|---|---|---|
| Independent events | Knowing one outcome does not change the probability of the other | A coin toss and another fair coin toss. |
| Dependent events | An earlier outcome changes the chance of a later outcome | Two counters drawn without replacement. |
| Mutually exclusive events | They cannot both happen on the same trial | One die shows an even number and the same die shows 5. |
| Complementary events | One event occurs or its exact opposite occurs | At least one red versus no red. |
| Ordered outcomes | The sequences are recorded distinctly | Red then blue is a different branch from blue then red. |
| Exhaustive outcomes | Every possible outcome is accounted for | The probabilities on each set of branches sum to 1. |
Did you know? “Mutually exclusive” does not mean “independent”. Two mutually exclusive events with non-zero probabilities cannot be independent, because if one occurs the other cannot occur. Mixing those terms leads to wrong addition or multiplication choices.
Eighteen original worked combined-event examples
1. A fair coin’s single event
A fair coin has possible outcomes H and T. Probability of heads on one toss = 1/2. The sample space has two equally likely outcomes.
2. Two fair coin tosses, two heads
Outcomes HH, HT, TH, TT are equally likely. Probability of HH = 1/4. Or multiply 1/2 × 1/2 because the tosses are independent.
3. Exactly one head
In two fair tosses, HT and TH meet the requirement. Probability = 2/4 = 1/2. “Exactly one” is not the same event as “at least one”.
4. At least one head
Complement of “at least one head” is TT. Probability = 1 − 1/4 = 3/4. Complements are often shorter than drawing every success.
5. Sum of 7 on two dice
Two fair six-sided dice have 36 ordered outcomes. Six ordered pairs sum to 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Probability = 6/36 = 1/6.
6. Sum of at least 10 on two dice
Sums 10, 11 and 12 occur in 3, 2 and 1 ordered ways respectively. Probability = (3 + 2 + 1)/36 = 1/6. Check the counted pairs.
7. Even number or 5 on one die
Even outcomes are 2, 4 and 6; the outcome 5 is separate. There are four successes out of six: 4/6 = 2/3. Here the events are mutually exclusive.
8. Even number or greater than 4
Even = {2,4,6}; greater than 4 = {5,6}. The union is {2,4,5,6}, so probability = 4/6 = 2/3. Adding 3/6 and 2/6 without removing the double-counted 6 would be wrong.
9. Odd number and greater than 3
The odd outcomes are {1,3,5}; greater than 3 gives {4,5,6}. Only 5 satisfies both, so probability = 1/6.
10. Two reds with replacement
Bag: 3 red, 2 blue. Replace the first counter before drawing again. Probability RR = (3/5) × (3/5) = 9/25. The denominator and composition stay unchanged.
11. Two reds without replacement
Bag: 3 red, 2 blue. After a red is removed, there are 2 red among 4 counters. RR = (3/5) × (2/4) = 3/10. The second branch changes.
12. Red then blue without replacement
Bag: 3 red, 2 blue. After red, there are 2 blue among 4 counters. P(RB) = (3/5) × (2/4) = 3/10. Write the ordered sequence first.
13. Blue then red without replacement
Bag: 3 red, 2 blue. P(BR) = (2/5) × (3/4) = 3/10. The branches have different first-stage probabilities but give the same final value here.
14. Exactly one red in two draws without replacement
Exactly one red means RB or BR. Add these separate pathways: 3/10 + 3/10 = 3/5. Do not multiply two alternative complete pathways together.
15. At least one red without replacement
The opposite event is two blues. BB = (2/5) × (1/4) = 1/10. Therefore P(at least one red) = 1 − 1/10 = 9/10.
16. At least one red with replacement
With replacement, BB = (2/5) × (2/5) = 4/25. So at least one red = 1 − 4/25 = 21/25. Compare with the preceding without-replacement result; the condition matters.
17. Two independent stages with different probabilities
Suppose a hypothetical first stage succeeds with probability 0.8 and an independent second stage succeeds with probability 0.7. Both succeeding has probability 0.8 × 0.7 = 0.56.
18. The complement of two independent successes
With the same independent stages, probability that both do not succeed together as a complete pair is 1 − P(both succeed) = 0.44. This includes first failing, second failing or both failing; state the event precisely.
Why a 3-red, 2-blue tree changes with replacement
| Two-draw event | With replacement | Without replacement |
|---|---|---|
| Red then red | (3/5)(3/5) = 9/25 | (3/5)(2/4) = 3/10 |
| Red then blue | (3/5)(2/5) = 6/25 | (3/5)(2/4) = 3/10 |
| Blue then red | (2/5)(3/5) = 6/25 | (2/5)(3/4) = 3/10 |
| Blue then blue | (2/5)(2/5) = 4/25 | (2/5)(1/4) = 1/10 |
| Check all paths | 9/25 + 6/25 + 6/25 + 4/25 = 1 | 3/10 + 3/10 + 3/10 + 1/10 = 1 |
That last row is one of the strongest self-checks in probability: the complete list of distinct outcomes must add up to 1. If it does not, a pathway is missing or a branch probability is wrong.
