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How to Perform in the new G3 SEC Examinations | Learner’s Guide Vol 0095 | Mathematics: Probability Without Replacement — Update the Sample Space After Every Draw

G3 SEC Mathematics · Learner’s Guide Vol 0095. A step-by-step workshop with original worked examples, practice questions and full answer explanations.

A bag contains three red counters and two blue counters. A student chooses one counter, keeps it, and chooses a second. What is the probability that both are red? Many learners know how to calculate a probability. The harder step is noticing that the second choice is made from a different bag. There are now four counters, not five, and possibly two red counters rather than three.

This G3 SEC Mathematics workshop is about updating the sample space after an event. You will learn to recognise a situation without replacement, draw a clear possibility tree, use changing fractions, identify when a complement helps, and explain why a plausible-looking shortcut is wrong. We will solve two-stage and three-stage questions, including ordered and unordered outcomes. Every numerical case below is original practice, not a forecast of any particular examination question.

The broader skills are continuity and care. In Volume 0087 on successive transfers, you updated percentages when people moved between groups. In Volume 0091 on multi-stage journeys, you updated time and distance after each stage. Here the population of possible choices changes. The common discipline is to calculate from the present state, not a state that no longer exists.

Chapter index

1. First understand what changed

A probability is a number between zero and one that describes the chance of an event under a given model. If five counters are equally likely to be chosen, three red and two blue, then the chance that the first counter is red is 3/5. This fraction refers to the initial bag. The denominator says five outcomes are possible on the first draw; the numerator says three of them meet the condition “red”.

Now imagine the first chosen counter was red and was not put back. The bag contains two red and two blue counters. The probability of red on the second draw, given that first red result, is 2/4. If instead the first counter was blue, the bag contains three red and one blue, so the chance of red on the second draw is 3/4. The second probability is not one fixed number independent of what happened first. It changes with the branch.

This is why a tree diagram helps. One branch says first red, then red or blue; another says first blue, then red or blue. Each second-stage branch begins from its own updated bag. You can explain the correct method aloud without formal notation: “If red was removed, there are four counters including two red. If blue was removed, there are four counters including three red.” That sentence contains the essential mathematical reasoning.

With replacement means returning the chosen counter before the next selection. The original composition then stays the same. Without replacement means keeping it out. It is not enough to look at the phrase “two counters are chosen”; read what the question says happens to the first. If the procedure is unspecified, do not silently impose one interpretation. Examination problems normally supply the relevant conditions.

A short state table is useful when drawings become crowded. Write “Before: R=3, B=2, total=5.” Below one branch write “After R: R=2, B=2, total=4.” Below the other write “After B: R=3, B=1, total=4.” The table and tree express the same model in two different ways.

2. Check the actual G3 SEC Mathematics scope

The official 2027 G3 school-candidate syllabus list identifies Mathematics as K310. Its 2027 K310 syllabus includes probability of single events, simple combined events using possibility and tree diagrams where appropriate, and addition and multiplication for mutually exclusive and independent events. This chapter uses simple changing-sample-space situations to strengthen that syllabus skill; it does not present advanced formal conditional-probability notation as a separate compulsory K310 topic.

K310 Paper 1 lasts 2 hours 15 minutes, has approximately 26 compulsory short-answer questions and carries 90 marks, worth 50% of the examination. Paper 2 also lasts 2 hours 15 minutes, carries 90 marks and has 9 to 10 compulsory questions. Its last question applies mathematics to a real-world scenario. An approved calculator may be used in both papers. Essential working matters because the syllabus notes that omitting it can lead to lost marks.

Keep exact fractions while solving probability questions when possible. If a non-exact numerical answer is needed, the syllabus generally asks for three significant figures unless the question specifies otherwise. Do not convert 2/3 into 0.67 halfway through a multi-stage calculation. Early rounding can change a final probability. An exact fraction is often easier to check than a string of approximate decimals.

This workshop is not a replacement for the Mathematics Hub or the site’s wider G3 SEC learner-guide route. The broader owners cover statistics, graphs, algebra and planning. This volume concentrates on one narrow difficulty: changing possible outcomes after a selection.

3. Build a two-draw tree from first principles

Return to three red and two blue counters, with two draws without replacement. Begin with the first draw. Red has probability 3/5; blue has probability 2/5. On the red branch, there are now two red and two blue counters, so second red has probability 2/4 and second blue 2/4. On the blue branch, there are three red and one blue remaining, so second red has probability 3/4 and second blue 1/4.

