A bag contains some coloured counters. You know the chance of drawing a red one, but the number of red counters is missing. Ordinary probability asks you to calculate a chance from known counts. Reverse probability asks you to recover counts from a chance, then prove that the recovered quantities are possible. This small reversal tests algebra, part-to-whole reasoning, sample spaces and the discipline of checking what the question actually tells you.
This chapter is an original workshop for the 2027 Singapore–Cambridge Secondary Education Certificate, G3 Mathematics K310, earlier O-Level reference code 4052. See the SEAB 2027 subject directory and official K310 syllabus. Section S2 covers simple and combined probabilities, possibility and tree diagrams, and addition or multiplication for mutually exclusive and independent events. The paper format is Paper 1 (90 marks, 2 hours 15 minutes, 50%) and Paper 2 (90 marks, 2 hours 15 minutes, 50%). Both papers permit an approved calculator, but essential working must still be shown. This guide’s algebraic reconstruction is a teaching application of K310 probability, ratio, equations and reasoning; it is not a separately named examination component.
Continue from Volume 0359 on cumulative frequency and quartiles if your difficulty is reading the correct total and denominator. For basic ratio repair use the PSLE learning guide and for the canonical combined-events owner use the Secondary 4 probability learning guide. This volume deliberately avoids repeating the earlier tree-diagram primer. Its own task is to move backwards from a probability to a missing integer count and to audit whether the solution makes sense.
The first rule: a probability is a ratio, not a count
If every counter has an equal chance of being selected, probability of an event is favourable outcomes divided by all possible outcomes. That denominator is the number of equally likely possibilities, not necessarily the number of red counters, the number of colours or the number of categories. If a bag has 4 red and 6 blue counters, the probability of red is 4/10, which simplifies to 2/5. The fraction describes a proportion of the whole. It does not assert that only two red counters exist.
To reverse the process, assign a letter to the missing count. If there are x red and 6 blue, the probability of red is x/(x+6). Every additional red counter changes both the numerator and the denominator. Writing x/6 is the familiar but serious error: it compares red with blue instead of red with the entire bag. The variable belongs in both places because adding favourable outcomes also changes the number of total outcomes.
Consider x/(x+6)=1/4. Cross-multiplication gives 4x=x+6, hence 3x=6 and x=2. Check it by reconstructing the bag: 2 red and 6 blue give 2/8=1/4. The check is more than arithmetic. It confirms that x is a non-negative whole number and that the total count is the same quantity described by the question.
Why the integer condition matters
Counting questions have a stricter answer space than many algebra exercises. A real-number solution such as x=4.5 may be perfectly valid for a distance but impossible as the number of individual counters. If a model produces a fractional count, investigate whether the provided probability was rounded, whether the model used the correct denominator or whether the scenario is inconsistent under the stated equal-likelihood assumptions.
Always state what x represents and where its values are allowed. An unknown number of tickets must be a whole number that is zero or greater. If the bag must contain at least one red counter, x must be positive. An algebraic manipulation may produce a number outside those conditions; such an answer is not rescued by being mathematically neat. The final statement should return to the real object counted.
A ratio such as 3:5 creates a different job. It may specify red-to-blue, red-to-total or one category to another. Only a red-to-total ratio corresponds directly to the probability of red. For red:blue=3:5, red is three of eight equal parts, so the chance is 3/8. For red:total=3:5, the chance is 3/5. The symbols look similar, but their denominators carry different meanings.
Build the equation before touching the calculator
Translate each sentence into a model. ‘There are 8 blue and x red’ gives total x+8. ‘The probability of red is 3/7’ gives x/(x+8)=3/7. Multiply through carefully: 7x=3x+24, so 4x=24 and x=6. Return to the original bag: 6 red and 8 blue give 6/14=3/7. A calculation may take only a few lines, but those lines must be attached to the correct noun.
A useful method has four steps. First list the favourable category and the whole sample space. Second represent the missing count with one variable. Third form an equation from the stated probability. Fourth check that the recovered counts reconstruct the original chance and satisfy any restrictions. Do not skip the fourth step: it catches both arithmetic slips and models based on the wrong group of objects.
