The second draw is different because the first draw changed the bag
A bag holds three blue counters and two yellow counters. A student draws a counter and keeps it outside the bag, then draws another. What is the probability of drawing two blue counters? The first blue has probability three fifths. After that draw, only two blue counters remain among four counters. The second probability is therefore two quarters, not three fifths again. Multiply along that particular route: three fifths times two quarters equals three tenths. The arithmetic is small. The reasoning is the real test: the first event changes what remains possible in the second.
Earlier volumes established the language of probability trees and mutually exclusive and independent events. Volume 0178 introduced probability trees, branch totals and route products; Volume 0182 distinguished mutually exclusive from independent events. This chapter narrows the focus: two or more selections without replacement, the shifting denominator, ordered routes and ways to check a result against direct counting. It does not repeat the basics for their own sake. It teaches the change in the sample space that makes the second step easy to misread.
What the 2027 SEC Mathematics syllabus says
SEAB lists G3 Mathematics for the 2027 Singapore-Cambridge Secondary Education Certificate as syllabus K310, replacing O-Level reference code 4052. Its Statistics and Probability strand includes probabilities of single events, simple combined events including possibility and tree diagrams where appropriate, and the addition and multiplication of probabilities for mutually exclusive and independent events. Without-replacement examples are a useful way to reason through simple sequential outcomes, but this guide does not treat advanced conditional-probability formulas as a separate K310 requirement. We use clear counts at each stage rather than assuming the syllabus demands a formal theorem beyond its published scope.
Paper 1 is 2 hours 15 minutes with about 26 compulsory short-answer questions, worth 90 marks or 50%. Paper 2 is also 2 hours 15 minutes, worth 90 marks or 50%, and has 9–10 questions, with the last focused on applying mathematics in a real-world scenario. Essential working matters: the 2027 syllabus explicitly warns that omitting it can lose marks. Use the official G3 syllabus index and SEAB K310 PDF for current details. Every number in the worked problems below is an original teaching example.
Start by recognising the selection rule
Three phrases signal different processes. With replacement means the selected object returns before the next selection, so the starting population is restored. Without replacement means the chosen object stays out, so the total declines and the composition can change. Simultaneously means objects are selected together; you can still use a hypothetical order for counting, but the final event may not care which came first. A question might also say ‘the drawn ticket is not put back’ rather than using technical vocabulary. Mark this instruction before doing any calculation.
The denominator on a second draw without replacement falls from n to n minus one, provided one item was removed from a starting group of n. The numerator depends on what was drawn. If the first item belonged to the category you want again, that category’s count also falls by one. If the first item belonged to a different category, the desired category’s count may stay the same. That is the entire mechanism. It is more powerful than memorising an expression because it applies to counters, names, cards and tickets.
Ordered routes and unordered events are not the same question
Suppose the bag contains three blue and two yellow counters. The ordered route blue then yellow has probability three fifths times two fourths, or three tenths. The reverse route yellow then blue has probability two fifths times three fourths, also three tenths. If the question asks for one blue and one yellow in either order, add those two disjoint routes to obtain three fifths. A student who gives three tenths has answered only the first route. A student who multiplies by two without checking the paths is relying on a shortcut that might fail when the paths are not equally likely.
Draw the possible colour routes: BB, BY, YB, YY. They are mutually exclusive outcomes of the two-draw experiment. Their probabilities must add to one. For this bag, BB is three tenths, BY is three tenths, YB is three tenths and YY is one tenth. The total is ten tenths. This sum check exposes an incorrectly repeated denominator immediately. If your paths total more than one, you have overcounted, failed to adjust a branch, or confused one ordered path with a collection of paths.
Why you multiply along a route and add alternative routes
A route such as blue THEN yellow contains both requirements in order. Its probability is the first-stage fraction multiplied by the relevant fraction of the remaining counters. Alternative complete routes such as blue–yellow OR yellow–blue are mutually exclusive: a single two-draw sequence cannot have both orders. You therefore add their route probabilities. This is a practical path method grounded in counting and the structure of the situation. Do not blindly multiply probabilities of events as if every selection were independent; when an item stays out, the second stage depends on the first outcome.
Many questions can be checked by counting ordered pairs directly. With five distinct counters and two draws without replacement, there are five possible first selections and four possible second selections, or 20 equally likely ordered selections. To get two blue counters, the first can be one of three blue counters and the second one of the remaining two blue counters: six favourable ordered selections out of 20, or three tenths. This method reproduces the branch product and provides an independent check. Count specific objects when labels matter, even if the final question asks only for colours.
