Find your next learning step
Choose the question closest to what you need today, or read the guide from the beginning.
Understand the idea · Compare the examples · See a full solution · Practise and check · Choose the next step
G3 Mathematics is a secondary school Mathematics subject level under Singapore’s Full Subject-Based Banding. This guide explains a common probability-tree difficulty: using the original bag total again after an item has been drawn and not replaced. The second draw takes place from the remaining items, so its denominator and relevant colour counts must describe that new situation.
The quick reasoning is to pause at the end of the first branch and state what remains. If a bag begins with three red counters and two blue counters, drawing a red without replacement leaves two red and two blue. The next red chance is two out of four, not three out of five. Drawing a blue first would leave three red and one blue, giving a different next red chance.
Those are conditional branch chances: they depend on the earlier result. The overall chance of red on the second draw is a separate question and can still equal the first-draw chance. Keeping those two ideas apart helps students understand a probability tree rather than copy changing fractions without knowing what each branch describes.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–19
For each branch, identify the number of suitable remaining items and the total number remaining. Divide the first by the second under the stated equally likely selection model. Multiply successive conditional branch probabilities along a complete route. Add the probabilities of separate routes when they are mutually exclusive ways to meet the required event.
Do not assume colours are equally likely simply because two colour names appear. Their counts matter. Do not assume replacement and no replacement are interchangeable. Read exactly what is returned, when it is returned and whether the original selection conditions are restored. The tree represents the process the question actually describes.
For wider context, use What Is G3 Mathematics?. The Mathematics Mastery probability-skills guide explains the broader aim of reasoning under uncertainty. This companion focuses on two draws without replacement, using original teaching examples appropriate to the student’s taught work and the 2027 SEC G3 Mathematics syllabus K310.
For a uniformly random selection from a well-mixed bag of individual counters, the probability of a colour is its counter count divided by the total counter count. With three red and two blue, red has chance three fifths and blue two fifths. The denominator five describes the five individual items available at that stage.
It does not count the number of colour names. There are two colours, but they are not equally represented. Assigning one half to each would treat three red items as though they collectively had the same chance as two blue. The selection model gives equal chances to individual counters, not automatically to categories.
After one counter is removed and not returned, four remain. The second draw’s denominator is therefore four in this example. The colour numerator depends on which counter was removed. Both parts of the fraction must be read from the current branch state rather than copied from the original bag.
Ask the learner to name what the denominator represents in words. “All counters still available for this draw” is a useful answer. It connects the number to the stage. A fraction written without that interpretation may look familiar while describing a different process from the one in the question.
The denominator can change differently in another task. Two items might be removed, one returned or new items added. There is no universal instruction to subtract one regardless of the story. The general habit is to reconstruct the available set at the current stage, then form the probability from that set.
Contents · Previous section · Next section
Begin with three red and two blue counters. If the first draw is red and that counter is not replaced, the remaining counts are two red and two blue. The second-draw probabilities on this branch are therefore two fourths for red and two fourths for blue. They sum to one.
If the first draw is blue, the remaining counts are three red and one blue. The second-draw probabilities on that branch are three fourths for red and one fourth for blue. They also sum to one. The denominator is the same because one item was removed in either case, while the colour counts differ.
This is why a tree has separate second-stage branches. Each describes a different condition inherited from the first result. The fractions beside one branch cannot automatically be moved to the other. The earlier event determines which current bag the next draw uses.
Draw or list the remaining items if needed. A learner can cross out one red in one copy of the original bag and one blue in another copy. Keep the two cases separate. Crossing both out in the same bag would describe two removals, not the alternative states after a single first draw.
Once the state update is clear, write the fractions. This order is important. A memorised instruction to reduce the numerator may fail when the first item had a different colour from the one sought next. Removing a blue does not reduce the red count; it changes the total and the blue count.
