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Secondary 2 G3 Mathematics Tutor Bukit Timah | Systematic Outcome Tables for Probability

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by

eduKate Asia
Three learners review open books together at a classroom table, with stacks of textbooks, stationery and a whiteboard in the bright room.

Secondary 2 G3 Mathematics Tutor Bukit Timah | Systematic Outcome Tables for Probability. Did you know? Tossing two coins creates four equally likely ordered outcomes, not three. ‘One head and one tail’ can happen as HT or TH, and leaving out either order changes the probability even though the final two coins look similar.

This Secondary 2 G3 Mathematics guide for Bukit Timah families teaches students how to list sample spaces systematically, build two-way outcome tables, count ordered pairs, avoid repeated or missing outcomes and calculate simple probabilities. We begin with a quick answer, then work through original coins, dice, spinners and counting tasks. Later-stage without-replacement examples are marked as extensions when they go beyond the child’s immediate test scope.

The objective is not to turn every simple probability question into a huge diagram. It is to develop a reliable way of ensuring that every possible result has been considered exactly once. If a student can explain the dimensions of an outcome table and justify why its entries are or are not equally likely, the numerical fraction becomes much more dependable.

Find the outcome-table explanation you need

Quick answer: list distinct ordered outcomes before counting

A sample space is the complete set of possible outcomes for a defined experiment. For two distinguishable fair coins, or a first and second independent fair toss, the ordered outcomes are HH, HT, TH and TT, each probability 1/4. Exactly one head occurs in HT and TH, so its probability is 2/4=1/2. The words ‘first’, ‘second’, ‘and’, ‘or’ and ‘exactly’ decide which outcomes qualify.

For two fair six-sided dice distinguished as die A and die B, a 6-by-6 table has 36 equally likely ordered pairs. Count the cells satisfying the event and divide by 36. When outcomes are not equally likely—such as a spinner whose sectors have unequal areas—an outcome count alone is insufficient; use the specified probability weights instead.

What makes an outcome ordered?

If you toss a coin twice, heads then tails and tails then heads are different sequences, even though both have one of each result. If you draw two coloured counters one after another, red-blue and blue-red may describe different time orders. A table labelled ‘first result’ by rows and ‘second result’ by columns makes this distinction visible.

Not every question asks for ordered outcomes. Choosing two members to form an unordered pair is different from appointing a first and second member. Read whether the roles or time positions are distinct before counting. Otherwise you may double-count or undercount the sample space.

How to build a complete row-and-column outcome table

Put every possible result of the first stage along the left side and every second-stage result along the top. Each intersection represents exactly one ordered pair. For two distinguishable fair dice, the cell at row 2 and column 5 represents (2,5), while the cell at row 5 and column 2 represents (5,2). These are different outcomes although they have the same sum.

A 6-by-6 grid contains 36 cells. Under the model of independent fair dice, each ordered pair has probability 1/36. A total of seven appears in six cells: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1), so its probability is 6/36=1/6.

Why the multiplication principle counts outcomes

If the first stage has m possible choices and, for every first-stage choice, the second has n available choices, there are m×n ordered outcomes. This is the same rectangular count shown by the table. A spinner with four equally likely labelled sectors followed by a three-sector spinner has 4×3=12 distinct ordered sector pairs.

When a choice cannot repeat, the second-stage count changes. Selecting two different letters in order from A,B,C gives three choices first and two remaining choices second, producing six ordered pairs. The multiplication principle remains useful, but the allowed number of choices must follow the actual conditions.

Not all equally labelled outcome types are equally likely

Suppose a bag contains three red counters and two blue, and a counter is drawn uniformly, replaced and drawn again. The colour sequences RR,RB,BR,BB are all possible, but they are not equally likely: their probabilities are 9/25,6/25,6/25 and4/25. Counting one colour sequence as one of four equally likely options would therefore give false answers.

To use equally likely elementary outcomes, temporarily distinguish the physical counters by labels R1,R2,R3,B1,B2. With replacement the outcome table has 5×5=25 equally likely ordered labelled pairs. Nine have two red counters, six have red then blue, six blue then red and four two blue. That table supports the unequal colour probabilities correctly.

