A probability fraction describes a particular experiment at a particular stage. Change what is in the bag, how the next object is selected or what is already known, and the fraction may have to change. When a quantity is unknown, the same fraction becomes an equation that must still respect the experiment’s rules.
This Secondary 3 Mathematics Learning Guide develops reverse-probability questions, changing sample spaces and algebraic constraints. You will find unknown numbers of counters, distinguish adding from removing, update probabilities along a tree, investigate multiple possible answers and recognise when given probability information cannot describe a whole-number collection.
For the first introduction to sample spaces and combined events, use Probability and Statistical Reasoning. This guide starts where routine forward calculation stops: the probability may be given while the collection is unknown, or the collection may change according to the outcome of an earlier draw.
Scope and source: the 2027 SEC G3 Mathematics syllabus includes single and simple combined events. The reverse problems below connect that content with algebra. They are original consolidation questions, not reproduced examination questions or a separate official topic list.
Choose your route: unknown counts · adding and removing · dependent draws · quadratic equations · different selection rules · practice · answers.
Return to the Secondary Mathematics Hub. Useful algebra support: Algebraic Fractions and Quadratic Equations.
Specify the Experiment Before Counting Outcomes
In the counter examples, assume each counter is equally likely to be selected from the stated well-mixed bag at each draw. That assumption allows favourable-counter count divided by total-counter count. It does not mean that colours are equally likely. A bag with one red and nine blue counters does not give a half chance of each colour merely because two colour names are possible.
Also identify whether order matters. “Red then blue” selects one ordered path. “Exactly one red in two draws” includes both red-then-blue and blue-then-red. “At least one red” also includes two reds. These are different events in the same experiment, so they usually have different probabilities.
Finally, state what happens between draws. A counter can be replaced, left out, exchanged or followed by an additional counter. The word “replaced” describes one action; it does not by itself guarantee that nothing else changes. The entire selection rule determines the next sample space.
A Four-Part Diagnostic
A bag contains three red and five blue counters. Find the probability of red on the first draw. After a blue is removed and not replaced, find the probability of red on the next draw. After a red is removed instead, find the probability of red on the next draw. Then name the two ordered outcomes that make “exactly one red in two draws”.
The answers are 3/8, 3/7, 2/7, and RB together with BR. Removing a blue changes the denominator but not the red numerator. Removing a red changes both. The final answer tests event interpretation rather than fraction arithmetic.
A useful self-check is to describe the physical collection before writing the fraction: “Three red remain among seven counters.” This sentence makes it harder to retain a stale denominator or remove the wrong colour. It also makes later algebra easier because the meaning of an expression such as n − 1 remains visible.
Worked Example 1: Find an Unknown Initial Count
A bag contains n red counters and five blue counters. The probability of selecting a red counter is 3/8. Find n. The total is n + 5, not five. Therefore n/(n + 5) = 3/8.
Multiply by the nonzero denominator: 8n = 3(n + 5). Then 8n = 3n + 15, so n = 3. Check the experiment: three red among eight counters gives probability 3/8.
The denominator depends on the unknown. Writing n/5 = 3/8 would compare red counters with blue counters, not red counters with the whole collection. That incorrect fraction is a part-to-part ratio. Probability here needs a part-to-whole ratio.
The answer must be a nonnegative integer because n counts physical counters. The algebraic domain also requires n + 5 ≠ 0, but the counting condition is stronger in this context. Stating the counting condition first often makes the denominator condition automatic.
Worked Example 2: Add Counters to Reach a Target Probability
A bag initially contains five red and seven blue counters. Some red counters are added. The probability of red becomes 2/3. How many were added? Let k be the added red count, where k is a nonnegative integer.
The new red count is 5 + k and the new total is 12 + k. Set (5 + k)/(12 + k) = 2/3. Cross-multiplication gives 15 + 3k = 24 + 2k, so k = 9.
The resulting bag has fourteen red and seven blue counters, giving 14/21 = 2/3. Notice that nine counters are added, but the final red count is fourteen. A question asking how many were added is not asking for the final numerator.
A common wrong model is (5 + k)/12. It changes the red count while pretending the total remains fixed. That could describe recolouring existing counters, but it does not describe adding new ones. Similar-looking wording can encode different conservation conditions; the model must follow the actual action.
Worked Example 3: Remove the Other Colour
A bag contains eight red and twelve blue counters. Blue counters are removed until the probability of red is 2/3. How many blue counters are removed? Let k be the removed blue count, with 0 ≤ k ≤ 12.
The red numerator remains eight. The new total is 20 − k. Thus 8/(20 − k) = 2/3, giving 24 = 40 − 2k and k = 8. Four blue counters remain, so the final probability is 8/12 = 2/3.