Contents · Previous section · Next section
The chance two fourths on the second red branch means the chance of red given that the first draw was red. It does not mean that the whole two-red event has probability one half. To reach that second branch, the first red draw must already have occurred, and its probability is three fifths.
The probability of red followed by red is therefore three fifths multiplied by two fourths, giving six twentieths, or three tenths. The multiplication accounts for the successive stages of that route. The second factor is appropriate because it describes the remaining bag after the first event in that route.
Read a route aloud. “First red, then red from the remaining bag” helps the student connect both fractions to the event. A bare product can be calculated correctly without understanding what it represents. Naming the ordered outcomes gives the result a clear meaning before any routes are combined.
The multiplication is not based on assuming the two draws are independent. Here the second factor explicitly changes with the first result. A tree can represent dependent stages by using the correct conditional chance after each preceding branch. Avoid teaching that multiplication is reserved only for independent events.
Use the language at the level taught in school. The learner need not introduce advanced probability notation to understand the idea. They need to know which previous result is being assumed on a branch and which current items are available. That is the meaning behind the changing fraction.
Contents · Previous section · Next section
There are four colour routes for two draws: red-red, red-blue, blue-red and blue-blue. Under the stated three-red, two-blue model without replacement, their probabilities are three tenths, three tenths, three tenths and one tenth respectively. The routes describe ordered outcomes, so red-blue and blue-red are separate cases.
Calculate each route from its own branch state. Red-blue is three fifths times two fourths. Blue-red is two fifths times three fourths. Blue-blue is two fifths times one fourth. Similar-looking products should still be attached to their meanings; equality of two numerical results does not make the events identical.
| Ordered colour route | Probability calculation | Route probability |
|---|---|---|
| Red, then red | 3/5 × 2/4 | 3/10 |
| Red, then blue | 3/5 × 2/4 | 3/10 |
| Blue, then red | 2/5 × 3/4 | 3/10 |
| Blue, then blue | 2/5 × 1/4 | 1/10 |
Check that the route probabilities sum to one. Three tenths plus three tenths plus three tenths plus one tenth gives one. The four routes cover all possible colour outcomes and do not overlap. This total check can reveal a missing route or incorrect fraction, although a sum of one alone does not prove every entry is correct.
Contents · Previous section · Next section
The event “different colours” occurs through red-blue or blue-red. These routes are mutually exclusive: a particular pair of ordered draws cannot be both at once. Add their probabilities, three tenths plus three tenths, to obtain six tenths or three fifths.
A student who calculates only red-blue has found one valid route but not the complete event. The missing blue-red route matters unless the question specifically requests red first and blue second. Read whether order is part of the condition. “One of each colour” and “red followed by blue” are different events.
This is a good place to ask for an event description before calculating. The learner can list the acceptable colour sequences in words. Once the routes are clear, the products and sum become easier to organise. Starting with arithmetic may conceal that only half the requested outcomes were included.
Do not add branches merely because the word or appears. Check whether the routes overlap. In this small tree, complete ordered paths are disjoint, so adding their probabilities is appropriate. Other descriptions may overlap and need different handling. The sample space and event definitions justify the operation.
As a check, the same-colour probability is red-red plus blue-blue, or four tenths. Different and same exhaust the possibilities here, so their probabilities sum to one. This complement comparison supports the result while reinforcing that the event must be defined before its numerical value can be trusted.
Contents · Previous section · Next section
For two draws, “at least one red” includes red-red, red-blue and blue-red. Adding the three route probabilities gives nine tenths. It does not mean exactly one red. Exactly one red includes only the mixed-colour routes and has probability three fifths in this example.
The complement is often a useful alternative. The only way to have no red in two draws is blue-blue, with probability one tenth. Therefore at least one red has probability one minus one tenth, or nine tenths. The complement method works because the two events are mutually exclusive and together cover all outcomes.