Watch for changes between the first draw and a later draw. Replacement leaves the bag composition unchanged; drawing without replacement changes the contents. This article revisits that distinction only when a reverse problem needs it. The detailed owner for tree diagrams and dependence is already available through the Mathematics hub. The fresh challenge here is to infer the missing original composition, not simply to multiply two known branch fractions.
Follow the language of a probability statement
‘At least one’ includes one or more successes. ‘Exactly one’ excludes zero successes and two successes. ‘Neither’ is the complement of ‘at least one’. ‘Either’ is sometimes inclusive in mathematical contexts; read the event definition before applying an addition rule. ‘Given that’ restricts the relevant sample space. When event words are unclear, write the outcomes that would count, then calculate.
Mutually exclusive events cannot happen together in the same trial. Independent events do not affect each other’s probabilities. These conditions are different. On a single draw, red and blue are mutually exclusive. On two independent draws with replacement, a red result on the first need not change the chance on the second. The usual multiplication and addition laws are powerful only when their conditions have been checked.
The official K310 syllabus expects students to reason with mathematics in context, not just execute a memorised formula. In a reverse problem, that means using words to explain why a count must be an integer, why a denominator changes after adding counters and why an observed frequency is not automatically a theoretical probability.
A complete worked diagnostic
Imagine 5 blue counters and an unknown number of red counters. The probability of red is 3/8. Let r be the number of red counters. Then r/(r+5)=3/8. Multiply both sides by 8(r+5) to obtain 8r=3r+15. Thus 5r=15 and r=3. The bag has 3 red and 5 blue, so the original chance is indeed 3/8. If you wrote 3/5 because the red count is 3 and the blue count is 5, your fractions are comparing red with blue rather than red with the whole.
Now the bag is enlarged by adding 3 red counters. Its composition becomes 6 red and 5 blue, so the new probability of red is 6/11. The event has become more likely, but not because the earlier probability was multiplied by two. The sample space has changed. That is why copying a fraction before and after an alteration creates incorrect results. Update the counts first; only then form the new probability.
Finally, the 6/11 fraction cannot tell you how frequently red must appear in exactly eleven trials. It describes the chance under the given model, not a guarantee that the outcomes will match the proportion in a small number of independent draws. Separating theoretical chance from observed frequency will matter in mixed English, Mathematics and Science evaluation tasks.
Twenty-nine varied worked problems
Each case uses an original question, a compact calculation and a specific wrong turn to avoid. The challenge is not to memorise every fraction. It is to decide what represents the whole, what changes when the situation changes and whether an algebraic answer can represent a real number of discrete objects. Cover the answers before testing yourself, then change one number and solve the modified question.
While practising, label the line containing your first equation. If a later answer is wrong, return to that line before checking multiplication. A well-calculated solution to the wrong model is still wrong, and model repair often produces faster improvement than more calculator practice.
Worked problem 01: Red count from blue count
Try it. A bag has five blue counters and r red counters; P(red)=3/8.
Working. Set r/(r+5)=3/8; then 8r=3r+15, so r=3. Check: 3 red out of eight total is 3/8. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. The numerator is not three because the fraction displays the number 3; it is three because the equation recovers that integer. In another problem the simplified numerator may not equal the actual red count. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 02: Blue count from red count
Try it. There are 12 red counters and b blue counters, with P(blue)=2/5.
Working. Write b/(12+b)=2/5. Then 5b=24+2b, so b=8. The resulting total is 20; 8/20 reduces to 2/5. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. Do not begin with 2/12. The probability compares blue with the full bag, not blue with the known red category. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 03: A probability is not a raw count
Try it. Twenty equally likely tickets are numbered; the chance of winning is 0.35.
Working. There are 0.35×20=7 winning tickets. The count must be a whole number. Check: 7/20=0.35 exactly. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. The shortcut works because the total is known and every ticket is equally likely. For unequal weights, multiplying a total count by a probability may not recover a number of winning physical tickets. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 04: Recover the total
Try it. Four special tickets have a selection probability of 1/6 in a uniform draw.
