Thirty-six small boxes, and one large idea
Two fair six-sided dice are rolled. A question asks for the probability that their total is seven. Many students remember that seven is ‘common’ but cannot show why. Others list the six pairs that total seven without first saying how many equally likely outcomes exist. The answer becomes clear when one die labels the rows of a six-by-six diagram and the other labels the columns. Each box records one ordered pair: first die, second die. There are thirty-six boxes because the first die has six possible results and so does the second.
Exactly six boxes have total seven: (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1). If the dice are fair and their results independent, each ordered pair has probability 1/36. The event therefore has probability 6/36 = 1/6. The important thing is not the numerical answer but the method: define an outcome precisely, construct the complete sample space, identify the event and count the event’s outcomes once.
This chapter builds beyond Volume 0262 on the ‘at least one’ complement and Volume 0258 on changing denominators. Those chapters own their particular probability decisions. This volume owns the systematic possibility diagram: how to construct ordered-pair cells, translate a sum or condition into a region, and recognise when cells are not equally likely.
The official 2027 G3 Mathematics examination context
The Singapore–Cambridge Secondary Education Certificate begins in 2027. G3 Mathematics uses SEAB syllabus K310; the previous O-Level reference code is 4052. K310 has two compulsory papers, each lasting 2 hours 15 minutes and carrying 90 marks. Paper 1 and Paper 2 each contribute 50% of the subject grade. Paper 1 contains approximately twenty-six short-answer questions. Paper 2 contains about nine to ten questions, with its last question a real-world application of mathematics.
The K310 Probability content includes probabilities of single events and simple combined events, including use of possibility diagrams and tree diagrams where appropriate. It also covers reasoning about mutually exclusive and independent events. Check the 2027 SEC G3 syllabus directory and SEAB’s K310 syllabus PDF for the official specifications. The numerical questions and worked solutions below are original exercises, not past SEC questions or predictions about a specific paper.
A diagram is a tool for representing simple outcomes, not a new compulsory method for every probability calculation. Some events are easier to solve by a tree, a complement or a direct rule. The examination goal is to choose the representation that keeps the sample space accurate and makes the reasoning easy to inspect. Show enough working to justify a probability; writing the final fraction alone may not communicate how you counted it.
Begin with the experiment, not the grid
Before drawing anything, ask what is being repeated or paired. With two dice, the outcome can be written (a,b), where a records the first die and b records the second. The ordered pair (2,5) differs from (5,2). They may have the same sum, but they are two distinct outcomes. The grid has six rows for a and six columns for b. There are thirty-six cells. All cells are equally likely only when both dice are fair and the rolls are independent.
Compare a different experiment: one fair spinner has three equal sectors numbered 1, 2 and 3; a second independent spinner has four equal sectors numbered 1, 2, 3 and 4. The diagram has three rows and four columns, or twelve equally likely cells. The size of a possibility diagram comes from the number of outcomes of each device, not from how many sums or colours you can name after the event.
The total probability of the entire sample space must be one. If a grid omits an outcome or counts a cell twice, the whole model becomes unreliable. For a dice grid, each row has six cells and each column has six. This simple structural check is often faster than checking every written pair individually.
Ordered pairs are not an optional detail
When the first die shows 2 and the second shows 5, the result is recorded as (2,5). Switching the values gives (5,2), a different outcome unless the experiment expressly says that order is irrelevant and the probabilities are adjusted accordingly. This distinction becomes important in conditions such as ‘the first die shows a larger number than the second’. The event is not symmetric: (6,2) qualifies, while (2,6) does not.
Even if two identical-looking dice are rolled, they can be distinguished for modelling, perhaps by colour or by the order in which results are recorded. A list of twenty-one unordered numerical combinations is not a list of twenty-one equally likely outcomes. A double such as {3,3} corresponds to one ordered pair; a mixed pair such as {3,5} corresponds to two. Treating them as equal gives incorrect probabilities.