For at least one blue, the complement is red-red, with probability three tenths. The answer is therefore seven tenths. The two at-least-one probabilities need not sum to one because they overlap: a mixed-colour pair contains at least one red and at least one blue.
This is a useful language distinction. A student may understand every branch fraction but select the wrong set of routes from the event wording. Our How Vocabulary Really Works guide offers a wider route into precise meaning. In probability, words such as at least, exactly, both and neither identify different events.
Use one tree with several event questions rather than redrawing a new tree each time. The underlying process stays fixed while the selected routes change. This isolates event interpretation from state updating and gives the teacher a clearer view of which part the learner finds difficult.
Contents · Previous section · Next section
Compare a process in which each counter is returned before the next draw, the bag is mixed again and every counter is equally likely at each selection. The original counts are then restored. The second red probability is three fifths regardless of the first colour, under this stated model.
The red-red probability becomes three fifths times three fifths, or nine twenty-fifths. Without replacement it was three tenths. The different second-stage bag produces a different combined probability. The first-draw probability alone does not tell you which product belongs to the process.
For the replacement model, the four route probabilities are nine twenty-fifths, six twenty-fifths, six twenty-fifths and four twenty-fifths. They sum to one. Different colours has probability twelve twenty-fifths, which differs from three fifths without replacement. The selection rules affect the outcome distribution.
Do not treat the words “with replacement” as permission to ignore every other condition. The task may specify additional changes or a selection mechanism. The examples here explicitly restore and mix the same bag. The general principle is to describe the actual next-stage state and selection rule before assigning probabilities.
Ask the learner to compare the two models in words before calculating. Without replacement, the first selected colour has one fewer representative next time. With the restored bag, the counts return to their starting values. That explanation gives the fractions a reason instead of turning replacement into an isolated keyword rule.
Contents · Previous section · Next section
Without replacement from three red and two blue, the second red branch chance is two fourths after a first red and three fourths after a first blue. Yet the overall probability of red on the second draw is still three fifths. Add the routes ending red: red-red plus blue-red gives three tenths plus three tenths.
There is no contradiction. A branch probability assumes a particular earlier outcome. The overall probability includes all earlier routes that lead to the requested second outcome. Different conditions are being used, so the numbers need not match. Keeping those questions separate is a major part of reading a probability tree well.
The matching first- and second-draw overall red probabilities do not make the draws independent. Knowing that the first was red changes the chance of a second red from the overall three fifths to two fourths. The dependence is visible in the branch states even though the overall single-position chances are equal.
Use this as an extension after the basic tree is understood. A learner who is still uncertain about updating the bag may need simpler practice first. The aim is to deepen the meaning of a branch, not to introduce a surprising result before the current fractions have become understandable.
A useful checking question is, “Are we assuming the first colour is already known?” If yes, read the appropriate second-stage branch. If not, identify every route ending in the requested second colour. The same wording difference can change whether a single conditional fraction or a sum of path probabilities is needed.
Contents · Previous section · Next section
Label the three red counters R1, R2 and R3, and the blue counters B1 and B2. For two ordered draws without replacement, there are five possible first counters and four possible second counters, giving twenty equally likely ordered pairs under the stated random-selection model.
Count red-red pairs. There are three choices for the first red and two remaining choices for the second, giving six pairs out of twenty. This is three tenths, matching the tree. Red-blue has three times two pairs, blue-red two times three, and blue-blue two times one.
The labels distinguish physical counters without changing their colours. This explains why a colour route can have a probability different from one quarter even though there are four named colour routes. The routes contain different numbers of equally likely individual outcomes. Categories are not automatically equally likely.
Use labelled counting as a check, not as an extra compulsory method for every tree question. For small bags it can make the probability structure concrete. For larger problems, a well-constructed tree may be more efficient. The methods should agree because they represent the same selection process.
If they disagree, compare the sample spaces. Did the counting method allow the same physical counter twice despite no replacement? Did the tree forget the second-stage update? Finding that mismatch teaches more than selecting whichever final number resembles the answer key. The process definition is the authority for both representations.
