SECONDARY 4 MATHEMATICS CLASSROOM · CHAPTER 2 · PROBABILITY OF COMBINED EVENTS · SEC G3 K310
Probability of Combined Events: Draw the Chance Structure Before You Calculate
In this classroom, you will not begin by asking whether to add or multiply. You will begin by asking what can happen, in what order, and which routes satisfy the event.
Combined probability becomes reliable when you make the chance process visible. A possibility diagram shows a grid of outcomes. A tree diagram shows a sequence of stages. A complement lets you replace many successful routes with one easier opposite event. Addition and multiplication then become consequences of the structure rather than rules chosen from keywords.
Classroom rule: draw or describe the outcome structure before choosing the arithmetic.
This classroom follows the current Singapore-Cambridge SEC G3 Mathematics syllabus, K310, where probability is listed under S2. The assessed content includes probability as a measure of chance, probability of single events, probability of simple combined events using possibility or tree diagrams where appropriate, and addition and multiplication of probabilities for mutually exclusive and independent events.
Reference: 2027 SEC G3 Mathematics K310 syllabus.
Featured Answer: What Is Combined Probability?
Combined probability studies events built from more than one condition or stage. A coin may be tossed twice. Two dice may be rolled together. A counter may be selected and then another selected. A question may ask for one event or another, one event and another, at least one success, exactly one success, or a particular ordered route.
The central skill is not calculation speed. It is knowing how the event is assembled from possible outcomes.
The Simple Classroom Answer
Every combined-probability question asks some version of:
Which routes through the chance process satisfy the event, and what is the probability of those routes?
- Sample space tells you what outcomes are possible.
- Possibility diagrams organise outcomes in a grid.
- Tree diagrams organise outcomes in stages.
- Multiplication usually combines probabilities along one complete route.
- Addition usually combines separate successful routes.
- Complement replaces an event by “everything except that event”.
- Checking confirms that probabilities remain between 0 and 1 and branches account for the whole next stage.
How to Use This Classroom
- Read the chance process before reading the final request.
- Pause at every Your Turn prompt.
- Draw the structure yourself before opening the worked answer.
- If you are wrong, identify whether the error came from the sample space, the route, the arithmetic or the interpretation.
- Repeat with one condition changed.
- Move to mixed exam transfer only after the basic route is stable.
You need paper, a pen and enough space to draw grids and branching trees. A calculator may help with arithmetic in some questions, but it cannot decide what the event means.
1. Start With the Probability Scale
A probability lies from 0 to 1 inclusive.
- 0 means impossible.
- 1 means certain.
- 1/2 represents an event with probability halfway between impossible and certain.
Probabilities may also be expressed as decimals or percentages when appropriate. The representation can change; the chance does not.
Teacher: Ask the student whether 1.2 can be a probability. The answer must be no. If a calculation produces 1.2, the error is already proved even before the exact correction is found.
2. Separate Outcomes From Events
An outcome is one possible result of an experiment. An event is a collection of outcomes satisfying a condition.
If a fair die is rolled, the outcome 4 is one result. The event “roll an even number” contains three outcomes: {2, 4, 6}.
This connects directly to Chapter 1 Sets: an event behaves like a set of outcomes.
Your Turn 1
A fair die is rolled. Let E be the event “the number is greater than 4”. List E.
Check your answer
E = {5, 6}. The event contains the outcomes that satisfy the condition.
3. Build the Sample Space Before Counting
The sample space is the collection of all possible outcomes. If the outcomes are equally likely, probability can be found by counting favourable outcomes and dividing by the total number of outcomes.
For a fair six-sided die:
S = {1,2,3,4,5,6}.
If A = “roll a factor of 6”, then A = {1,2,3,6}. Hence:
P(A) = 4/6 = 2/3.
The calculation is correct because the six die outcomes are equally likely.
4. Do Not Assume Equal Likelihood Without Evidence
A list of outcomes does not automatically mean the outcomes have equal probabilities. If a spinner has unequal sectors, counting colour names alone is not enough. The physical or stated chance structure matters.