A good examination habit is to label axes before marking qualifying cells. Write ‘first die’ at one edge and ‘second die’ at the other. When you read the condition, translate it into a statement about a and b. That one sentence prevents many mistakes that are otherwise hidden inside a neat-looking grid.
Sums form diagonal patterns
With two fair six-sided dice, totals range from two to twelve. The counts of outcomes with each total follow the sequence 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1. The sequence can be derived from the diagram rather than memorised. There is only one way to total two: (1,1). There are two ways to total three: (1,2) and (2,1). The number rises until the total of seven, then falls because the faces cannot exceed six.
The counts add to thirty-six. This is a useful check that each cell has been allocated to exactly one sum. To calculate the probability of an event such as ‘the total is at least ten’, identify the sums ten, eleven and twelve. Their counts are three, two and one. Six cells qualify, so the probability is 6/36 = 1/6. The wording tells you which totals belong, while the diagram tells you how many outcomes those totals represent.
Do not count the number of acceptable totals and divide by eleven. The totals are not equally likely; seven has six ordered pairs, while two and twelve each have one. This is the central purpose of a possibility diagram in such problems: it preserves the number of elementary outcomes behind each summary value.
Conditions are regions, not separate guessing games
A condition such as ‘first die greater than second die’ describes a triangular region of the grid. In the row where the first die is one, there are no smaller second-die results. In row two there is one. Then two, three, four and five. The count is 0 + 1 + 2 + 3 + 4 + 5 = 15, so the probability is 15/36 = 5/12. The corresponding region with first die less than second also has fifteen cells. Six cells are doubles. Together the three mutually exclusive regions account for all thirty-six outcomes.
An event may be defined by several restrictions at once. ‘The first die is even and the total is odd’ means a is in {2,4,6} and a+b is odd. Because a is even, b must be odd, giving three choices for each of the two coordinates: nine cells. The probability is 9/36 = 1/4. The word ‘and’ makes the qualifying region an intersection.
When a condition contains ‘or’, take care not to count an overlap twice. For example, ‘the first die is six or the second die is six’ covers six cells in the final row and six in the final column, but (6,6) belongs to both. The event has eleven cells, not twelve. This is ordinary counting of a union; see Volume 0103 on set overlap when a Venn diagram provides a better representation.
Even and odd conditions make careful counting easier
A fair die has three odd faces and three even faces. With two fair dice, the sum is even when both results have the same parity: odd–odd or even–even. There are 3 × 3 + 3 × 3 = 18 qualifying cells, so the probability is one half. The sum is odd when one die is odd and the other even, also eighteen cells. This derivation is more reliable than a vague memory that ‘half the sums look even’.
For products, the logic changes. A product is odd only when both factors are odd. That is nine cells out of thirty-six. The complement—at least one even factor—has twenty-seven cells, so the probability of an even product is 27/36 = 3/4. These events may sound similar, but ‘even sum’ and ‘even product’ describe different regions of the same grid.
A good transfer exercise changes one word in the question. Find the event for ‘both results even’, then ‘exactly one even’, then ‘at least one even’. Their counts are nine, eighteen and twenty-seven respectively. The correct calculation depends on what the wording includes, not just which number looks familiar.
Absolute difference requires checking both orders
Suppose the dice must differ by exactly one. The qualifying results are (1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6) and (6,5). There are ten outcomes, so the probability is 10/36 = 5/18. The word ‘differ’ concerns absolute difference, so both directions count.
Contrast this with ‘the first die is one greater than the second’. Only five cells qualify: (2,1), (3,2), (4,3), (5,4), (6,5). That probability is 5/36. The two questions use much of the same arithmetic but have different event sets. Writing the condition as |a−b| = 1 or a = b+1 makes the distinction visible.
The useful exam technique is to convert descriptive language into a simple rule and check that the listed cells satisfy it. If you can explain why one pair belongs and its reverse does not, you probably understand the event.