Contents · Previous section · Next section
Suppose a bag contains one red and four blue counters, and two are drawn without replacement. After a first red, no red counters remain. The second red chance on that branch is zero out of four. Therefore red-red has probability zero. Two red counters cannot be drawn when only one was present initially.
A learner who mechanically subtracts one from every numerator may write a negative count on some branch. That is a sign to return to the available items. Counts cannot become negative. The fraction should represent a possible state, including zero suitable items where an event is impossible.
The denominator can still be four even when the sought colour count is zero. The bag is not empty; it contains four blue counters. Distinguish “no favourable item” from “no item available.” Zero probability and an undefined empty selection are different situations and should not be described interchangeably.
For the stated two-draw example, every first selection leaves enough items for the second draw. A task asking for more draws than available without replacement would need different handling or indicate an impossible procedure. Read the process before extending a memorised pattern of denominators.
These boundary cases are useful checks on understanding. They show whether the learner is thinking about the bag or only changing digits in fractions. Once the counts are meaningful, impossible routes become straightforward to identify and do not require a separate collection of unexplained rules.
Contents · Previous section · Next section
An original exercise uses five red and three blue counters. Two counters are selected uniformly at random without replacement. Find the probability of two red counters, different colours and a red counter on the second draw. The question deliberately asks for three events from one unchanged process.
For two red, use five eighths for the first red and four sevenths for the next red after that removal. The product is twenty fifty-sixths, or five fourteenths. A denominator of eight on the second stage would describe the wrong available total unless the process included replacement.
For different colours, red-blue has probability five eighths times three sevenths, or fifteen fifty-sixths. Blue-red has probability three eighths times five sevenths, also fifteen fifty-sixths. Add them to obtain thirty fifty-sixths, or fifteen twenty-eighths. Both orders are included because the event does not specify which colour comes first.
For red on the second draw, add red-red and blue-red. The total is twenty fifty-sixths plus fifteen fifty-sixths, giving thirty-five fifty-sixths, or five eighths. That matches the overall first-draw red chance, while the individual second-stage red branches remain different conditional probabilities.
Check the omitted route: blue-blue is three eighths times two sevenths, or six fifty-sixths. All four path probabilities sum to fifty-six fifty-sixths. This confirms the total coverage. Also explain each selected event in words so a correct sum is not credited to an accidental choice of routes.
Contents · Previous section · Next section
At first, keep fractions such as two fourths visible because their numerator and denominator describe actual counts. Simplifying to one half is mathematically correct, but it can hide the bag state from a learner who is still building the model. Meaningful unsimplified fractions can make checking easier.
After the state is clear, simplify as appropriate for efficient calculation and the requested final form. There is no rule that all working must remain unsimplified. The point is to preserve enough information to understand why the probability was assigned, especially during explanation and correction.
A student who sees only one half may forget whether it came from two out of four, three out of six or a genuinely fair two-outcome mechanism. Those situations can have the same current chance while producing different later states. The probability number alone does not describe every feature of the process.
Use a short state note beside each first branch if needed: “two red, two blue remain” or “three red, one blue remain.” This is not extra decoration. It records the information from which the next fractions follow. Once the learner can update reliably, the note can become shorter or be omitted.
This connects to the wider habit in checking valid Mathematics solutions: a convenient simplified form should not erase the conditions that justified it. The tree is dependable when every branch still corresponds to the stated process, not merely when its fractions look neat.
Contents · Previous section · Next section
If the denominator stays at five after one unreplaced draw, ask how many counters remain. If the red numerator falls after a first blue, ask which colour was actually removed. If different colours includes only one route, list both orders before calculating. Each error concerns a different part of the model.