There are four colour sequences: red-red, red-blue, blue-red and blue-blue. Their probabilities are respectively (3/5)(2/4)=3/10, (3/5)(2/4)=3/10, (2/5)(3/4)=3/10 and (2/5)(1/4)=1/10. Adding them gives 3/10+3/10+3/10+1/10=1. The total of one is a valuable check that the outcomes have been accounted for properly.

The chance of two red counters is 3/10. The chance of one of each colour in any order is 3/10+3/10=3/5. The words “in any order” matter. They require red-blue and blue-red. If the question asks for red followed by blue, only one branch qualifies. If it asks for at least one red, red-red, red-blue and blue-red all qualify.

Why multiply along a branch? The first fraction represents the chance of reaching that branch. The second fraction represents the chance of the next outcome once that branch has been reached. For this simple equally likely draw model, the product gives the chance of that complete sequence. Why add branches? They describe distinct colour sequences that cannot both occur in the same two-draw experiment. They are alternative successful ways to meet an “or” condition.

One common mistake is to calculate red-red as (3/5)(3/5)=9/25. That repeats the initial red probability after red has already been removed. Another is to calculate (3/5)(2/5), reducing red without reducing the total. The correct second fraction must update both numerator and denominator. The bag is now smaller.

Check an answer by imagining actual counters R1, R2, R3, B1 and B2. There are 5 choices for the first and 4 for the second, giving 20 equally likely ordered counter-pairs. Six are red-red: 3 choices for first red and 2 for second. Thus 6/20=3/10. The counting check and probability tree agree.

4. With replacement is a different experiment

Suppose the same bag is used, but after the first choice the counter is returned and mixed before the second draw. Now the bag contains three red and two blue before each draw. The chance of red-red is (3/5)(3/5)=9/25. The chance of red-blue is (3/5)(2/5)=6/25; blue-red is another 6/25; blue-blue is (2/5)(2/5)=4/25. The four outcomes add to 25/25.

Compare red-red in the two experiments: 9/25 with replacement, 3/10 without. The values differ because keeping a red counter out reduces the chance of another red. This is not a general rule saying all probabilities decrease without replacement. If blue was removed, the chance of the next red rises from 3/5 to 3/4. Which probability changes depends on what was removed.

With replacement, the second draw has the same probabilities regardless of the first. In the specified model, the draws are independent. Without replacement, the first result changes the composition of the bag and therefore changes relevant second-draw probabilities. The second draw depends on the first. Use “independent” with care: it describes events in a probability model, not whether people performed the actions separately.

A student might say “The drawings are on different turns, so they are independent.” Different times do not make events independent. The condition that matters is whether knowing the first outcome changes the probability of the second. Our trees reveal that difference directly.

At G3 level, it is useful to know the named idea without rushing into advanced notation. Draw the appropriate branches, label the updated counts and show the multiplication. The examination rewards correct reasoning within the specified syllabus, not importing extra formulas that conceal how the numbers were chosen.

5. Exactly one, at least one and both

Three common phrases describe different groups of successful outcomes. Both red means red-red. Exactly one red means red-blue or blue-red. At least one red means red-red, red-blue or blue-red. These are not interchangeable. A word such as “exactly” can change which branches belong in the answer.

Using the three-red, two-blue bag without replacement, “exactly one red” has probability 3/10+3/10=3/5. “Both red” is 3/10. “At least one red” is 3/10+3/10+3/10=9/10. Notice that the last answer can also be found by considering the only way to fail: blue-blue has probability 1/10. Therefore at least one red has probability 1−1/10=9/10.

The complement method is especially effective when a question asks for “at least one”. Instead of adding many successful branches, identify the single or smaller collection of branches that fail. But check the event carefully. The complement of “at least one red” is “no red”, not “exactly one blue”. In a two-draw case, no red means both blue.

Now change the bag to four green and one yellow. For two draws without replacement, the chance of at least one yellow can be found by calculating one minus the chance of no yellow. No yellow means two green: (4/5)(3/4)=3/5. Therefore at least one yellow is 2/5. Since only one yellow counter exists, “at least one yellow” and “exactly one yellow” describe the same successful outcome set in this particular experiment.

The word “at least” is often easier to understand through its opposite. At least two successes fails when there are zero or one. At least one success fails when there are none. The opposite set depends on how many draws occur and what counts as success.

Practice speaking in ordinary English before multiplying. “I need either green then yellow or yellow then green.” Once the successful sequences are clear, the arithmetic follows. A correct calculation for the wrong event is still a wrong answer, so interpretation comes first.

6. Three-stage probability: keep the counts alive

A longer question might ask for three draws without replacement from a bag containing three red and two blue counters. Suppose the task is to find the chance of exactly two red. The successful colour sequences are red-red-blue, red-blue-red and blue-red-red. These are three distinct ordered ways to obtain two red and one blue.