Ask:
- Is the die stated to be fair?
- Are the cards or counters selected randomly?
- Are spinner sectors equal?
- Does the problem give probabilities directly?
Probability calculation begins with the model, not with a formula.
5. Teach the Complement Early
If A is an event, the complement A′ means that A does not occur. Since either A occurs or it does not:
P(A′) = 1 − P(A).
If P(rain tomorrow) = 0.35 under a stated model, then P(no rain tomorrow) = 0.65.
Complements become especially powerful in combined events when phrases such as at least one create many direct routes but a single simple opposite route.
6. Possibility Diagrams Are Grids of Combined Outcomes
A possibility diagram is useful when two results combine in a rectangular grid. Two dice, two spinners or two independent selections from simple finite sets can often be displayed this way.
If Coin 1 and Coin 2 are tossed, the four ordered outcomes are:
- HH
- HT
- TH
- TT
For fair independent coins, these four outcomes are equally likely.
Your Turn 2
Two fair coins are tossed. Find the probability of exactly one head.
Worked answer
The successful outcomes are HT and TH, two out of four equally likely outcomes. P(exactly one head) = 2/4 = 1/2.
7. Ordered Pairs Matter in Two-Dice Problems
If a red die and a blue die are rolled, the outcome (2,6) differs from (6,2). The first number belongs to the red die and the second to the blue die.
A 6 × 6 possibility diagram therefore contains 36 ordered outcomes. This makes questions about sums, differences, products and comparisons much easier to organise.
8. Teacher Model 1: Sum of Two Dice
A red die and a blue die are fair. Find the probability that the sum is 8.
The favourable ordered pairs are:
(2,6), (3,5), (4,4), (5,3), (6,2).
There are 5 favourable outcomes out of 36 equally likely ordered outcomes.
P(sum = 8) = 5/36.
Teacher question: Why are (2,6) and (6,2) both counted? Because the dice are distinguishable by colour.
9. Possibility Diagrams Prevent Missing Cases
When a question asks for “sum greater than 9”, mental listing can easily omit a pair. A grid lets you mark every cell satisfying the condition and count systematically.
The diagram is not extra working. It is an error-control device.
10. Tree Diagrams Are for Stages
A tree diagram is useful when one event happens, then another, and the second stage may depend on the first.
Each branch represents one possible next outcome. A complete route from the start of the tree to an endpoint represents one combined outcome.
Tree rule: branch probabilities leaving the same node must add to 1.
This gives an immediate check before any route probabilities are calculated.
11. Multiply Along One Complete Route
Suppose a fair coin is tossed twice. The probability of Head then Tail is:
P(HT) = 1/2 × 1/2 = 1/4.
The multiplication corresponds to one route requiring both stages to occur in sequence.
Say the route aloud: “Head on the first toss and then Tail on the second.” The word “and then” makes the multiplication meaningful.
12. Add Separate Successful Routes
Exactly one head in two tosses can happen through HT or TH. These are separate complete routes.
P(exactly one head) = P(HT) + P(TH) = 1/4 + 1/4 = 1/2.
Multiply along a route. Add between alternative successful routes.
Do not memorise the sentence without drawing the routes. The routes are the reason the arithmetic works.
13. Replacement Resets the Composition
A bag contains 3 red and 2 blue counters. One counter is selected, replaced, and a second counter is selected.
Because the first counter is replaced, the bag returns to 3 red and 2 blue before the second selection. Therefore the second-stage probabilities remain:
- P(R) = 3/5
- P(B) = 2/5
The second draw uses the same probability structure as the first.
14. Without Replacement Changes the Next Stage
Now use the same bag, but do not replace the first counter.
If the first counter is red, the bag now contains 2 red and 2 blue counters. The second-stage probabilities become 2/4 and 2/4.
If the first counter is blue, the bag now contains 3 red and 1 blue. The second-stage probabilities become 3/4 and 1/4.
The first outcome changes the composition, so the second-stage probability must be recalculated on each branch.