Replacement changes what the second stage can contain

Without replacement, a five-counter bag has five choices on the first draw and four remaining labelled-counter choices on the second, giving 20 ordered labelled outcomes under uniform drawing. If three are red and two blue, red-red has 3×2=6 ordered outcomes, red-blue has3×2=6, blue-red has2×3=6 and blue-blue has2×1=2. The probabilities are 6/20,6/20,6/20 and2/20.

The resulting colour probabilities are 3/10,3/10,3/10 and1/10, summing to one. This is a systematic counting alternative to drawing a probability tree. It is a useful extension; the core lesson remains listing all outcomes fairly and tracking what choices are permitted.

Complement events can shorten a table calculation

For two fair dice, ‘at least one six’ qualifies whenever the first die is six or the second is six, including the double-six pair. Instead of listing all favourable cases, count the complement ‘neither die shows six’. There are 5×5=25 such pairs, leaving 36−25=11 favourable pairs. The probability is 11/36.

The complement method is not a trick that changes probabilities. It partitions the complete sample space into an event and its non-event. Their probabilities sum to one. A student should still define both events carefully before subtraction.

How to avoid counting a shared outcome twice

If event A is ‘first die shows six’ and event B is ‘second die shows six’, each includes six ordered pairs. But (6,6) belongs to both. Their union therefore contains 6+6−1=11, not 12. This is the same inclusion-exclusion reasoning used in Venn diagrams, now applied to pairs of dice.

With a simple one-die sample, even results {2,4,6} and values above three {4,5,6} also overlap at {4,6}. Their union is {2,4,5,6}: four favourable results out of six. Recognising overlap matters even before pupils learn a formal probability addition rule.

A reliable six-point outcome-table check

  • Write the trial sequence and label what the first and second results mean.
  • List all possible stage-one outcomes and all stage-two possibilities permitted after each stage-one result.
  • Keep the same row/column orientation, and do not silently treat unordered selections as ordered pairs.
  • Check that the elementary cases are equally likely before using favourable/total counts directly.
  • Mark each favourable cell once, remembering overlaps when conditions use ‘or’.
  • Confirm the probability lies between zero and one and that the sum of complete mutually exclusive outcomes is one.

Forty-eight original sample-space and outcome-table clinics

One fair coin

Systematic counting question. A fair coin is tossed once. List possible outcomes and P(heads).

Complete result. Sample {H,T}; P(H)=1/2.

Why this counting method applies. Two elementary results are equally likely in the assumed fair model. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. A probability of one would mean heads is certain.

Independent table check. T is the complementary result with probability1/2. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Two fair tosses

Systematic counting question. A fair coin is tossed twice, first and second. List all ordered outcomes.

Complete result. HH,HT,TH,TT; each has probability1/4.

Why this counting method applies. The two time positions are distinguishable. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Only HH,HT,TT misses TH.

Independent table check. Four probabilities sum one. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Exactly one head

Systematic counting question. Two fair independent tosses: probability exactly one head?

Complete result. HT and TH qualify, giving2/4=1/2.

Why this counting method applies. Exactly one head can occur in either time order. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Choosing only HT halves the true probability.

Independent table check. Neither head is TT with probability1/4. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

At least one head

Systematic counting question. Two fair independent tosses: P(at least one head)?

Complete result. HH,HT,TH qualify; 3/4.

Why this counting method applies. Only TT is excluded, so complement also gives1−1/4. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Exactly one is a smaller event.

Independent table check. P(no heads)=1/4. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Same results on two tosses

Systematic counting question. Two fair coins: probability both outcomes match?

Complete result. HH or TT, so2/4=1/2.

Why this counting method applies. Matching outcomes include both double heads and double tails. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Considering only HH omits a matching case.

Independent table check. Different outcomes HT or TH also total1/2. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

First head, second tail

Systematic counting question. Two fair tosses require heads first and tails second.

Complete result. Only HT, probability1/4.

Why this counting method applies. The condition specifies the order exactly. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Adding TH would answer one head of either order.

Independent table check. Compare with P(exactly one head)=1/2. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Three fair tosses

Systematic counting question. How many ordered results in three fair coin tosses?

Complete result. 2×2×2=8 equally likely sequences.

Why this counting method applies. Each position has two possibilities. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. 2+2+2=6 is not a counting product.