For red-red-blue, the probability is (3/5)(2/4)(2/3)=1/5. For red-blue-red, the probability is (3/5)(2/4)(2/3)=1/5. For blue-red-red, it is (2/5)(3/4)(2/3)=1/5. Adding the three branches gives 3/5. Each branch updates its counts after each draw; the products happen to be equal in this symmetrical example, but do not assume they always will.

A visual way to follow red-blue-red is to write the bag composition after every step. Start R=3, B=2. After removing red: R=2, B=2. After removing blue: R=2, B=1. For the final red: 2 of 3 remaining counters are red. The fractions 3/5, 2/4 and 2/3 now have transparent meanings.

An alternative counting check lists combinations of which physical counters appear among three draws. There are 10 equally likely unordered selections of three distinct counters from five. To choose exactly two reds and one blue, there are three choices of two reds from the three, and two choices of blue, giving six successful selections. Thus 6/10=3/5. You do not need to introduce general combination formulas to understand this small counting argument.

If the problem changes to three red and three blue counters, recalculate every fraction. Do not preserve 3/5 merely because the earlier example had a neat answer. The success counts, denominators and complement all depend on the starting composition.

You can also stop an argument when an impossible branch appears. If the bag contains only two red counters, a sequence requiring three reds without replacement has probability zero. Do not multiply fractions that suggest a third red exists. Before calculations, count the available objects and check feasibility. This connects probability to Volume 0071 on constraints and feasibility.

7. Conditional branches do not need advanced notation

A question may say, “Given that the first counter drawn was blue, what is the probability that the second counter is red?” This is a request to examine the blue-first branch. In our original bag, after removing one blue, the remaining bag has three red and one blue. The answer is 3/4. You do not multiply by 2/5 because the question tells you the first result has already occurred.

Contrast it with, “What is the probability of drawing blue first and red second?” That asks for the full two-stage path. Its probability is (2/5)(3/4)=3/10. The two questions share a branch but do not ask for the same event. A learner who gives 3/10 to both has not noticed the difference between a condition already given and a sequence whose probability must be found.

A useful reading rule is to circle words such as “given that”, “after”, “if the first” and “then”. Ask whether the opening event is assumed or is part of the chance to be calculated. Reword the question: “I know the first was blue. What is still in the bag?” or “How likely is the whole blue-then-red sequence?” The difference becomes clear.

This is one reason a tree is better than a memorised multiplication slogan. It shows the total experiment and the smaller sample space reached under each condition. It also keeps you from using an unconditional fraction after a specific earlier result has changed the population.

In an examination answer, a short line often suffices: “After a blue is removed: 3 red out of 4 remaining, so P(second red given first blue)=3/4.” Write the words if the notation is unfamiliar. Clarity about what the denominator measures matters more than advanced symbols.

8. Repeated draws, fair games and misleading intuition

Imagine a simple game: a player wins if two drawn counters, without replacement, are both yellow. A bag contains two yellow and three black counters. The chance of a win is (2/5)(1/4)=1/10. If the prize is large, a student may think the game is favourable because “there are two yellow counters.” The probability says only one in ten trials is a win under the stated model.

Suppose a second game pays a smaller prize if exactly one yellow is drawn. Yellow-black has probability (2/5)(3/4)=3/10. Black-yellow has probability (3/5)(2/4)=3/10. Exactly one yellow therefore has probability 3/5. The event has six times the probability of two yellow counters. Whether either game is financially fair also depends on the entry cost and prize, not merely the chance of winning.

Consider a choice between two bags. Bag A has 2 red and 3 blue; Bag B has 3 red and 7 blue. Which bag gives the higher chance of first red? A: 2/5=0.4. B: 3/10=0.3. More red counters in absolute number does not necessarily mean a higher proportion. This is a denominator lesson like Volume 0085 on reference auditing.

If you choose two counters, the relative attraction of the bags can change. For two reds without replacement, A gives (2/5)(1/4)=1/10; B gives (3/10)(2/9)=1/15. Bag A remains better for this target, but that conclusion comes from calculation, not from assuming the first-draw ranking automatically settles every multi-draw event.

When a probability is used for a decision, separate the event, the probability and the reward or cost. A 1/10 win chance is not itself an average financial return. Different prizes or losses could alter the decision. For K310 practice, calculate the stated probability first and interpret it in the actual context given.

9. Detect and repair six common errors

The first error is keeping the denominator unchanged. “Three red out of five, then two red out of five” forgets that one counter was removed. The repair is to write the new total at the start of every stage. A denominator is not a decoration: it counts the available choices in that state.