Your Turn 3
A bag contains 4 green and 3 yellow counters. Two are drawn without replacement. Find P(green then yellow).
Answer
P(GY) = 4/7 × 3/6 = 12/42 = 2/7.
15. Denominators Tell the Story of Replacement
In a without-replacement question, the total number of objects usually decreases after the first selection. That should appear in the denominator.
If a bag starts with 7 objects, the second draw without replacement is from 6 objects. If your denominator is still 7, pause and ask whether you accidentally modelled replacement.
Denominator changes are not a cosmetic detail. They record the new sample space.
16. Teacher Model 2: Different Colours Without Replacement
A box contains 4 green and 3 yellow cards. Two cards are drawn without replacement. Find the probability that the colours are different.
There are two successful routes:
- G then Y;
- Y then G.
P(GY) = 4/7 × 3/6 = 2/7.
P(YG) = 3/7 × 4/6 = 2/7.
Therefore:
P(different colours) = 2/7 + 2/7 = 4/7.
The route structure determines both operations: multiplication inside each path, then addition between the two successful paths.
17. “Exactly One” Means One Success and One Failure
For two trials, “exactly one success” usually has two orderings:
- success then failure;
- failure then success.
Do not count the routes with two successes or zero successes.
Teacher: Ask the student to list the route patterns before inserting any probabilities. This separates event meaning from arithmetic.
18. “At Least One” Often Calls for a Complement
“At least one” means one or more. Direct calculation may require several successful routes. The complement is often “none”.
If a fair die is rolled twice, find the probability of at least one 6.
The complement is no 6 on either roll.
P(no 6) = 5/6 × 5/6 = 25/36.
Therefore:
P(at least one 6) = 1 − 25/36 = 11/36.
The important step was recognising the simpler opposite event.
19. Do Not Use Complements Automatically
A complement is useful only when the opposite event is simpler. If a question asks for exactly one red in two draws, the complement includes zero red and two red, so it may not be shorter than calculating the two direct routes.
Method choice should reduce complexity, not merely follow a keyword.
20. Mutually Exclusive Events Cannot Happen Together
Two events are mutually exclusive if they cannot occur in the same trial.
On one die roll:
- A = “roll a 2”;
- B = “roll a 5”.
A and B cannot both occur on the same roll. Therefore:
P(A or B) = P(A) + P(B) = 1/6 + 1/6 = 1/3.
Addition works directly here because the events do not overlap.
21. Non-Mutually-Exclusive Events Have an Overlap
On one die roll, let A = “even number” and B = “number greater than 3”.
A = {2,4,6} and B = {4,5,6}. The outcomes 4 and 6 belong to both events.
These events are not mutually exclusive. Blindly adding 3/6 + 3/6 double-counts 4 and 6.
This is the probability version of the union-counting issue from Sets. Overlap matters.
22. Independent Events Do Not Change Each Other’s Probabilities
Two events are independent if the occurrence of one does not change the probability of the other.
Two tosses of a fair coin are independent. The first result does not change the probability of Head on the second toss.
If P(A) = 1/2 and P(B) = 1/3 for independent events, then:
P(A and B) = 1/2 × 1/3 = 1/6.
Multiplication reflects that both independent conditions must occur.
23. Independent Is Not the Same as Mutually Exclusive
| Idea | Question to ask |
|---|---|
| Mutually exclusive | Can both events happen in the same trial? |
| Independent | Does one event change the probability of the other? |
Mutually exclusive events with positive probabilities are not independent, because if one occurs, the other becomes impossible in that trial.
Independent events can occur together. Their defining feature is lack of probability influence, not separation.
24. Replacement Often Creates an Independent Repeat
If a counter is chosen at random, replaced, mixed and then another counter is chosen under the same conditions, the original composition has been restored. This often creates identical stage probabilities.
Do not merely assume independence because the word “replace” appears. Read the full process. But replacement is a strong structural signal that the chance mechanism has reset.
25. Without Replacement Usually Creates Dependence
If the first selected object is not returned, the composition changes. The probability on the second draw can therefore depend on the first result.