When the grid is not equally likely
A possibility diagram does not guarantee equal probability merely because it has tidy rows and columns. Suppose a spinner is red on half its surface, blue on one quarter and green on one quarter. A second independent spinner has four equal numbered sectors. The twelve colour–number combinations are not equally likely: each red–number combination has probability (1/2) × (1/4) = 1/8, while each blue–number or green–number combination has probability (1/4) × (1/4) = 1/16. Counting red cells as two out of twelve would be wrong.
The difference is a distinction between labelled categories and equal-probability elementary outcomes. When probabilities vary, record the probability of each row, column or outcome and combine them correctly. Alternatively, use a tree diagram to make branch probabilities explicit. The earlier Volume 0178 on probability trees is the primary home for that representation.
This also explains why four possible colour routes when drawing counters without replacement need not each have probability one quarter. The numbers of red and blue counters and the removal of the first counter affect the second draw. Return to Volume 0258 for the detailed changing-denominator method rather than pretending every colour route is equally likely.
Do not confuse an event with a result summary
A sum of seven is an event that gathers six ordered pairs. The number seven is not itself one of the thirty-six elementary outcomes; it is a value calculated from them. Similarly, the event ‘at least one six’ includes many pairs with different totals. When a question presents a summary such as sum, product, maximum or difference, first translate the summary into a rule on the underlying pair.
This matters for sums that are prime. The prime totals among two dice are 2, 3, 5, 7 and 11. The number of ordered pairs with those totals is 1 + 2 + 4 + 6 + 2 = 15. The probability is 15/36 = 5/12, not five elevenths. The correct answer depends on the counts behind each total.
A quick reasonableness check is that every probability lies between zero and one and each region belongs to the complete grid. For a condition involving ‘or’, check overlap; for ‘and’, check that both requirements hold; for ‘exactly’, make sure the other possibilities are excluded. These are mathematical reading habits as much as arithmetic habits.
How to decide between a diagram, tree and complement
Use a possibility grid when two small sets of outcomes form a natural row-and-column structure, especially if the event is a sum, a comparison or a property of both results. Use a tree when the second event depends on the first, probabilities change along branches or several stages must be followed. Use a complement when the opposite event is shorter to describe and calculate. None of the tools is automatically superior; the right choice is the one that preserves the sample space with the least risk of error.
If the question asks ‘at least one six’ on two independent fair dice, a complement is efficient: the chance of no six on either die is (5/6) × (5/6), so the desired probability is 11/36. A grid gives the same result by counting the union of row six and column six. The complement method belongs in Volume 0262; here we use the grid to audit that count.
If two coloured counters are selected without replacement, equal-sized cell counting may be inappropriate. A tree of conditional branches helps preserve the changing composition. A student’s aim should not be to draw the longest diagram. It should be to make an answer checkable, with labelled outcomes, correct probability weights and a clear event.
Workings that let a marker follow your reasoning
A concise response to the total-seven question could be: ‘There are 6 × 6 = 36 equally likely ordered outcomes. Six ordered pairs give total seven, namely (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Therefore P(total seven) = 6/36 = 1/6.’ The sample-space size, event count and probability conclusion are all visible. In a longer problem you may draw and mark a table instead of listing all pairs.
A response is weaker when it writes only ‘1/6’ or when it gives ‘six ways’ without explaining why the denominator is thirty-six. The arithmetic might be right by luck. Show the modelling choice. In situations where event cells have unequal probability, display the probability calculation rather than counting categories as though they are equal.
Before submitting, test the boundaries. A sum on two ordinary six-sided dice cannot be one or thirteen. A product may be odd only with two odd values. A first result cannot be smaller than zero. These checks catch event misinterpretation quickly. They are not substitutes for the full method, but they can expose a stray cell before it affects the probability.
From primary-school possibility listing to G3 proof
At the PSLE stage, learners often begin by listing outcomes carefully in patterns or small tables. The secondary-school development is to formalise the complete sample space, distinguish elementary outcomes from summary values, reason about independence and explain the event’s probability. You are not expected to abandon clear primary-school methods; you are expected to know why those methods work and where they fail.