Working. Let N be the total. Then 4/N=1/6, so N=24. Check that 4/24 simplifies to 1/6. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. Do not invent twenty-four special tickets. The known count belongs only to the favourable category. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 05: Find a hidden green count
Try it. There are 9 white counters and g green counters; P(green)=1/4.
Working. Solve g/(g+9)=1/4: 4g=g+9, so g=3. The whole bag contains 12 counters. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. The answer 3 is plausible because the green portion is a quarter of twelve; confirm this before ending. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 06: Translate a ratio before probability
Try it. The red-to-blue ratio in a bag is 2:3. What is P(red)?
Working. Two red parts and three blue parts make five total parts. The chance of red is 2/5, not 2/3. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. A part-to-part ratio is not the same as a favourable-to-whole probability. Add the ratio parts when they cover the entire bag. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 07: Recover the ‘not’ category
Try it. Six black counters are in a bag, and P(not black)=3/5.
Working. The black share is 2/5. If six counters represent two-fifths, the total is 15 and nine counters are not black. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. The event ‘not black’ can include several other colours. It is not necessarily a single alternative colour. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 08: Use a complement to find missing counts
Try it. A bag has 4 red, 5 blue and 3 green counters. What is P(not green)?
Working. There are 12 counters and nine are not green. The chance is 9/12=3/4. Equivalently, 1−3/12=3/4. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. Students may omit one of the two categories in ‘not green’. Define the complement before counting. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 09: Two linked unknown groups
Try it. A bag contains x red and x+2 blue counters. P(red)=3/8.
Working. The total is 2x+2. Solve x/(2x+2)=3/8: 8x=6x+6, hence x=3. So the bag has three red and five blue. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. The plus two belongs to the blue count, not to the entire bag a second time. Expand the denominator once. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 10: Three groups and a fixed category
Try it. There are x red, 2x green and five yellow counters. P(yellow)=1/4.
Working. The total is 3x+5, giving 5/(3x+5)=1/4. Therefore 20=3x+5 and x=5. The recovered group sizes are 5, 10 and 5. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. A final check must use all three groups. Dropping the green category would change the denominator and the conclusion. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 11: Solve backwards after removal
Try it. A bag originally had n red and six blue. Two red are removed. P(red) is now 1/2.
Working. After removal, red=n−2 and total=n+4. Solve (n−2)/(n+4)=1/2 to get 2n−4=n+4, so n=8. The original chance was 8/14=4/7. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. The answer n=6 comes from ignoring the changed denominator. Write the after-removal inventory before writing an equation. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 12: Solve backwards after addition
Try it. A bag begins with four red and eight blue. Some red are added until P(red)=1/2.
Working. Let a be the number added. Then (4+a)/(12+a)=1/2. Solving gives a=4; the final bag has eight red and eight blue. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. When only red counters are added, both the favourable count and total increase. Updating just the numerator is wrong. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 13: Check changed chance against counts
Try it. A bag has six red and six blue. Four blue are added. Find the new P(red).
Working. The updated total is 16 and the red count remains six, so P(red)=6/16=3/8. The chance falls because the same favourable count is now a smaller part of the total. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. Probabilities change with the sample space. They should not be increased or decreased by directly adding the number of new counters to a probability fraction. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 14: Complement a fixed total
Try it. A box contains 18 equally likely cards. The probability of blue is 1/3. How many cards are not blue?
Working. One-third of 18 is six blue cards. The complement contains 18−6=12 cards, with chance 2/3. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. The word ‘not blue’ includes every other card, regardless of the names or number of other categories. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 15: Decide whether a count is possible
Try it. A model says eight green counters have probability 0.7 in an equal-likelihood bag. Test consistency.
Working. The implied total would be 8/0.7=80/7, not a whole number. If 0.7 is exact and all physical counters are equally likely, the statement is inconsistent. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. Do not round an impossible total to 11 or 12 without being told that the probability was approximate. State the contradiction clearly. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 16: Recognise rounded information
Try it. A student reports P(red)=0.33 after rounding. There are eight red counters. Can you recover a unique total?