That is why a tree diagram is especially useful: different first branches can lead to different second-stage probabilities.
26. Branch Probabilities Must Complete the Next Stage
Suppose one node of a tree has only two possible next outcomes, A and B. If P(A) = 0.37, then P(B) = 0.63.
The outgoing branches must add to 1 because one of the possible next outcomes must occur.
This lets you reconstruct missing branch probabilities before calculating complete routes.
Your Turn 4
A tree node has two branches: Success and Failure. P(Success) = 0.42. Find P(Failure).
Answer
P(Failure) = 1 − 0.42 = 0.58.
27. Teacher Model 3: With Replacement
A bag contains 2 red and 3 blue counters. A counter is selected, replaced, and then a second counter is selected. Find the probability of two red counters.
Each draw has P(R) = 2/5 because replacement restores the bag.
P(RR) = 2/5 × 2/5 = 4/25.
Now find exactly one red:
P(RB) = 2/5 × 3/5 = 6/25.
P(BR) = 3/5 × 2/5 = 6/25.
Therefore P(exactly one red) = 12/25.
28. Teacher Model 4: Without Replacement
A bag contains 2 red and 3 blue counters. Two counters are selected without replacement. Find the probability of two red counters.
The first red has probability 2/5. After one red is removed, one red remains among four counters.
P(RR) = 2/5 × 1/4 = 1/10.
Compare this with the replacement answer 4/25. The process changed, so the probability changed.
29. Route Order Matters When the Question Specifies Order
“Red then blue” is one route. “One red and one blue in any order” is two routes.
Read these phrases carefully:
| Question wording | Routes included |
|---|---|
| red then blue | RB only |
| blue then red | BR only |
| one red and one blue | RB and BR |
| same colour | RR and BB |
| different colours | RB and BR |
The arithmetic is easy once the route language is correct.
30. Use a Route Table Before a Tree if the Tree Feels Crowded
For a two-stage problem, you can first list route labels such as RR, RB, BR and BB. Then attach probabilities. This is useful when a student loses track of which endpoints satisfy the event.
The route table does not replace the tree. It helps the student see the logical structure before drawing.
31. Decide Between a Possibility Diagram and a Tree Diagram
| Use | When it is especially useful |
|---|---|
| Possibility diagram | Two simple outcome sets combine into a grid and outcomes are easiest to inspect together. |
| Tree diagram | The experiment happens in stages, especially when later probabilities can change. |
The correct representation is the one that makes the chance structure easiest to see without omitting cases.
32. Teacher Model 5: Choosing the Representation
Question A: Two dice are rolled and the sum is examined. Use a possibility diagram because a 6 × 6 grid makes the sums easy to inspect.
Question B: Two counters are drawn without replacement from a bag. Use a tree diagram because the second-stage probabilities depend on the first draw.
Representation choice is part of problem solving. It should be deliberate.
33. Language Drill: Words That Change the Probability Model
| Phrase | What it should trigger |
|---|---|
| at least one | consider the complement “none” |
| exactly one | include one-success routes only |
| without replacement | change later-stage probabilities |
| with replacement | restore the original composition |
| both | one route or intersection requiring both conditions |
| either A or B | check whether the events are mutually exclusive |
| independent | one event does not alter the other’s probability |
| mutually exclusive | the events cannot occur together |
| in any order | include all valid route orderings |
34. Do Not Choose Operations From Single Keywords
The word “and” does not always mean “multiply immediately”. The word “or” does not always mean “add immediately”.
Instead:
- identify the event;
- draw or list the valid routes;
- calculate each complete route;
- combine the successful routes.
The operation follows the structure. Keywords only help you notice possible relationships.
35. Misconception Clinic: Adding Along a Route
If P(R on first draw) = 3/5 and P(B on second draw) = 2/5 under replacement, the probability of R then B is not 3/5 + 2/5 = 1.
The route requires both stages to occur in sequence, so multiply:
3/5 × 2/5 = 6/25.
Addition would combine alternative complete routes, not successive stages of one route.