The second error is updating the total but not the colour. “Three red out of five, then three red out of four” is wrong after a red first draw. The repair is to record the removed colour as well as the number removed. Count what is left rather than subtracting from the denominator alone.

The third is treating two sequences as one. “Exactly one red” includes red-blue and blue-red. A student who multiplies only one path will often get half the answer in a symmetrical case. The repair is to list success sequences in words before calculating.

The fourth is assuming that every branch has the same probability. In a bag of three red and two blue, blue-blue is less likely than red-red. Tree ends represent different colour sequences but not equally likely sequences. Calculate the product along each branch instead of dividing by the number of drawn branches unless equal likelihood has been justified.

The fifth is confusing “given” with “and”. The probability of red second given blue first is 3/4 in our original model. The probability of blue first and red second is 3/10. The repair is to identify whether the first event is a known condition or part of the success path.

The sixth is forgetting a complement or overlap. “At least one red” is not the same as “exactly one red”. Draw successful events explicitly or subtract a clearly defined failure set from one. Check that no successful path has been counted twice and none left out.

After correcting any error, re-explain the bag in a sentence. The purpose is to repair a decision rule, not simply replace an incorrect number with a correct one.

10. Original practice set: from basic to demanding

A. A bag has four red counters and one blue counter. Two are selected without replacement. Find the probability of both being red.

B. A bag has two green counters and three yellow counters. Two are selected without replacement. Find the probability of green followed by yellow.

C. In Question B, find the probability of exactly one green counter in the two draws.

D. A bag has three white and two black counters. Two are selected without replacement. Find the probability of at least one black.

E. A bag contains two red and three blue. The first draw is known to be red and the counter was not replaced. What is the chance of blue on the next draw? Explain why this differs from the chance of red followed by blue in two random draws.

F. A bag holds three orange and two purple counters. Two are drawn with replacement. Find the chance of two orange counters, then calculate the same event without replacement.

G. A bag contains three red, two blue and one green counter. Two are selected without replacement. Find the probability of red followed by green.

H. From Question G, find the probability of one red and one green in any order.

I. A bag contains four red and two blue counters. Three are drawn without replacement. Find the probability that all three are red.

J. A bag has two yellow and three black counters. Two are selected without replacement. A pupil says the chance of exactly one yellow is 2/5 × 3/4 = 3/10. Identify what is missing and give the correct answer.

K. In a bag of one white and four black counters, two are drawn without replacement. Find the chance of at least one white counter using a complement. Then explain why there cannot be two white counters.

L. A bag has four red and three blue counters. The first two counters drawn without replacement are both blue. Find the probability that the third is red, and state the contents of the bag at that moment.

Try A–D with a tree or short state table, E–H with event-language checks, and I–L without looking at examples. If you cannot start a question, write only the bag before and after the first selection. That is a productive first step.

11. Full worked answer clinic

A. Four red out of five on the first draw, three red out of four on the second: (4/5)(3/4)=3/5. The second numerator and denominator both change. The exact fraction is the final answer.

B. First green is 2/5. After a green is removed, three yellow remain out of four counters. Therefore (2/5)(3/4)=3/10. “Green followed by yellow” is one ordered path, not both colour orders.

C. Exactly one green means green-yellow or yellow-green. Each path has probability 3/10, so the answer is 3/5. The second path is (3/5)(2/4)=3/10. List both before adding.

D. The opposite of “at least one black” is “both white”. Both white is (3/5)(2/4)=3/10. Therefore the required probability is 1−3/10=7/10. A full tree gives the same result.

E. Given that first red has occurred, the bag has one red and three blue among four counters. Chance of blue next is 3/4. For the full red-then-blue event, multiply (2/5)(3/4)=3/10. The latter includes the chance of the first event.

F. With replacement, the chance of orange-orange is (3/5)(3/5)=9/25. Without replacement it is (3/5)(2/4)=3/10. The second experiment has one fewer orange and one fewer counter after the first orange.

G. First red probability is 3/6. After red, the one green remains among five counters. Result: (3/6)(1/5)=1/10. Neither the blue nor the green count changes after removing red.

H. Add red-green and green-red. Red-green is 1/10. Green-red is (1/6)(3/5)=1/10. So one of each in any order is 1/5. The two sequences cannot occur together in one two-draw trial, so they may be added.

I. First three red probabilities are 4/6, 3/5 and 2/4. Product is 24/120=1/5. A useful physical-count check is that four red were available initially, so drawing three red is possible; after the second red there are two left among four.