Independent table check. List HHH,HHT,HTH,HTT,THH,THT,TTH,TTT. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Exactly two heads in three

Systematic counting question. Three fair tosses: P(exactly two heads)?

Complete result. HHT,HTH,THH are favourable, so3/8.

Why this counting method applies. There are three positions for the sole tail. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Counting only HHT forgets permutations.

Independent table check. P(all heads)=1/8. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Exactly one head in three

Systematic counting question. Three fair tosses: P(exactly one head)?

Complete result. HTT,THT,TTH are favourable; 3/8.

Why this counting method applies. The single head can occupy any of three positions. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. One head means no second head anywhere else.

Independent table check. P(no head)=1/8. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

At least one head in three

Systematic counting question. Three fair tosses: P(at least one head)?

Complete result. 1−P(TTT)=1−1/8=7/8.

Why this counting method applies. The complement is one easily counted sequence. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Only counting exactly one head gives3/8.

Independent table check. The favourable event also includes two or three heads. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

One fair die

Systematic counting question. A fair six-sided die is rolled once. P(even)?

Complete result. Even set {2,4,6}; probability3/6=1/2.

Why this counting method applies. Three of six faces satisfy the event. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. The even count three is not itself a probability.

Independent table check. Odd complement also has3/6. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

A prime face

Systematic counting question. One fair six-sided die: P(prime face)?

Complete result. Prime faces2,3,5; probability3/6=1/2.

Why this counting method applies. One is not a prime number. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Including one would overcount.

Independent table check. Composite face set4,6 has probability2/6. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Two fair dice table

Systematic counting question. Roll two distinguishable fair dice. How many ordered pairs?

Complete result. 6×6=36.

Why this counting method applies. Die A and die B provide the row and column results. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Treating (1,2) and(2,1) as the same pair undercounts.

Independent table check. All pairs have probability1/36. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Dice sum seven

Systematic counting question. Two fair dice: P(sum seven)?

Complete result. Six outcomes (1,6),(2,5),(3,4),(4,3),(5,2),(6,1); 6/36=1/6.

Why this counting method applies. A sum is shared by several ordered cells. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Only three unordered combinations loses half.

Independent table check. Two dice can produce sum7 in either order. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Dice sum eight

Systematic counting question. Two fair dice: P(sum eight)?

Complete result. (2,6),(3,5),(4,4),(5,3),(6,2), five results, 5/36.

Why this counting method applies. A diagonal across the outcome table contains the five cells. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Counting (4,4) twice is incorrect.

Independent table check. Sum seven is more likely than sum eight. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Dice sum two

Systematic counting question. Two fair dice: P(sum two)?

Complete result. Only (1,1), so1/36.

Why this counting method applies. The smallest possible sum has one ordered realization. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Two is not two favourable cases.

Independent table check. P(sum twelve) is also1/36. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Dice sum at least ten

Systematic counting question. Two fair dice: P(sum10,11 or12)?

Complete result. Counts3+2+1=6; probability6/36=1/6.

Why this counting method applies. High sums occupy six distinct cells. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Treating each sum as one equally likely outcome is false.

Independent table check. Sum9 has four ordered outcomes. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Dice sum less than five

Systematic counting question. Two fair dice: P(sum2,3 or4)?

Complete result. Counts1+2+3=6; probability1/6.

Why this counting method applies. Count the ordered combinations for each possible sum. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Only summing numbers2+3+4 does not count outcomes.

Independent table check. The distribution is symmetric about sum7. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Double dice faces

Systematic counting question. Two fair dice: probability both faces show same number?

Complete result. Pairs (1,1) through(6,6), six cells; 1/6.

Why this counting method applies. Six diagonal table cells represent doubles. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Counting one category ‘double’ out of seven categories is wrong.

Independent table check. P(different faces)=5/6. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Difference exactly one

Systematic counting question. Two fair dice: absolute face difference is one.

Complete result. Ten ordered pairs, so10/36=5/18.

Why this counting method applies. Five consecutive adjacent number pairs can occur in two orders. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Counting five unordered pairs misses orientations.

Independent table check. List (1,2),(2,1) through(5,6),(6,5). Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

At least one six

Systematic counting question. Two fair dice: P(at least one die six)?