If every second-stage fraction is copied from the first stage, compare replacement and no replacement through actual counter lists. If a learner treats colour names as equally likely, label the individual counters and count them. A small physical or drawn model can clarify the selection mechanism better than another unexplained tree.
If the result exceeds one, inspect whether incompatible quantities were added or an event was counted more than once. A probability within zero and one can still be wrong, so the numerical bound is only one check. The sample space and selected routes must also match the question.
After explanation, change the bag counts or the requested event. Keep the process manageable so the repaired decision can be observed. A learner who reproduces one completed tree may still need help on a new one. Independent state updating is stronger evidence than copying the final products neatly.
A useful correction note states the action. “After each first branch, recount the remaining bag” is specific. “Remember probability rules” is too broad. The note should prepare the learner to handle a changed situation rather than attach only to the exact red-blue example used during teaching.
Contents · Previous section · Next section
These original tasks are informal learning checks, not placement tests. Use them after the relevant methods have been taught. Ask the learner to state the remaining counts and accepted routes before calculating. This separates process understanding from multiplication and fraction simplification.
One: four red and two blue counters are drawn twice without replacement. Two red has probability four sixths times three fifths, or two fifths. Two: from that same bag, blue-red has probability two sixths times four fifths, or four fifteenths. Removing a blue leaves all four red counters available.
Three: different colours in that bag includes red-blue and blue-red. Each route has probability four fifteenths, so the total is eight fifteenths. Four: at least one red is one minus the probability of two blue, giving one minus two sixths times one fifth, or fourteen fifteenths.
Five: with one red and three blue, two reds without replacement are impossible. The second red branch after a first red has numerator zero. Six: if the same bag is fully restored and mixed before the second uniform draw, two reds has probability one fourth times one fourth, or one sixteenth.
Seven: in the three-red, two-blue example, the chance of a second red given a first blue is three fourths. The overall chance of a second red is three fifths. Explain why these are different questions. Eight: a tree’s four paths sum to one; that is a useful coverage check, but their branch states still need to be correct.
Contents · Previous section · Next section
The probability-skills guide is useful for the broader aim: identifying outcomes and reasoning under uncertainty. Return there when the learner needs to understand what probability represents before working through a particular tree. This article supplies a narrower model update within that larger skill.
For assessment application, use Probability and Statistics Exam Questions Explained. Read it for the examination context rather than treating a single bag example as the whole topic. Current school instructions and the applicable syllabus still determine what is being assessed.
For language, connect to How Vocabulary Really Works. The useful link is the ability to distinguish exactly one from at least one, and given from an overall event. For learning routines, How Education Works provides a broader route beyond the individual worksheet.
These links have different jobs. They should help a reader choose the next explanation, not create a long list to read before attempting any question. If the current gap is the second denominator, begin with the remaining bag. If the gap is event wording, compare selected routes. Choose the route that addresses the actual barrier.
The main Mathematics hub connects this series with the wider subject. Keeping those links clear helps students and parents move from a specific mistake to the appropriate foundation, explanation or practice habit without assuming that every difficulty needs a new set of unrelated questions.
Contents · Previous section · Next section
Ask the learner to describe the bag after the first draw before writing a fraction. The answer reveals whether the process is understood. If the learner needs help, identify the smallest missing decision: the remaining total, the colour count or the event being sought next. Avoid supplying the whole tree immediately.
In a small group, students can compare the two first-result states after individual attempts. One branch may have two red remaining and another three. A tutor can explain why both are correct under different conditions. Every learner should then construct a fresh tree alone so discussion does not hide dependence.
Our Clementi Secondary 1 Mathematics tutorial guide explains eduKateSG’s small-group teaching rationale. Confirm current subject-level and learning fit before arranging support. The useful teaching outcome is a learner who can update and explain the next state, not merely reproduce a familiar arrangement of fractions.
Notice whether the child can answer several event questions from one tree. That checks reading and route selection separately from construction. A correct two-red answer does not establish that at-least-one or second-position questions are understood. Use a few targeted contrasts rather than assuming one successful product proves the whole topic secure.