The numerator changes differently on different branches
With three blue and two yellow counters, follow a first blue draw. The remaining bag has two blue and two yellow. On the branch following a first yellow draw, the bag instead has three blue and one yellow. A tree diagram with identical second-stage fractions on both branches is wrong. In the first branch, the next blue probability is two quarters; in the second branch, it is three quarters. The denominator four is the same across both second branches because one item has been removed, but the numerator records which category lost an item.
If a student starts with four red and one white card and wants one of each colour, the ordered paths are not equal because of a broad rule about symmetry alone; they are equal here because each is based on four possible red–white ordered pairs out of the same 20 pairs. In other experiments, selection rules, additional restrictions, or nonuniform probabilities can remove that symmetry. The robust procedure is always to write and evaluate both routes. If they happen to match, the equality becomes a check rather than an assumption.
The complement can simplify ‘at least one’
An event such as ‘at least one yellow’ includes many routes. For two draws from three blue and two yellow, the complement is ‘no yellow’, meaning two blues. We already know the probability of two blues is three tenths, so the probability of at least one yellow is one minus three tenths, or seven tenths. This is usually faster than adding BY, YB and YY, though both methods should agree. The complement strategy is safe when you identify the exact logical opposite, not a merely convenient-looking event.
Be especially careful with ‘exactly one’, ‘at least one’ and ‘both’. Exactly one yellow means BY or YB, three fifths. At least one yellow adds YY, giving seven tenths. Both yellow means YY, one tenth. These three phrases sound similar but describe three different sets of outcomes. Under timed examination conditions, underline the quantifier before drawing a tree. A wrong interpretation can lead to perfect arithmetic for the wrong event.
When to use a possibility diagram instead
A possibility diagram is useful when each selection has a small labelled set of possible outcomes and the question asks for sums, pairs or combinations. For two distinct cards labelled 1, 2, 3 and 4, drawn without replacement, construct a grid whose rows represent the first card and columns represent the second. The diagonal entries where the same card would be drawn twice are impossible, because the card is not returned. There are twelve permitted ordered pairs. If instead the card is returned between draws, all sixteen entries are possible. The difference lies in a rule, not in the drawing style.
Choose the representation that reduces errors. A tree diagram is helpful when the probability on a later branch depends on the earlier colour. A possibility diagram is helpful when you must enumerate the combinations of two labels. Direct counting may be shortest when the favourable outcomes can be counted systematically. The exam does not award elegance for a diagram that hides a mistaken sample space. It rewards sound mathematical reasoning and correct results with enough working for the examiner to follow.
There is a real difference between probability and expectation
If the probability of two blue counters is three tenths, that does not promise exactly three successes in every ten trials. Random outcomes vary. The result describes the chance for one run of the experiment under the stated assumptions. If many independent repetitions are made after restoring the original bag each time, the relative frequency may approach the theoretical probability, but an individual small sample can differ. Do not confuse what happens within a two-draw experiment—dependent stages without replacement—with what may happen between separately reset experiments.
This distinction also matters for classroom demonstrations. A group might try the experiment ten times, resetting the bag to three blue and two yellow before each two-draw trial. Their observed frequency of two blues could be two or four rather than exactly three. A sensible conclusion compares observation with theory and acknowledges sampling variation. See Volume 0186 on expected and relative frequencies for that distinct question.
Twenty-six worked clinics with one changing rule at a time
Clinic 1: A bag has three blue and two yellow counters. Two are removed.
Question: Find P(blue then blue). Worked answer: The first blue chance is 3/5. Two blue remain out of four after that draw. Therefore P(BB) = (3/5) × (2/4) = 3/10.
What to check: The second numerator and denominator both change after a first blue. Repeating 3/5 would model replacement instead. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 2: The same bag is used without replacement.
Question: Find P(yellow then yellow). Worked answer: The first yellow chance is 2/5; after one yellow is removed, the next yellow chance is 1/4. The product is 2/20 = 1/10.
What to check: The event has only one ordered colour route, YY. There is no factor of two. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 3: The same bag is used without replacement.
Question: Find P(blue then yellow). Worked answer: The first blue chance is 3/5; after it is removed, both yellows remain among four counters, giving 2/4. The product is 6/20 = 3/10.
What to check: The yellow numerator stays two on a first blue branch. Only the total decreases. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 4: The same bag is used without replacement.
Question: Find P(exactly one yellow). Worked answer: Blue–yellow has probability 3/10, and yellow–blue also has probability (2/5) × (3/4) = 3/10. The routes do not overlap, so their sum is 3/5.
What to check: Exactly one yellow requires two ordered routes. Do not confuse it with two yellows. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 5: The same bag is used without replacement.
Question: Find P(at least one yellow). Worked answer: Use the complement of no yellow: P(BB) = 3/10. Therefore P(at least one yellow) = 1 − 3/10 = 7/10.
What to check: A complement is often shorter; it must be the exact opposite of the requested event. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 6: A pouch contains four red and one white card, and two are removed.