Complete result. 36−5×5=11 favourable cells; probability11/36.

Why this counting method applies. Complement counts no six on either die. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. 6+6=12 double-counts (6,6).

Independent table check. P(neither six)=25/36. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Exactly one six

Systematic counting question. Two fair dice: P(exactly one six)?

Complete result. First six and second not six gives5; reverse gives5; total10/36=5/18.

Why this counting method applies. Double six is explicitly excluded. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. At least one six has eleven cells, not ten.

Independent table check. Exactly one and two sixes are disjoint. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

No six

Systematic counting question. Two fair dice: P(neither six)?

Complete result. 5×5=25 cells; P=25/36.

Why this counting method applies. Each die has five allowable faces in the event. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. 5/6 alone describes only one die.

Independent table check. Complement11/36 checks. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Dice product odd

Systematic counting question. Two fair dice: P(product odd)?

Complete result. Both faces must be odd:3×3=9 cells, 9/36=1/4.

Why this counting method applies. Odd times odd is odd; any even factor produces even. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Treating one odd die as sufficient is wrong.

Independent table check. P(product even)=3/4. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Dice product even

Systematic counting question. Two fair dice: P(product even)?

Complete result. 1−P(both odd)=1−9/36=27/36=3/4.

Why this counting method applies. Even product needs at least one even factor. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Counting exactly one even omits even-even.

Independent table check. The complement is odd product. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Two dice sum prime

Systematic counting question. Two fair dice: P(sum is prime)?

Complete result. Prime sums2,3,5,7,11 have1+2+4+6+2=15 cells, so15/36=5/12.

Why this counting method applies. List valid sums and count each sum’s ordered pairs. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Using five favourable prime values out of eleven sums assumes equal sum likelihood.

Independent table check. Check the five counts add15. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Die and coin

Systematic counting question. Toss a fair coin and roll a fair die once. How many pairs?

Complete result. 2×6=12 equally likely outcomes.

Why this counting method applies. The coin has two results and die six. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Adding2+6 gives eight, not twelve.

Independent table check. P(head and even)=3/12=1/4. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Coin H and even die

Systematic counting question. From independent fair coin and die, P(H and even)?

Complete result. H2,H4,H6 are three of twelve pairs, so1/4.

Why this counting method applies. Three die faces combine with specified H. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. P(H)=1/2 alone is not sufficient.

Independent table check. Count (H,2),(H,4),(H,6). Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Three shirts four trousers

Systematic counting question. A wardrobe has three distinct shirts and four distinct trousers. Count outfits.

Complete result. 3×4=12 distinct shirt-trouser combinations.

Why this counting method applies. Each shirt can pair with each trouser. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. 3+4 counts garments, not outfits.

Independent table check. A 3-by-4 grid lists twelve. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Two spinners of different sizes

Systematic counting question. A fair four-sector spinner and fair three-sector spinner are spun independently.

Complete result. 4×3=12 equally likely sector pairs.

Why this counting method applies. Different stage sizes multiply for ordered combinations. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. 4+3 is not total combinations.

Independent table check. One specified pair has chance1/12. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Two-colour and three-number spinner

Systematic counting question. First equally likely R/B; second equally likely1/2/3. P(red and odd)?

Complete result. Pairs (R,1),(R,3) qualify, giving2/6=1/3.

Why this counting method applies. Only the selected red row and two odd columns are favourable. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Two red outcomes out of three would omit total six.

Independent table check. P(blue and even)=1/6. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Unequal sector probabilities

Systematic counting question. A spinner produces R with probability1/2, B with1/4, G with1/4. Another fair spinner is1 or2.

Complete result. A specific R-and-1 pair has probability(1/2)(1/2)=1/4; B-and-1 has1/8.

Why this counting method applies. Six labelled pairs exist but are not equally likely. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Treating all six as1/6 is false.

Independent table check. Sum all six pair weights to one. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

With replacement colour sample space

Systematic counting question. Bag has three labelled red and two labelled blue counters; draw, replace, draw.

Complete result. Five choices at each stage, giving25 equiprobable labelled pairs under uniform selection.

Why this counting method applies. Labelling physical counters creates correct elementary outcomes. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Four colour sequences are not equally likely.