36. Misconception Clinic: Multiplying Alternative Routes
If “different colours” can occur by RB or BR, those routes are alternatives. You do not need RB and BR to happen simultaneously. Calculate each route, then add them.
Teacher phrase: “Along = multiply; alternatives = add.”
37. Misconception Clinic: Keeping the Same Denominator Without Replacement
If one object leaves a bag and is not returned, the next sample space is smaller. A student who keeps the original denominator has ignored the changed experiment.
Correction routine:
- cross out the first selected object physically in a sketch or tally;
- recount what remains;
- write the new numerator and denominator before multiplying.
38. Misconception Clinic: Independent Means “Different”
Two events can be different but still dependent. Drawing a red counter first can change the probability of drawing red again when there is no replacement.
Independence asks whether one event changes the probability of the other, not whether the event labels are different.
39. Misconception Clinic: Mutually Exclusive Means Independent
If two positive-probability events are mutually exclusive, the occurrence of one makes the other impossible in that trial. That is the opposite of “no influence”.
Keep the definitions separate by asking the two diagnostic questions from Section 23.
40. Misconception Clinic: “At Least One” Means Exactly One
At least one includes one, two, three or more successes depending on the number of trials. Exactly one includes one success only.
With two coin tosses:
- exactly one head: HT, TH;
- at least one head: HH, HT, TH.
Make the student list routes before calculating until this distinction becomes automatic.
41. Misconception Clinic: Forgetting an Ordering
“One red and one blue” usually includes RB and BR unless the question specifies order. Missing one route often halves the correct probability in symmetric situations.
A route list written before calculation prevents this.
42. Misconception Clinic: Probabilities Do Not Add to 1 at a Node
If a tree node has only two possible next outcomes and their probabilities are 0.6 and 0.5, the branch model is impossible because the total is 1.1.
Check each node before calculating route probabilities. This catches transcription errors early.
43. Misconception Clinic: Final Probability Outside 0 to 1
A final probability below 0 or above 1 proves that something is wrong. Do not merely round or continue. Inspect:
- whether routes were double-counted;
- whether mutually exclusive alternatives were identified correctly;
- whether branch probabilities add to 1;
- whether a complement was subtracted in the correct direction;
- whether the arithmetic was entered correctly.
44. Guided Practice Set A: Single Events and Complements
A fair die is rolled.
- Find P(odd number).
- Find P(number greater than 2).
- Find P(not a multiple of 3).
- Find P(number equal to 7).
Solutions
1 odd outcomes {1,3,5}: 3/6 = 1/2. 2 outcomes {3,4,5,6}: 4/6 = 2/3. 3 multiples of 3 are {3,6}, so not a multiple of 3 has probability 4/6 = 2/3. 4 impossible on a six-sided die: 0.
45. Guided Practice Set B: Two Fair Coins
Two fair coins are tossed. Use the sample space HH, HT, TH, TT.
- Find P(two heads).
- Find P(exactly one head).
- Find P(at least one head).
- Find P(no heads).
Solutions
1 1/4. 2 2/4 = 1/2. 3 3/4. 4 1/4.
46. Guided Practice Set C: Two Dice
A red and a blue fair die are rolled.
- Find P(sum = 7).
- Find P(sum = 12).
- Find P(both dice show the same number).
- Find P(red die greater than blue die).
Solutions
1 Sum 7 has six ordered pairs: 6/36 = 1/6. 2 Only (6,6): 1/36. 3 Six doubles: 6/36 = 1/6. 4 There are 15 ordered pairs with red greater than blue, so 15/36 = 5/12.
47. Guided Practice Set D: With Replacement
A bag contains 3 red and 2 blue counters. One counter is selected, replaced, and a second is selected.
- Find P(RR).
- Find P(RB).
- Find P(exactly one red).
- Find P(at least one red).
Solutions
1 3/5 × 3/5 = 9/25. 2 3/5 × 2/5 = 6/25. 3 RB + BR = 6/25 + 6/25 = 12/25. 4 Complement of BB: 1 − (2/5 × 2/5) = 1 − 4/25 = 21/25.