The G3 guide to mutually exclusive and independent events, Volume 0182, explains important related distinctions. A possibility diagram helps make them concrete: mutually exclusive complete outcomes occupy different cells, while independence describes probabilities of the paired results. Those are different ideas. Confusing them can produce the wrong calculation even when a diagram appears perfectly neat.
The next time a question asks for a probability, write three short labels before calculating: experiment, elementary outcome, event. If you can fill them accurately, a diagram is usually straightforward. If you cannot, the confusion is in the interpretation, and extra arithmetic will not repair it.
Twenty-eight worked possibility-diagram clinics: build, count, check
01. Total seven on two dice
Question. Two independent fair six-sided dice are rolled. Find P(sum = 7).
Full answer. There are 36 equally likely ordered pairs. The six satisfying pairs are (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1). Thus P = 6/36 = 1/6. The two orders for different numbers count separately because the dice are distinguishable in the sample space.
Transfer. Check that (3,4) and (4,3) occupy different boxes in the diagram. Explain why seven is much more frequent than two.
02. Smallest possible total
Question. Find the probability of a total of two.
Full answer. Only the pair (1,1) produces two. Therefore P = 1/36. There are eleven distinct totals from two to twelve, but they are not equally likely. The denominator must be the number of elementary ordered outcomes, not the number of different sums.
Transfer. Change the target to total three and show that two ordered pairs, (1,2) and (2,1), qualify.
03. At least ten as a total
Question. Find P(sum ≥ 10).
Full answer. The qualifying sums are 10, 11 and 12. Sum ten has three ordered pairs, sum eleven has two, and sum twelve has one. Therefore P = (3+2+1)/36 = 6/36 = 1/6. A sum of nine must not be included because ‘at least ten’ includes ten and larger values only.
Transfer. List all six qualifying ordered pairs to check the diagonal count directly.
04. At most five as a total
Question. Find P(sum ≤ 5).
Full answer. Sums 2, 3, 4 and 5 correspond to 1, 2, 3 and 4 ordered pairs. The total number of qualifying cells is 10, so P = 10/36 = 5/18. The maximum value in the event is five, but that does not mean only five cells qualify.
Transfer. Describe which cells fail the event and use the complement as a numerical check.
05. Equal numbers
Question. Find the probability that both dice display the same face.
Full answer. The diagonal pairs (1,1) through (6,6) form six cells. The probability is 6/36 = 1/6. ‘Same number’ does not mean that the total must be even in the opposite direction: doubles always have even sums, but many non-double pairs also have even sums.
Transfer. Explain why the reverse assertion ‘every even sum is a double’ is false using (1,3).
06. Different numbers
Question. Find the probability that the dice show different values.
Full answer. The event is every cell except the six doubles. That leaves 36−6 = 30 cells, giving P = 30/36 = 5/6. The complement is convenient, but the diagram remains the proof of what has been excluded.
Transfer. Check that the probabilities for ‘same’ and ‘different’ add to one.
07. The first result is larger
Question. Find P(first die greater than second die).
Full answer. For first-die values 1, 2, 3, 4, 5 and 6, the numbers of permitted second-die values are 0, 1, 2, 3, 4 and 5. The total is 15. Thus P = 15/36 = 5/12. This is a triangular half of the grid excluding the diagonal.
Transfer. Use symmetry to show that ‘first smaller’ also has 15 outcomes, and reconcile the remaining six.
08. Difference exactly one
Question. Find P(|first − second| = 1).
Full answer. The adjacent-number pairs (1,2) and (2,1), continuing up to (5,6) and (6,5), give ten ordered outcomes. Hence P = 10/36 = 5/18. The absolute value symbol means either result may be larger; omitting reverse pairs halves the count incorrectly.
Transfer. Compare this with the one-direction condition ‘first is one greater than second’, which has only five cells.
09. Difference exactly two
Question. Find P(|first − second| = 2).
Full answer. The increasing-order pairs are (1,3), (2,4), (3,5), (4,6). Each has a reverse ordered pair. Eight cells qualify, so P = 8/36 = 2/9. This is a pair of bands parallel to the diagonal of doubles.