J. The pupil included yellow-black but omitted black-yellow. The first path is (2/5)(3/4)=3/10. The second is (3/5)(2/4)=3/10. Exactly one yellow therefore has probability 3/5.

K. The chance of two black is (4/5)(3/4)=3/5. So at least one white has probability 1−3/5=2/5. Drawing two white without replacement is impossible because only one white counter exists. An impossible event has probability zero.

L. After two blue counters have been removed from four red and three blue, the bag has four red and one blue, total five. The chance of red on the third draw, given the first two were blue, is 4/5. Do not multiply by the chances of the first two blues; they are stipulated as known.

For every incorrect answer, locate the first branch at which the written composition no longer matches the actual bag. A new attempt should start there. Rewriting every correct stage wastes time and hides the specific misunderstanding.

12. An examination-quality way to show working

A good solution has five visible stages. Name the event you want. Write the starting counts. Identify the successful sequences. Update counts for each branch. Calculate, combine and check. On a short question, this may fit into three lines. On a longer one, the same stages prevent confusion.

Consider exactly one green from two green and three yellow, without replacement. Write: “Success = GY or YG.” Then “P(GY)=(2/5)(3/4)=3/10; P(YG)=(3/5)(2/4)=3/10.” Then “Required probability = 3/10+3/10=3/5.” There are no unnecessary paragraphs, yet every decision is traceable.

Before committing an answer, check whether it lies between zero and one. A value above one signals that paths may have been added incorrectly or counted more than once. A zero may be valid for an impossible event, but not for a sequence that clearly exists in the bag. A quick physical check can help: imagine drawing the counters in the stated order.

Also check whether the result behaves sensibly when the bag changes. With fewer red counters initially, a red-red event should not mysteriously become more likely under the same draw procedure unless another relevant condition changes too. This is a plausibility check, not a substitute for arithmetic. Volume 0075 on independent verification gives a wider method for such checks.

In a wordy real-world problem, isolate the mathematics from the setting. A raffle, card draw, quality-control sample or box of objects may use the same changing-sample-space structure. Be careful about assumptions: random selection and equal likelihood may be specified or implied by the problem; if not, do not invent equal probabilities.

Remember the scope. This is a workshop on simple combined events. You do not need a sophisticated statistical theory to make sense of a bag with five counters. Reliable reasoning with small counts is the foundation for later mathematics.

13. Seven-day progression: from the bag to the unfamiliar question

Day 1: one draw. Use five bags with different compositions. Name the total and favourable counts, express probability as a fraction, and say which event the fraction describes. Do not rush to multiple draws before a one-draw denominator feels natural.

Day 2: state updates. Write before-and-after tables for removing a red, a blue or a differently coloured object. Do not calculate path probabilities yet. Check that the total falls by one each time in a without-replacement model.

Day 3: two-step trees. Draw two-stage colour trees and label the correct remaining counts. Multiply along one complete path, then add two alternative success paths. Use small numbers so you can physically enumerate choices if a result looks wrong.

Day 4: replacement contrast. Solve the same problem once with replacement and once without. Explain why the second-stage probabilities differ. Practise the language of dependence and independence without treating those labels as substitute calculations.

Day 5: event interpretation. Solve three tasks each using the terms “exactly one”, “both” and “at least one”. Draw successful paths in words first. Use a complement where it genuinely reduces the work.

Day 6: three draws. Solve one three-stage question with a tree or state record. Keep exact fractions and verify by a second counting route when feasible. Avoid introducing advanced techniques if a clear enumeration already works.

Day 7: mixed rehearsal. Attempt Questions A–L independently and mark both arithmetic and interpretation. Record only the errors that reveal a skill gap: incorrect denominator, unupdated numerator, omitted order, false independence, wrong event or unjustified rounding. Revisit those three days later with new counter counts.

Parents can help by asking, “What is still in the bag?” rather than immediately telling a child the correct fraction. That question rebuilds the state the next calculation depends on. The same habit benefits algebra, percentages and scientific experiments when a condition changes.

14. Continue the sequence with one reliable habit

The most useful sentence in this chapter is “The next probability belongs to the next state.” It protects you against memorised fractions that no longer describe the situation. When you can update a sample space correctly, a tree is not a mysterious diagram. It is simply a record of possible events and their changing conditions.

Continue through the G3 SEC learner-guide route and the Mathematics Hub for the full K310 programme. If your difficulty begins before Secondary School probability, revisit the PSLE-to-SEC subject study plan and rebuild fractions and sample spaces with smaller examples.

Official examination references: the SEAB 2027 G3 syllabus index and Mathematics K310 syllabus. The questions here are original educational practice. They are not presented as SEAB specimen or examination questions.