Working. No exact total follows from a rounded decimal alone. A total of 24 gives 8/24≈0.3333, which rounds to 0.33, but rounding leaves a range of possible underlying probabilities. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. Write ‘a possible total’ rather than ‘the total’ unless the exact fractional probability or sufficient extra information is supplied. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 17: Distinguish experimental from theoretical
Try it. Nine successes were observed in thirty trials of an unknown process.
Working. The relative frequency is 9/30=0.3. That is an observed proportion, not automatically the exact theoretical success probability of every future trial. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. A sample estimate can be useful, but it should not be used as an exact count equation without a model linking trials to equally likely outcomes. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 18: Recover a simple uniform space
Try it. Tickets numbered 1 to 12 are equally likely. What is P(a multiple of three)?
Working. The favourable tickets are 3, 6, 9 and 12: four of twelve, so the probability is 1/3. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. List the possible outcomes first. Do not use 3/12 simply because the relevant property mentions the number three. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 19: Two independent success events
Try it. Two independent tasks have success probabilities 1/3 and 2/5. What is P(both succeed)?
Working. Multiply the chances because independence is stated: (1/3)(2/5)=2/15. The result lies between zero and each individual probability. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. The multiplication is justified by the condition, not merely by the appearance of the word ‘and’. Without independence or other conditional information, a simple product may be unjustified. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 20: Neither of two independent events
Try it. Independent events A and B have probabilities 1/3 and 2/5. What is P(neither)?
Working. The failure chances are 2/3 and 3/5. Multiply to obtain (2/3)(3/5)=2/5. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. Taking one minus the sum of the two event probabilities is not correct here because the events can happen together. Use complements of each event with independence. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 21: Exactly one colour on a single draw
Try it. A bag has two red, three blue and two green. What is P(red or blue)?
Working. Red and blue cannot occur on the same single draw, so they are mutually exclusive. Add 2/7+3/7=5/7. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. On two different draws, red and blue may both occur across the pair. Define the trial before deciding that events cannot overlap. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 22: Avoid false exclusivity
Try it. Independent events A and B have probabilities 0.4 and 0.3. Find P(A or B).
Working. P(A and B)=0.4×0.3=0.12. Thus P(A or B)=0.4+0.3−0.12=0.58. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. Simply adding 0.4+0.3 treats overlapping events as mutually exclusive. Independence does not imply exclusivity. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 23: With replacement
Try it. A bag has three red and two blue. Two counters are drawn with replacement. Find P(two red).
Working. Each draw has P(red)=3/5, and replacing the first counter restores the bag, so (3/5)(3/5)=9/25. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. Do not use 2/4 for the second red chance. That would describe the bag without replacement. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 24: Without replacement
Try it. From three red and two blue, two counters are drawn without replacement. Find P(two red).
Working. The first red chance is 3/5. After it is taken, two red remain out of four. The product is (3/5)(2/4)=3/10. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. The denominator changes because the bag gets smaller. This is a short reinforcement of the earlier dedicated without-replacement owner. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 25: Exactly one of each in two draws
Try it. Three red and two blue counters are drawn twice without replacement. Find P(one red and one blue).
Working. Two orders work: red then blue is (3/5)(2/4)=3/10; blue then red is (2/5)(3/4)=3/10. Add to get 3/5. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. The product 3/10 covers only one order. ‘One of each’ does not prescribe an order unless the question says so. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 26: Solve from a fraction and a restriction
Try it. A uniform box contains three prize tickets, and the chance of drawing a prize is 0.2.
Working. The total must be 3/0.2=15. The remaining twelve tickets are not prizes. The result is consistent with whole-number counting. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. If a probability is expressed as a decimal, convert it to a reliable fraction or equation rather than guessing the total. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 27: Test a misleading percentage
Try it. A display says 40% of a set of cards are yellow, with seven yellow cards. Can the stated percentage be exact?