Transfer. Change the required difference to five. How many ordered outcomes survive? Check the extreme values.
10. Sum even
Question. Find P(sum is even).
Full answer. Two odd results produce an even sum, as do two even results. A fair die has three odd faces and three even faces. There are 3×3 + 3×3 = 18 qualifying pairs, so P = 18/36 = 1/2. A sum cannot be even merely because at least one result is even.
Transfer. List the parity categories that fail: odd–even and even–odd.
11. Sum odd
Question. Find P(sum is odd).
Full answer. An odd sum requires one odd and one even result. There are 3×3 odd–even pairs and 3×3 even–odd pairs, for 18 outcomes. Therefore P = 18/36 = 1/2. The reasoning is structural and works without writing all eighteen ordered pairs.
Transfer. Verify that even-sum and odd-sum events are mutually exclusive and exhaustive.
12. Product even
Question. Find P(product is even).
Full answer. A product is odd only if both dice show odd numbers. Nine ordered pairs have both values odd. The other 27 yield an even product, so P = 27/36 = 3/4. This differs from the even-sum probability because the underlying parity condition is different.
Transfer. Explain why (2,3) has an odd sum but an even product.
13. Product odd
Question. Find P(product is odd).
Full answer. Both numbers must be odd: each die can show 1, 3 or 5. There are 3×3 = 9 qualifying ordered pairs. Thus P = 9/36 = 1/4. The property of the product depends on both factors; one odd factor is insufficient if the other is even.
Transfer. Check that product-even and product-odd probabilities total one.
14. Both even
Question. Find the probability that both dice show even numbers.
Full answer. Each die has even faces 2, 4 and 6, so there are 3×3 = 9 qualifying cells and P = 9/36 = 1/4. This is not the same as ‘at least one even’, which includes two additional strips beyond the square of both-even outcomes.
Transfer. Describe the 3-by-3 block on the diagram.
15. Exactly one even
Question. Find the probability that exactly one die shows an even number.
Full answer. The qualifying parity patterns are even–odd and odd–even. Each has 3×3 = 9 outcomes, giving 18/36 = 1/2. Both-even pairs are excluded because ‘exactly one’ disallows two even results. Both-odd pairs are also excluded.
Transfer. Compare the event with ‘at most one even’ and explain which additional cells would be included.
16. A prime total
Question. Find the probability that the sum is a prime number.
Full answer. Prime totals possible are 2, 3, 5, 7 and 11. Their ordered-pair counts are 1, 2, 4, 6 and 2, summing to 15. The probability is 15/36 = 5/12. Do not confuse prime face numbers with prime totals: a pair of composite or non-prime faces can still give a prime sum.
Transfer. List one pair producing prime total seven that contains neither a prime face greater than three nor two prime faces.
17. Multiple of three total
Question. Find P(sum is divisible by three).
Full answer. The possible sums are 3, 6, 9 and 12. They correspond to 2, 5, 4 and 1 ordered pairs. Hence P = 12/36 = 1/3. The counts matter because there are four eligible totals but these are not four equally likely categories.
Transfer. Check the result another way by grouping faces by remainders modulo three.
18. Two restrictions at once
Question. Find P(first die even and sum odd).
Full answer. The first die can be 2, 4 or 6. To make the sum odd, the second die must be odd: 1, 3 or 5. These choices form 3×3 = 9 ordered pairs, giving P = 9/36 = 1/4. Both parts of the word ‘and’ must hold.
Transfer. Change ‘sum odd’ to ‘sum even’ and explain which second-die faces qualify.
19. A number comparison and a threshold
Question. Find P(first die odd and second die greater than four).
Full answer. There are three choices for the first result, 1, 3 and 5, and two choices for the second, 5 and 6. Thus 3×2 = 6 ordered pairs qualify, giving P = 6/36 = 1/6. This rectangular event region is easy to see in a labelled grid.
Transfer. Explain why swapping ‘first’ and ‘second’ gives a different set of pairs even if the probability remains the same here.
20. A union with one overlap
Question. Find P(first die six or second die six).