48. Guided Practice Set E: Without Replacement
A bag contains 3 red and 2 blue counters. Two are selected without replacement.
- Find P(RR).
- Find P(RB).
- Find P(exactly one red).
- Find P(at least one red).
Solutions
1 3/5 × 2/4 = 3/10. 2 3/5 × 2/4 = 3/10. BR = 2/5 × 3/4 = 3/10, so exactly one red = 3/5. BB = 2/5 × 1/4 = 1/10, so at least one red = 1 − 1/10 = 9/10.
49. Guided Practice Set F: Missing Branch Probability
A machine produces an acceptable item with probability 0.92 and a defective item otherwise. Two independent items are selected from repeated production under the same model.
- Find P(defective).
- Find P(both acceptable).
- Find P(exactly one defective).
- Find P(at least one defective).
Solutions
1 0.08. 2 0.92² = 0.8464. 3 0.92(0.08) + 0.08(0.92) = 0.1472. 4 1 − 0.8464 = 0.1536.
50. Challenge Practice: Exactly One Success in Three Trials
A fair coin is tossed three times. Find the probability of exactly one head.
List the successful route patterns first: HTT, THT, TTH.
Worked solution
Each route has probability 1/2 × 1/2 × 1/2 = 1/8. There are three successful routes, so P(exactly one head) = 3/8.
51. Challenge Practice: At Least One Success in Three Trials
A fair coin is tossed three times. Find the probability of at least one head.
The complement is no heads, which is TTT.
Worked solution
P(TTT) = 1/8. Therefore P(at least one head) = 1 − 1/8 = 7/8.
52. Challenge Practice: Same Colour Without Replacement
A bag contains 5 red and 3 blue counters. Two are drawn without replacement. Find the probability that they are the same colour.
The successful routes are RR and BB.
Worked solution
P(RR) = 5/8 × 4/7 = 20/56. P(BB) = 3/8 × 2/7 = 6/56. Total = 26/56 = 13/28.
53. Challenge Practice: Probability From a Possibility Grid
Two fair six-sided dice are rolled. Find the probability that the sum is greater than 9.
Possible sums are 10, 11 and 12. Count the ordered pairs:
- sum 10: (4,6), (5,5), (6,4) → 3;
- sum 11: (5,6), (6,5) → 2;
- sum 12: (6,6) → 1.
Solution
There are 6 favourable ordered outcomes out of 36, so the probability is 6/36 = 1/6.
54. Examination Method: Write the Event Before the Numbers
For a tree problem, write a route statement such as:
P(different colours) = P(RB) + P(BR).
Then substitute branch probabilities. This separates translation from arithmetic and makes the reasoning easier to inspect.
55. Examination Method: Mark Replacement Explicitly
Write with replacement or without replacement near the tree before filling second-stage probabilities. This prevents the most common structural error in selection questions.
If no replacement occurs, cross out or adjust the first selected category in a small tally before writing the next fraction.
56. Examination Method: Label Endpoints
At the end of each tree route, write the outcome label: RR, RB, BR, BB or the corresponding event description.
Then circle the endpoints satisfying the question. This dramatically reduces route omission when several branches look similar.
57. Examination Method: Check the Tree Before Calculating
- Do outgoing branch probabilities at every node add to 1?
- Did replacement or non-replacement alter the correct branches?
- Are all possible next outcomes represented?
- Are the successful endpoints correctly identified?
Only after these checks should you multiply route probabilities.
58. Examination Method: Keep Exact Fractions Until the End
If the probabilities are given naturally as fractions, keep exact fractions through the route where practical. Premature decimal rounding can create avoidable error.
If decimals are given by the question, preserve enough digits and follow the required accuracy in the final answer.
59. Examination Method: Use Complements Strategically
When you see “at least one”, ask:
Is “none” a single simple route?
If yes, the complement is often faster and less error-prone. If the opposite event is equally complicated, calculate directly.
60. Examination Method: Interpret the Final Probability
A probability is not just a number. If the final result is 0.72, it represents a 72% chance under the stated model. If the context asks for a comparison or decision, attach the number to the event it describes.