Independent table check. RR has3×3=9 of25. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

With replacement exactly one red

Systematic counting question. In that 3R/2B bag with replacement, P(exactly one red)?

Complete result. RB has3×2=6 labelled pairs and BR another6, total12/25.

Why this counting method applies. Different time orders qualify. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. RR is not exactly one red.

Independent table check. P(RR)=9/25 and P(BB)=4/25; totals sum25/25. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Without replacement counted pairs

Systematic counting question. The same five labelled counters are drawn twice without replacement.

Complete result. There are5×4=20 equally likely ordered labelled pairs.

Why this counting method applies. Second draw cannot be the previously removed physical counter. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. 5×5=25 would be replacement.

Independent table check. Count RR6,RB6,BR6,BB2, sum20. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Without replacement RR

Systematic counting question. Three red and two blue, two uniform draws without replacement. P(RR)?

Complete result. RR labelled paths3×2=6 of20, giving3/10.

Why this counting method applies. Two red remain after first red. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Using 9/25 assumes replacement.

Independent table check. 3/5×2/4 also gives3/10. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Without replacement BB

Systematic counting question. Same bag: P(two blues)?

Complete result. 2×1=2 labelled pairs of20, giving1/10.

Why this counting method applies. Only one blue remains after first blue. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Counting BB as one of four equally likely colour paths is false.

Independent table check. 2/5×1/4=1/10. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Without replacement exactly one red

Systematic counting question. Same bag: P(exactly one red)?

Complete result. RB6 plus BR6 =12 labelled pairs, so12/20=3/5.

Why this counting method applies. Both orders must be included. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Counting just one order gives3/10.

Independent table check. Complement RR+BB=3/10+1/10=2/5. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Two distinct letters ordered

Systematic counting question. Choose two different letters in order from A,B,C.

Complete result. AB,AC,BA,BC,CA,CB: six choices.

Why this counting method applies. Three first choices and two second choices. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. AA cannot appear when repetition is forbidden.

Independent table check. There are only three unordered letter pairs. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Two letters with repetition

Systematic counting question. Choose two letters in order from A,B,C, repetition allowed.

Complete result. 3×3=9 sequences, including AA,BB,CC.

Why this counting method applies. Second position retains three options. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Using 3×2 assumes no repetition.

Independent table check. All nine can be listed in a 3-by-3 grid. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Unordered pair of letters

Systematic counting question. Choose two different letters without any first or second role from A,B,C.

Complete result. Pairs {A,B},{A,C},{B,C}: three choices.

Why this counting method applies. AB and BA describe the same unordered selection. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Treating the six ordered outcomes as six unordered pairs doubles count.

Independent table check. If roles are assigned, six ordered selections return. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

A three-digit code

Systematic counting question. A code has three positions, each selected from0,1,2 with repetition allowed.

Complete result. 3³=27 equally likely codes under uniform selection.

Why this counting method applies. Leading zeros are allowed in a code. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. A three-digit whole number cannot start zero, a different restriction.

Independent table check. 000 and012 are both codes here. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

A three-digit whole number

Systematic counting question. Form an ordinary three-digit number using digits0,1,2 with repetition allowed.

Complete result. First digit2 options (1 or2), other digits3 each: 2×3×3=18.

Why this counting method applies. Leading zero is excluded for a three-digit number. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. 3³=27 counts three-character codes including leading zero.

Independent table check. 100 is valid; 012 is not a three-digit whole number. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

Code with exactly one zero

Systematic counting question. From three-position codes with digits0,1,2 and repetition allowed, find codes containing exactly one zero.

Complete result. Choose zero position3 ways; other two each have2 choices, total3×2²=12 of27.

Why this counting method applies. Position choice and nonzero digit choices are independent steps. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Counting codes with two zeros includes unwanted outcomes.

Independent table check. Probability under uniform code selection12/27=4/9. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

A spinner pair sum

Systematic counting question. Two independent fair spinners each show1,2 or3. P(sum four)?

Complete result. Ordered pairs (1,3),(2,2),(3,1), 3 of9, so1/3.

Why this counting method applies. Three combinations from a 3-by-3 table give four. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Only listing (2,2) omits other pairs.