Question: Find P(two red). Worked answer: The first red probability is 4/5 and the second red probability is 3/4. Their product is 12/20 = 3/5.
What to check: The three remaining red cards are counted from the updated bag, not the original four. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 7: The same pouch is used.
Question: Find P(one red and one white, in any order). Worked answer: Red then white has probability (4/5) × (1/4) = 1/5. White then red has probability (1/5) × (4/4) = 1/5. Sum = 2/5.
What to check: Both orders are possible. Notice that after taking the sole white card, every remaining card is red. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 8: The same pouch is used.
Question: Why is P(two white) zero? Worked answer: There is only one white card. With no replacement, a second white card cannot be drawn after the first. The event has probability zero.
What to check: This is a logical impossibility, not merely an unlikely event. The numerator becomes zero. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 9: Six distinct badges include two gold and four silver. Two are chosen in order.
Question: Find P(gold then silver). Worked answer: The first gold probability is 2/6. Four silver remain among five, so the second chance is 4/5. Product = 8/30 = 4/15.
What to check: The second numerator remains four because the removed badge was gold. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 10: The same badges are used.
Question: Find P(exactly one gold). Worked answer: Gold then silver is 4/15. Silver then gold is (4/6) × (2/5) = 4/15. Total = 8/15.
What to check: The two routes are mutually exclusive and cover both possible orders for one gold. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 11: Five numbered cards are 1 through 5. Draw two without replacement.
Question: Find P(first 1, then 5). Worked answer: The first specific card has probability 1/5. After taking it out, the card labelled 5 remains among four. Product = 1/20.
What to check: Each numbered card is distinct. Do not treat this as a colour-category question with repeated indistinguishable objects. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 12: The same numbered cards are used.
Question: Find P(draw 1 and 5 in either order). Worked answer: The ordered routes (1,5) and (5,1) each have probability 1/20. Together the probability is 2/20 = 1/10.
What to check: The word ‘either order’ authorises adding two routes. The sample space has 5 × 4 = 20 ordered pairs. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 13: Four numbered cards are 1 through 4. Two are removed.
Question: How many equally likely ordered outcomes exist? Worked answer: There are four choices for the first card and three remaining choices for the second: 4 × 3 = 12 ordered outcomes.
What to check: Do not include the four diagonal repeated-card pairs as possible when replacement is forbidden. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 14: The same four cards are used.
Question: Find P(the sum of labels is five). Worked answer: The permitted ordered pairs are (1,4), (4,1), (2,3) and (3,2): four favourable pairs out of twelve. Therefore P = 1/3.
What to check: Listing ordered pairs is often the cleanest proof. If the event ignores order, both orientations still count in an ordered sample space. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 15: The same four cards are used.
Question: Find P(the first number is greater than the second). Worked answer: For each unequal pair, exactly one of its two orders has first greater than second. Six of twelve ordered pairs qualify, so probability = 1/2.
What to check: The symmetry is valid because all four numbers are distinct and both orders have equal probability. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 16: Three green and three orange tokens are in a bag.
Question: Find P(two same-colour tokens). Worked answer: Green–green has probability (3/6) × (2/5) = 1/5; orange–orange also equals 1/5. Add to obtain 2/5.
What to check: Two same-colour routes are alternatives. The event is not the same as drawing two green. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 17: Three green and three orange tokens are in a bag.
Question: Find P(two different colours). Worked answer: Green–orange contributes (3/6) × (3/5) = 3/10. Orange–green also contributes 3/10. Total = 3/5.
What to check: The same-colour probability 2/5 provides a complement check: 1 − 2/5 = 3/5. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 18: A bag holds two red and four blue counters. Two are removed.
Question: Find P(blue on second draw). Worked answer: Blue–blue has probability (4/6) × (3/5) = 2/5. Red–blue has probability (2/6) × (4/5) = 4/15. Sum = 2/3.
What to check: The second-draw event spans two possible first colours. It is not just the blue–blue route. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 19: The same bag holds two red and four blue counters.
Question: Find P(red then blue). Worked answer: The first red chance is 2/6. Four blue remain among five. Product is (2/6) × (4/5) = 8/30 = 4/15.
What to check: Do not use 4/6 for the second probability, because the first removal leaves five counters. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 20: A bag holds five red and two blue cards. Two are removed.
Question: Find P(at least one blue). Worked answer: The complement is two red cards. P(RR) = (5/7) × (4/6) = 20/42 = 10/21. Therefore P(at least one blue) = 11/21.
What to check: Choose the complement when the requested event has multiple routes. Keep fractions exact until the end. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 21: A bag holds one silver and five bronze cards.
Question: Find P(two silver). Worked answer: It is impossible because only one silver card exists, and the first card is not replaced. Hence the probability is 0.