If errors persist, bring original attempts to the teacher. “Keeps the original denominator after no replacement” is a clear observation. “Cannot do probability” is much broader. A precise description helps the next explanation address the actual stage where the model stops matching the process.
Test the state update before testing a long calculation
Give the learner an event description and ask only for the next bag, with no probability calculation yet. From four red and two blue, a first red leaves three red and two blue. A first blue leaves four red and one blue. Both states contain five counters, but their favourable counts for a next red differ.
Then ask for the next fractions. They are three fifths after a first red and four fifths after a first blue. This short contrast isolates the update decision. If it is secure but the full route answer is wrong, inspect event selection, multiplication or simplification next rather than repeating the same count explanation.
Reverse the check too. If a branch is labelled four fifths for a next red, ask which earlier removal could have produced that state under this original bag. The answer is a first blue. The fraction should lead back to a coherent remaining set, not exist as a detached number chosen to make the paths sum to one.
Finally change the process to replacement and ask what is restored. Under the stated full return and uniform mixing arrangement, both second-stage red chances become four sixths. The learner should explain the restored count before simplifying to two thirds. This comparison makes the process condition visible without needing another large tree or a new collection of arithmetic facts.
Keep the activity proportionate. Once the state update is reliable, move to independent event questions and appropriate school practice. The purpose is a precise diagnostic check, not a requirement to describe every counter aloud forever. Efficient working should grow from understood states rather than replace them with unexplained patterns.
Contents · Previous section · Next section
Should the denominator always decrease by one on a second draw?
Only when exactly one item has been removed and not returned, with no other change before the next selection. Read the process. Replacement, multiple removals or additions can produce another available total. The general rule is to count what is currently available, not apply a fixed subtraction blindly.
Does multiplying along a tree mean the events are independent?
No. A tree can multiply the first-event probability by the appropriate conditional next-event probability. In a no-replacement model, the second fraction depends on the earlier branch. Independence is a separate property and should not be assumed simply because a multiplication appears.
Why are red-blue and blue-red counted separately?
They are different ordered routes. If the event asks for one of each colour in either order, both are included. If it specifies red first and blue second, only that route is included. Read the event before selecting paths so the calculation matches the requested condition.
Can the second-draw overall chance equal the first without independence?
Yes, as in the stated uniformly random no-replacement example. The overall chance combines all relevant earlier routes. Knowing the first colour changes the appropriate branch chance, which shows dependence. Equal overall chances at two positions do not by themselves establish independence.
What should be checked first when the answer is wrong?
Check the remaining bag after each first branch, then the routes selected for the event. Only then inspect multiplication and simplification. A correct calculation using the wrong state or an incomplete event still answers the wrong question. The process meaning is the first priority.
Contents · Previous section · Next section
A probability tree is a record of changing possibilities. Without replacement, the second stage uses a different available set, and the appropriate fraction follows from that set. Once this meaning is clear, multiplication, addition and complements become connected tools rather than isolated instructions triggered by keywords.
Take one current school example and write the remaining counts after each first branch. Then answer one ordered-route question, one combined-event question and one overall second-position question. Explain the difference between them. This small set can reveal whether the learner understands the state, the event and the numerical calculation separately.
Continue through How Mathematics Works for the connected routes. Course context was checked against the linked SEAB syllabus on 16 October 2026. All bags, counter counts and exercises here are original teaching examples, to be used alongside the student’s taught scope and applicable examination year.
Contents · Previous section · Mathematics hub
Continue through the Mathematics learning hub
Browse all 16 G1, G2 and G3 Mathematics explainers for a guide matched to the next difficulty.
Why Square Unit Conversions Need Two Factors
Why a Letter Must Keep Its Meaning Within a Solution
Why Rotating a Triangle Does Not Change Its Base and Height Relationship