Independent table check. Sum two is only (1,1), probability1/9. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

One die union event

Systematic counting question. Fair die: A even and B greater than three. P(A or B)?

Complete result. A={2,4,6}, B={4,5,6}; union={2,4,5,6}, so4/6=2/3.

Why this counting method applies. Overlap at4 and6 must be counted once. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. 3+3=6 favourable counts the two overlaps twice.

Independent table check. Intersection has2/6=1/3. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

One die intersection

Systematic counting question. Fair die: even and greater than three simultaneously.

Complete result. Outcomes {4,6}, so2/6=1/3.

Why this counting method applies. The word ‘and’ restricts to the overlap. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. The union {2,4,5,6} answers ‘or’.

Independent table check. Both outcomes meet both conditions. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

A complement event

Systematic counting question. Fair die: probability not prime?

Complete result. Prime {2,3,5}; nonprime {1,4,6}, 3/6=1/2.

Why this counting method applies. The complement includes every face not prime. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. One is not prime and must be included.

Independent table check. Prime and nonprime sets partition all six faces. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

A table with missing outcome

Systematic counting question. A learner lists two tosses as HH,HT,TT. What is missing?

Complete result. TH is missing.

Why this counting method applies. Each first-stage result must pair with each second-stage result. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Treating HT and TH as the same when order matters is invalid.

Independent table check. A complete 2-by-2 table gives four entries. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

An apparently fair four-colour sequence

Systematic counting question. Three reds and one blue are drawn twice with replacement, but someone assigns all RR,RB,BR,BB probability1/4.

Complete result. This is wrong: with uniform counters P(RR)=9/16, RB=3/16, BR=3/16, BB=1/16.

Why this counting method applies. Colour outcomes have unequal numbers of underlying counter pairs. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Listing labels does not make labels equally likely.

Independent table check. The probabilities total16/16. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

A count with zero possibility

Systematic counting question. Bag has two red and one blue; draw two without replacement. P(BB)?

Complete result. Zero, because only one blue exists.

Why this counting method applies. A second blue cannot be drawn once the only blue was removed. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. Assigning BB one of four colour sequences invents a physically impossible outcome.

Independent table check. The possible colour sequences are RR,RB,BR. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

A sample-space size check

Systematic counting question. One fair die and one fair four-sector spinner. How many ordered outcomes?

Complete result. 6×4=24.

Why this counting method applies. Each die face has four spinner partners. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. 6+4=10 is an item count, not a combination count.

Independent table check. A specified ordered pair has chance1/24 under independence and fairness. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

An equal-likelihood assumption check

Systematic counting question. A first spinner has sectors of unequal sizes, and a second spinner is fair. Can we divide favourable labels by all labels?

Complete result. Not unless the probabilities of elementary pairs are equal; unequal sector sizes require weighted outcomes.

Why this counting method applies. Counting works directly only for equally likely elementary cases. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. A rectangular table is a layout, not proof of fairness.

Independent table check. Label probabilities before computing the event. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

At least two identical by pigeonhole

Systematic counting question. Five objects are each assigned one of four categories. Must two share a category?

Complete result. Yes, by the pigeonhole principle.

Why this counting method applies. Five assigned objects cannot occupy four categories with at most one in each. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. This counting conclusion does not say which category repeats.

Independent table check. A useful optional extension, not a probability calculation by itself. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

The total-one audit

Systematic counting question. A complete equally likely four-outcome model lists probabilities1/4,1/4,1/4,1/4.

Complete result. They sum to one.

Why this counting method applies. Complete mutually exclusive events exhaust the sample space. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. A fifth identical case may signal double-counting.

Independent table check. Check the number of allowed table cells. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

A final language change

Systematic counting question. Two coin tosses: compare ‘at least one head’ with ‘exactly one head’.

Complete result. At least one has HH,HT,TH =3/4; exactly one has HT,TH =1/2.

Why this counting method applies. Changing one word changes the marked cells. Name the ordered elementary outcomes and the relevant fairness assumptions before turning a favourable count into a probability.

First mistake to catch. A correct table can be reused, but the event changes.

Independent table check. Say explicitly why HH belongs in only the first event. Confirm that no permitted pair was omitted or counted twice, and distinguish a count of labelled cases from a probability-weighted colour category.