Do not write 0.72 and then discuss the opposite event by mistake. Keep the event label beside the value.
61. Oral Classroom Check
A teacher can test this chapter without a worksheet. Ask:
- What is the difference between an outcome and an event?
- When would you choose a possibility diagram?
- When would you choose a tree diagram?
- Why do probabilities multiply along a route?
- Why do separate successful routes get added?
- What changes when there is no replacement?
- What does mutually exclusive mean?
- What does independent mean?
- Why are those two ideas different?
- When is a complement especially useful?
The student should answer in complete sentences and use one example. If the definition can only be repeated but not applied, return to the relevant model.
62. Exit Ticket
A bag contains 4 red and 2 blue counters. Two counters are drawn without replacement.
- Find P(RR).
- Find P(BB).
- Find P(different colours).
- Find P(at least one red).
Exit-ticket solution
P(RR) = 4/6 × 3/5 = 2/5. P(BB) = 2/6 × 1/5 = 1/15. Different colours = RB + BR = 4/6 × 2/5 + 2/6 × 4/5 = 4/15 + 4/15 = 8/15. At least one red = 1 − P(BB) = 14/15. Check: 2/5 + 1/15 + 8/15 = 6/15 + 1/15 + 8/15 = 1.
63. Homework: Retrieval, Variation and Transfer
Use three layers rather than twenty identical exercises.
Layer 1 — Retrieval
- Define outcome, event and sample space.
- Write the complement rule from memory.
- Explain mutually exclusive and independent in your own words.
- Write “multiply along, add alternatives” and give one example.
Layer 2 — Variation
- One two-dice possibility-diagram question.
- One tree question with replacement.
- One tree question without replacement.
- One “exactly one” question.
- One “at least one” question solved by complement.
Layer 3 — Transfer
Create a simple chance game using coins, dice, cards or coloured counters. State the rules, draw the sample space or tree, and calculate the probability of at least two different events. Then change one rule—such as replacement—and explain how the probability model changes.
Designing a valid chance process is a strong test because every branch, route and probability must remain internally consistent.
64. The Full Chapter Routine
For a single-event question, use:
Sample space → event → favourable outcomes → probability → check 0 to 1.
For a possibility-diagram question, use:
Build grid → mark valid cells → count systematically → divide by total equally likely cells.
For a tree question, use:
Stages → branch probabilities → replacement check → endpoint labels → multiply routes → add successful routes → final check.
For “at least one”, add one extra question:
Is the complement simpler?
65. Why This Chapter Matters Beyond Probability
Combined probability trains several mathematical habits at once: enumerating possibilities, preserving conditions, tracking changing states, distinguishing alternative routes from sequential routes, and checking whether a model accounts for the whole process.
These habits reappear in statistics, decision-making, algorithms, reliability, risk, finance, science and everyday reasoning under uncertainty.
Probability does not remove uncertainty. It gives uncertainty a structure precise enough to reason about.
66. Connect Back to Sets
Chapter 1 gave you union, intersection and complement. Probability uses the same logical relationships on events:
- A ∪ B means A or B or both;
- A ∩ B means A and B;
- A′ means not A.
If event language feels confusing, return to the Chapter 1 Sets Classroom and practise translation before doing more probability arithmetic.
67. Ready for Chapter 3?
You are ready to move on when you can do all of the following without prompts:
- build a sample space for a simple experiment;
- distinguish outcomes from events;
- use complements correctly;
- choose between a possibility diagram and a tree diagram;
- fill branch probabilities so each node sums to 1;
- adjust probabilities correctly without replacement;
- multiply along one route;
- add separate successful routes;
- distinguish mutually exclusive from independent events;
- recognise “exactly one” and “at least one” structures;
- check that the final probability lies from 0 to 1; and
- explain why the calculation matches the chance process.
If one item is weak, return to that section and solve a changed example. If all are stable, continue to statistical data analysis, where uncertainty moves from possible outcomes to observed data.