What changes in the 2027 Paper 2 Section B choice?
In the 2027 K210 assessment scheme, Paper 2 Section B contains two questions; candidates answer one. One draws from specified, underlined Geometry and Measurement content and the other from specified, underlined Statistics and Probability content. The probability-tree and combined-event rules covered in this article are underlined S2 topics. A student should practise them alongside the rest of the prescribed syllabus and prepare an informed Section B decision—not assume all probability appears only in this section or discard the other mathematical strand.
Five first-step mistakes to correct, not just erase
| Observed error | Cause | Tutor’s corrective question |
|---|---|---|
| Second draw remains 3/5 red after removing a red | The student assumed replacement without reading the condition | What objects remain after the first event? |
| Probability for RB multiplied by probability for BR | An OR event was treated as a single AND pathway | Are these alternatives or stages along one path? |
| P(even or greater than 4) counted 6 twice | Overlapping events were treated as disjoint | Which outcomes belong to both sets? |
| Probability greater than 1 is left as final answer | Completeness and basic range checks were omitted | Could a probability be 1.2? Which path was double-counted? |
| At least one red treated as exactly one red | The wording was changed during calculation | Does RR also satisfy the requested condition? |
Five-question no-hint student mini-test
- A fair coin is tossed twice. Find the probability of HT in that exact order.
- A standard die is rolled once. Find P(1 or 6).
- A bag has 4 green and 1 yellow counter. Two counters are drawn without replacement. Find P(two green).
- For the same bag, find P(yellow first, green second).
- A bag has 3 red and 2 blue counters. Two are drawn without replacement. Find the probability of no red.
Answers: (1) 1/4; (2) 2/6 = 1/3; (3) (4/5)(3/4) = 3/5; (4) (1/5)(4/4) = 1/5; (5) (2/5)(1/4) = 1/10. Draw the branches and see if each denominator represents what remains at that stage.
An illustrative three-pupil, 90-minute probability tutorial
| Time | Instruction | Transfer check |
|---|---|---|
| 0–10 min | Recall independent, dependent and mutually exclusive | Student explains the terms using a new example. |
| 10–25 min | Short diagnostic on tree denominators and ordering | Tutor spots the first incorrectly labelled branch. |
| 25–50 min | Compare replacement versus no replacement side by side | Student explains which totals change and why. |
| 50–75 min | Individual unseen bag and dice problems | Learner selects multiplication, addition or complement appropriately. |
| 75–90 min | Check total probability, record errors and plan retest | Tutor verifies the method works without prompts. |
The original immutable eduKateSG Mathematics small-group reference describes a premium three-pupil model with 1.5-hour weekly lessons near Sixth Avenue MRT. This is an illustrative adaptation for Secondary 4 G2 probability, not confirmation of current group schedules, tutor assignments or vacancies.
FAQs for Bukit Timah parents
Do I need to memorise the entire probability tree?
No. The tree is a reasoning device. Start with the available outcomes, label each conditional probability and build every pathway. The critical habit is asking what can still happen after the earlier stage.
Is ‘with replacement’ a minor detail?
It is essential. With replacement returns the original composition after each draw; without replacement changes the next stage. In our 3-red, 2-blue example, P(two reds) changes from 9/25 to 3/10.
Are all these rules tested in every part of the 2027 SEC paper?
No. The combined-events and addition/multiplication probability items are among the underlined content marked for 2027 K210 Paper 2 Section B only. Single-event probability is separately listed. Candidates should use the paper instructions and actual school entry, especially if their graduating year is 2026.
What is good evidence of improvement?
Ask the student to explain a new without-replacement scenario and calculate two ordered paths before looking at a model answer. After several days, change the counter counts. Durable reasoning is more meaningful than a familiar tree copied neatly.
Continue across the G2 Mathematics series
- Statistics, probability and graphs overview
- Cumulative frequency, box plots and data spread
- Real-world mathematical modelling
- 2026 N(A) and 2027 SEC revision roadmap
Official references: 2027 SEC G2 syllabus directory · K210 Mathematics syllabus and Section B instructions · 2026 N(A) syllabus directory. For information about the original tuition format, use the unchanged three-student tutorial page and eduKate Singapore. The exercises are illustrative original questions, not official SEAB past-paper content.