A twenty-minute three-pupil experiment with one common sample space

Give three pupils a paper 6-by-6 grid for two fair dice. Ask one to highlight cells with a sum of seven, a second to mark doubles, and a third to mark at least one six. They should obtain six, six and eleven cells respectively, each out of 36. Now exchange the questions. The physical grid does not change, only the event being selected, demonstrating why a carefully built table is reusable across multiple questions.

After that, provide a three-red/two-blue counter bag model. Students should first list the four colour words RR,RB,BR,BB, then discuss why those four labels are not equally likely. A counter-labelled 5-by-5 table gives 25 outcomes for replacement and produces probabilities9/25,6/25,6/25,4/25. This comparison tests understanding of the probability model, not only arithmetic.

A two-week progression for sample-space reasoning

Week one: Build complete outcome sets for one die, two coins and a fair coin-plus-die experiment. Use row-and-column tables so every pair is accounted for. Practise changing ‘exactly’ to ‘at least’, and mark new favourable cells without redrawing the entire sample space. Check every calculated probability between zero and one.

Week two: Introduce separate counting questions involving clothing combinations, ordered letters and codes. Distinguish repetition allowed from not allowed. For pupils ready for extension, compare labelled counter selections with and without replacement and explain why colour words can have unequal probabilities. Finish with a no-heading problem in which the student must decide whether order matters.

Frequently asked: why do HT and TH count separately?

They are different ordered sequences: heads first then tails, versus tails first then heads. If the experiment consists of a first and second coin toss, those results are distinct even though both contain one head. If an entirely different question considers only the unordered collection of two coins, it must define that sample space and its probability weights appropriately.

Frequently asked: can I always use favourable cases divided by total cases?

Only when the elementary cases counted in the denominator are equally likely under the stated model. A fair die’s six faces are equiprobable; colour sequences drawn from an uneven mix of counters are generally not. If cases have different probabilities, add the appropriate probability weights or refine the outcomes to labelled equally likely cases.

Frequently asked: is a two-way table the same as a probability tree?

Both can organise multistage outcomes. A table is especially clear when two stages have a small fixed set of results, while a tree displays changing conditions and probabilities naturally. This guide emphasises systematic enumeration and cell counting; the separate probability-tree guide develops conditional branch probabilities in more detail.

Frequently asked: what if the question does not say whether repetition is allowed?

Look for language such as ‘with replacement’, ‘without replacement’, ‘different letters’ or ‘code may repeat digits’. If the condition is truly absent and changes the count, a student should not silently invent it. In school questions, the surrounding context or instructions typically specify which model to use.

Frequently asked: why does a double-six need special treatment?

When counting ‘at least one six’, the ordered pair (6,6) belongs to both ‘first die shows six’ and ‘second die shows six’. Adding both six-cell sets without subtracting their shared pair double-counts it. The correct total is 6+6−1=11 out of36.

Frequently asked: are advanced without-replacement questions compulsory in Secondary 2?

The current school course sequence determines assessment coverage. The official lower-secondary G3 material includes sample spaces and probability of single events; more elaborate multistage conditional calculations may be reserved for later. Use simple two-coin and die tables as the core and treat physical-counter extensions according to the pupil’s readiness.

Original teaching reference, official Mathematics and further reading

The unchanged eduKateSG small-group Mathematics reference describes three-pupil weekly tutorials of about 1.5 hours, close feedback and the published 8 Fourth Avenue, Singapore 268674 venue near Sixth Avenue MRT. The Bukit Timah title refers to nearby families, not additional premises or guaranteed places.

G3 is a Mathematics subject level under Full Subject-Based Banding. The official MOE G2/G3 Mathematics syllabus gives the broader probability foundation, while school chapter notices determine immediate tests.

For complementary topics read single-event probability, two-stage probability trees and sets and overlapping events. This article’s distinct focus is building and auditing complete outcome tables.

For a Secondary 2 G3 Mathematics consultation near Bukit Timah, review the original programme reference and bring one marked probability problem and the student’s next assessed chapters. Confirm current group fit, availability and timetable directly.

Reviewed October 2026. All dice, coins, counters and spinners are fictional mathematical models unless their fairness conditions are explicitly stated. The examples are original educational illustrations, not copied school examination questions.