Full answer. Six cells lie in the row with first result six and six in the column with second result six. Their intersection (6,6) was counted twice, so the number of distinct cells is 6 + 6 − 1 = 11. Thus P = 11/36. The diagram makes the shared cell visible.
Transfer. Compare the answer with the complement calculation (5×5 excluded cells) and name the distinct event owner in Volume 0262.
21. Three-by-four spinner diagram
Question. Spinner A has three equally sized colour sectors R, G, B. Independent spinner B has four equally sized sectors 1, 2, 3, 4. Find P(red and even number).
Full answer. There are 3×4 = 12 equally likely colour–number outcomes. Red with 2 and red with 4 are the two qualifying cells. P = 2/12 = 1/6. The second device may have numbered outcomes while the first has colours; the multiplication principle remains the same.
Transfer. Find P(green or numbered 1) by counting its row and column with the shared cell counted once.
22. Unequal spinner sectors
Question. Spinner A has a red half and blue and green quarters; spinner B has four equal numbered sectors. Find P(red and even number).
Full answer. The red probability is 1/2, while the second spinner gives an even number with probability 2/4 = 1/2. Assuming independent spins, the required probability is (1/2)(1/2) = 1/4. Simply counting two red-even categories out of twelve categories gives the wrong value because the colour categories have unequal probabilities.
Transfer. Compare P(blue and even) = (1/4)(1/2) = 1/8. Explain why a labelled three-by-four grid needs probabilities attached to cells in this experiment.
23. Unordered faces are not equally likely
Question. A student writes the unordered outcomes {1,1}, {1,2}, …, {6,6} and treats the twenty-one distinct combinations as equally likely. Explain the error.
Full answer. For two fair dice, {1,1} has one ordered realisation, (1,1), but {1,2} has two: (1,2) and (2,1). These unordered categories have different probabilities. The correct elementary sample space contains thirty-six equally likely ordered results. A category list of twenty-one can be useful only if its unequal weights are respected.
Transfer. Check that P(sum three) = 2/36, not 1/21, to show the practical consequence.
24. First result exceeds the sum
Question. A question asks for the probability that the first die exceeds the total of both dice. Identify the event.
Full answer. There are no qualifying outcomes. Since the second die is at least one, first + second is always greater than the first value. Therefore the probability is 0/36 = 0. A zero-probability event can be understood from the defining inequality without marking all cells.
Transfer. Explain why this is a logical impossibility under the stated faces, not merely an event that failed to appear in a small experiment.
25. First is exactly one more
Question. Find P(first die = second die + 1).
Full answer. The valid pairs are (2,1), (3,2), (4,3), (5,4) and (6,5). That is five cells, so P = 5/36. The reverse order does not satisfy the same directional statement. This is a useful test of whether the row and column labels were read correctly.
Transfer. Write a sentence distinguishing this event from an absolute difference of one.
26. Even results and high total
Question. Find P(both dice even and the sum greater than nine).
Full answer. Both-even pairs use faces 2, 4 and 6. Among the nine such pairs, only (4,6), (6,4) and (6,6) have sums above nine. There are three qualifying outcomes, so P = 3/36 = 1/12. An ‘and’ condition narrows the region to cells satisfying both restrictions.
Transfer. Change ‘greater than nine’ to ‘at least nine’; determine whether any additional even-even pairs qualify.
27. A fixed total and parity
Question. Find P(sum seven and first die even).
Full answer. The six sum-seven pairs are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Only (2,5), (4,3) and (6,1) have an even first result. Thus the probability is 3/36 = 1/12. Begin with the sum restriction, then apply the parity condition.
Transfer. Explain why the event has a different count from ‘sum seven or first die even’.
28. A bag instead of two dice
Question. A bag has three red and two blue counters. Two counters are drawn without replacement. Can a two-by-two colour diagram have four equal-probability cells?