What to check: Before calculating, test whether the physical inventory permits the event. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 22: A bag holds four striped and three plain cards. Two are removed.
Question: Find P(striped then plain). Worked answer: The first striped probability is 4/7; all three plain cards then remain among six. Product = (4/7) × (3/6) = 2/7.
What to check: The second denominator is six; the second numerator remains three after a striped card is removed. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 23: A bag holds four striped and three plain cards. Two are removed.
Question: Find P(exactly one plain). Worked answer: Striped–plain contributes 2/7. Plain–striped contributes (3/7) × (4/6) = 2/7. Total = 4/7.
What to check: The unordered description needs both orders. Equal terms arise from counting, not from a universal shortcut. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 24: A bag has three blue and two yellow counters, but the first is returned.
Question: Find P(two blue) and compare it with no replacement. Worked answer: With replacement, P(BB) = (3/5) × (3/5) = 9/25. Without replacement, it was 3/10. The results differ because the pool is restored in one experiment.
What to check: The wording ‘returned’ changes the sample space. Do not carry forward the wrong denominator from an earlier question. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 25: A bag has three blue and two yellow counters. Two are drawn at once.
Question: Find P(both blue) by pretending there is an order. Worked answer: Imagine drawing the first blue then the second blue, without replacement: (3/5) × (2/4) = 3/10. Both-blue does not care which individual blue was first.
What to check: The hypothetical ordered method is valid here because both blue selections are covered. Do not double the answer for swapping the two blues. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
Clinic 26: A student says ‘P(first red) is 2/5; therefore P(second red) must also be 2/5 even without replacement.’
Question: Explain the mistake carefully. Worked answer: The probability of red on the second draw depends on what happened first: after red it is 1/4, after a different colour it is 2/4, assuming the original bag had two red among five. The student has confused a conditional second-stage branch with an overall second-position probability.
What to check: Interestingly, when every item is equally likely to occupy each position in a random draw, the overall probability that the second is red is still 2/5. But it must be justified across branches, not copied into each branch. In your written working, name the event you calculated, show the changing count and decide whether complete routes should be multiplied or added. This habit is more dependable than memorising a shape for every question.
A stronger check than repeating the same calculation
After finding a probability, verify it with a different approach. If you used a branch product, count favourable ordered pairs. If you summed three alternatives, use a complement when one exists. If the question has two colours, check that every mutually exclusive two-draw colour route adds to one. A second calculation with the same mistaken fraction is not an independent check. Deliberately switch representation. This technique builds from the coordinate-geometry evidence checking in Volume 0254: a calculation deserves trust when a different property confirms it.
For the original bag with three blue and two yellow counters, the four colour outcomes have probabilities BB = 3/10, BY = 3/10, YB = 3/10 and YY = 1/10. Their total is exactly one. Count ordered pairs instead: there are 5 × 4 = 20, with six BB, six BY, six YB and two YY. Divide each by 20. The two approaches agree because they represent the same experiment. When your two results disagree, do not round away the difference. Locate which sample space each calculation actually used.
Training and exam-day decision routine
Practise in three short rounds. In the first round, identify whether each scenario uses replacement, no replacement or simultaneous selection. In the second, draw or list the full outcome space. In the third, replace the target event with a closely related one: exactly one blue becomes at least one blue; two of the same colour becomes two of different colours; the first draw changes from red to blue. These near-miss variations expose whether the student understands conditions or has memorised a single pathway. Check answers with exact fractions before converting to decimals.
In the examination, write the initial inventory first. On each branch, subtract what has been removed before calculating the next fraction. Mark a path only if every stage fits the requested order. Add separate complete paths only when the event includes those alternatives. Before finalising, check that a probability lies between zero and one and whether a complement offers a quick independent check. The score comes from identifying the correct event and making the sample-space adjustment visible. A tidy product of incorrect fractions is not a correct probability.
Follow the learner route without losing the question
Students who need a primary-level restart can use the PSLE Learning Guide for English, Mathematics and Science to rebuild reading instructions and basic fractions. The more advanced probability owner, Volume 0178, develops the full tree-diagram method; Volume 0182 keeps addition and multiplication rules straight. This chapter advances one narrower intent: counting correctly after the available objects change. Its practical interpretation also complements the evidence-safe claims in Volume 0256, where denominators, observed populations and assumptions govern decisions.
All examples are original teaching cases, and official assessment details reflect SEAB’s 2027 K310 syllabus, not predictions of particular questions. The underlying lesson is calm and repeatable: after an item leaves a group, the group’s counts have changed. Write those counts correctly, and the fractions usually become straightforward. Then check whether you have answered one ordered path or the entire event. It is a small distinction with a large effect on examination accuracy.