Working. An exact proportion would require total=7/0.4=17.5 cards, impossible for a finite equally likely card set. Either the percentage is rounded or some assumption is missing. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. A fractional implied total is a useful alarm. Do not quietly turn it into eighteen, because 7/18 is not exactly 40%. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 28: State the event before counting
Try it. A bag has ten numbered tiles from 1 to 10. Find P(an even number greater than six).
Working. Only eight and ten satisfy both conditions. There are two favourable tiles out of ten, giving 1/5. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. Some students count six as greater than six, or count all even tiles without applying the second condition. Translate each word into a membership rule. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 29: Read a restricted sample space
Try it. A tile from 1 to 12 is known to be even. What fraction of the remaining possibilities are multiples of three?
Working. Among the six even tiles 2, 4, 6, 8, 10 and 12, two are multiples of three, so the restricted proportion is 2/6=1/3. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. The denominator is six, not twelve, because the condition changed the eligible outcome set. This is a description of a restricted sample space, not a new required K310 topic name. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
Worked problem 30: Audit a two-step public claim
Try it. A game has a winning chance of 1/4 per independent play. A poster claims two plays guarantee one win.
Working. Even with independence, two misses remain possible. The probability of at least one win is 1−(3/4)²=7/16, not one. State the favourable outcomes and the full eligible sample space before writing the fraction. If a letter is used, say what physical count it represents. Finish by substituting the recovered quantity into the original condition rather than checking only your final arithmetic.
Why answers fail. Distinguish a possible outcome, an expected tendency and a guarantee. A probability less than one never guarantees occurrence in a short series of trials. To transfer this method, alter the question’s known total, replace one colour with a different category, or change ‘with replacement’ to ‘without replacement’. Then decide which parts of the original reasoning survive and which must change.
A stronger examination answer is also a stronger explanation
For Paper 1, make the model before the calculation and write an economical answer. For Paper 2, define the variable, include the equation, rearrange visibly and show the integer check. If the result is a number of counters, do not leave an unqualified decimal. If the result is a probability, give an exact fraction where appropriate. The official syllabus also specifies three significant figures for non-exact numerical answers unless a different accuracy is requested, and advises students to show a more accurate value first where the required rounding must be demonstrated.
An effective revision page is built from causes of error. One column records the missed word in a question, such as ‘not’, ‘both’, ‘exactly’, ‘after’ or ‘given’. Another records the false equation and its corrected form. A final column supplies a changed example solved without notes. This reveals whether a learner is repeatedly confusing favourable-to-total with part-to-part, whether updates to the sample space are being overlooked, or whether an integer restriction is being forgotten.
On an unfamiliar context, stay with the underlying model. A token, ticket, counter, seat label or prize card is still a discrete object if each one is equally likely. A survey percentage or experimental rate, however, may not supply such an equal-likelihood model. Do not move from ‘we observed thirty percent’ to ‘there must be exactly three winning tickets in a ten-ticket box’ unless the setup establishes that conclusion. This distinction becomes important in cross-subject evidence audits.
Check yourself before moving to the combined EMS case
Try these five prompts without notes. A box has four blue and x red counters with P(red)=1/3: find x. A class has twenty-four response cards, with probability 3/8 of selecting an A card: how many A cards are there? A poster reports seven green counters as exactly forty percent of a uniformly sampled bag: decide whether the claim is possible. Two independent events have probabilities one-half and one-quarter: find the chance of neither. Finally, explain in a sentence why the probability of a label being selected cannot prove the chemical identity of the item behind that label.
The first answer is x=2 because x/(x+4)=1/3. The second is nine because twenty-four times 3/8 equals nine. The third exact claim is inconsistent with a whole-number total. The fourth is (1/2)(3/4)=3/8. The last answer is a matter of evidence: a selection rule describes outcomes in a defined model, while chemical identity needs separate chemical information. If you can explain all five without copying the structure of the worked cases, your reasoning is becoming transferable.
Keep the Mathematics probability owner as your main topic route, use the PSLE learner route for part-to-whole repair, and check SEAB’s 2027 K310 syllabus when confirming formal content. The next integrated exercise will use simple selection probabilities alongside chemical labels and English editing, with each claim kept inside its proper evidence boundary.