Full answer. No. The four possible colour orders have unequal probabilities: red–red 3/5×2/4 = 3/10; red–blue 3/5×2/4 = 3/10; blue–red 2/5×3/4 = 3/10; blue–blue 2/5×1/4 = 1/10. They sum to one, but the four categories are not equiprobable. A tree can display the changing counts more naturally.
Transfer. Explain which denominator changes after the first draw and how this differs from independent fair-dice results.
A ten-point audit before you hand in a probability answer
Audit 1: Is the experiment defined?
Write down the devices and how they produce results. Two independent fair dice are not the same experiment as two counters removed from a bag. If you still have time, test the result against one simple boundary case or count the opposite region. A probability answer becomes trustworthy when its sample space and event can be explained without relying on a memorised fraction.
Audit 2: Is an outcome ordered?
In a row-and-column grid, (a,b) records which device produced which value. Unless the model justifies a different treatment, reversed mixed pairs count separately. If you still have time, test the result against one simple boundary case or count the opposite region. A probability answer becomes trustworthy when its sample space and event can be explained without relying on a memorised fraction.
Audit 3: Are the cells equally likely?
Equal-looking categories need not be equiprobable. For a spinner with unequal sector sizes, weight cells by their actual probabilities. If you still have time, test the result against one simple boundary case or count the opposite region. A probability answer becomes trustworthy when its sample space and event can be explained without relying on a memorised fraction.
Audit 4: Have you stated the sample-space size?
For two independent fair six-sided dice there are 36 equally likely ordered pairs. State the basis before dividing a count. If you still have time, test the result against one simple boundary case or count the opposite region. A probability answer becomes trustworthy when its sample space and event can be explained without relying on a memorised fraction.
Audit 5: Does the event include its boundaries?
At least ten includes ten; greater than ten excludes ten. Exactly one differs from at least one. Translate the words before counting. If you still have time, test the result against one simple boundary case or count the opposite region. A probability answer becomes trustworthy when its sample space and event can be explained without relying on a memorised fraction.
Audit 6: Did you count sums by outcome, not by label?
A total of seven arises from six pairs, but a total of two from only one. The eleven possible totals are not equally likely. If you still have time, test the result against one simple boundary case or count the opposite region. A probability answer becomes trustworthy when its sample space and event can be explained without relying on a memorised fraction.
Audit 7: Is an ‘and’ a true intersection?
Each counted cell must meet every stated requirement. A single correct condition is not enough when the question has two restrictions. If you still have time, test the result against one simple boundary case or count the opposite region. A probability answer becomes trustworthy when its sample space and event can be explained without relying on a memorised fraction.
Audit 8: Was an ‘or’ counted once?
When two regions overlap, avoid adding their common cells twice. Mark the intersection visually before forming the probability. If you still have time, test the result against one simple boundary case or count the opposite region. A probability answer becomes trustworthy when its sample space and event can be explained without relying on a memorised fraction.
Audit 9: Can a complement or symmetry check the count?
The same event can often be counted another way. A check is valuable when it is logically independent, not merely the same mistaken step repeated. If you still have time, test the result against one simple boundary case or count the opposite region. A probability answer becomes trustworthy when its sample space and event can be explained without relying on a memorised fraction.
Audit 10: Is the final fraction in a sensible range?
Probability lies between zero and one. Ensure the counted event is a subset of the sample space and simplify only after the reasoning is complete. If you still have time, test the result against one simple boundary case or count the opposite region. A probability answer becomes trustworthy when its sample space and event can be explained without relying on a memorised fraction.
The wider learning sequence
For the elementary idea that a set of outcomes must be complete, revisit early G3 statistics and probability foundations, Volume 0019. For route-by-route probability, use Volume 0178. For independence and mutually exclusive events, follow Volume 0182. For the new owner’s close neighbours, see Volume 0258 and Volume 0262.
The 2027 SEAB G3 Mathematics syllabus K310 remains the authority for examination content. This guide’s original examples teach how to use possibility diagrams for clearly defined two-stage sample spaces. The transferable habit is simple: know what a cell means, know whether it has the same probability as its neighbours, and count the event rather than guessing from the answer’s appearance.