The same target probability as the previous example came from a different action. Adding red changes numerator and denominator upward. Removing blue changes only the denominator downward. A solution that begins with a before-and-after count record can separate those cases before algebra starts.
| Action on a bag with R red and B blue | New probability of red |
|---|---|
| Add k red counters | (R + k)/(R + B + k) |
| Remove k blue counters | R/(R + B − k) |
| Recolour k blue counters red | (R + k)/(R + B) |
Each row has its own permitted values of k. Removed or recoloured blue counters cannot exceed the available blue count. A formula without those conditions could describe an impossible collection.
Worked Example 4: Build Every Path Without Replacement
A bag contains three red and two blue counters. Two counters are drawn one after the other without replacement. After an initial red, the remaining bag has two red and two blue counters. After an initial blue, it has three red and one blue counter.
| Ordered path | Calculation | Probability |
|---|---|---|
| RR | (3/5)(2/4) | 3/10 |
| RB | (3/5)(2/4) | 3/10 |
| BR | (2/5)(3/4) | 3/10 |
| BB | (2/5)(1/4) | 1/10 |
The four probabilities total one. Exactly one red includes RB and BR, giving 3/10 + 3/10 = 3/5. Two red counters correspond only to RR, giving 3/10. The event description selects the relevant rows after each path has been modelled correctly.
Why multiply along a path? The second fraction gives the chance of the next result within the collection produced by the first result. It is a conditional branch probability. Multiplication along a tree does not require pretending that the draws are independent. Independence would allow the second probability to remain the same regardless of the first result; here it does not.
There is a subtle check. The probability that the second counter is red is RR + BR = 3/10 + 3/10 = 3/5, equal to the original red proportion. That equality of overall probabilities does not make the draws independent. The probability after a known first red is 2/4, whereas after a known first blue it is 3/4. The branch information still matters.
Worked Example 5: A Probability Equation That Becomes Quadratic
A bag contains r red counters and three blue counters. Two counters are drawn without replacement. The probability of two red counters is 2/7. Find r. Since two reds must be possible, r is an integer at least two.
The first red probability is r/(r + 3). After drawing one red, the next red probability is (r − 1)/(r + 2). Therefore r(r − 1)/[(r + 3)(r + 2)] = 2/7.
Clear the denominators: 7r(r − 1) = 2(r + 3)(r + 2). Expanding gives 7r² − 7r = 2r² + 10r + 12. Rearranging produces 5r² − 17r − 12 = 0, which factorises as (r − 4)(5r + 3) = 0.
The algebraic candidates are r = 4 and r = −3/5. Only r = 4 is a possible count. Check the original experiment: (4/7)(3/6) = 12/42 = 2/7. The rejected value fails the counting domain; it is not discarded merely because fractions are inconvenient.
The distinctive work happened before factorisation. Both the red count and the total changed after the first draw. Writing r²/(r + 3)² would describe independent draws with replacement, a different experiment. A polished quadratic solution cannot repair the wrong experiment.
Worked Example 6: Reject a Positive but Non-Integer Candidate
A bag contains five red counters and b blue counters. Two are drawn without replacement. The probability of red followed by blue is 15/56. Find b. The event is ordered, so there is one path: [5/(b + 5)] × [b/(b + 4)].
Set 5b/[(b + 5)(b + 4)] = 15/56. Multiplication gives 280b = 15(b² + 9b + 20). Rearranging and dividing by five gives 3b² − 29b + 60 = 0. Hence (b − 3)(3b − 20) = 0.
The candidates are 3 and 20/3. Both are positive, but only b = 3 is an integer. Substitution gives (5/8)(3/7) = 15/56. A positive root is not automatically valid in a count problem.
If the wording instead said “one red and one blue”, the event would include RB and BR. Doubling the one-path expression would then be appropriate for this experiment. It is not appropriate for the explicitly ordered event. Read the word “followed” as carefully as any algebraic sign.
Worked Example 7: Two Different Collections Can Fit the Same Probability
A bag contains ten counters, each red or blue. Two are drawn without replacement. The probability of exactly one red is 8/15. Find the possible numbers of red counters. Let r be the red count, so blue count is 10 − r.
The two favourable paths give [r/10][(10 − r)/9] + [(10 − r)/10][r/9] = 8/15. Thus 2r(10 − r)/90 = 8/15, and r(10 − r) = 24.
Rearranging gives r² − 10r + 24 = 0, or (r − 4)(r − 6) = 0. Both four red counters and six red counters are possible. Interchanging red and blue preserves the probability of one of each colour, so the symmetry of the answers is meaningful.
An extra condition such as “there are more red than blue” would select r = 6. Without that condition, selecting six simply because it is the larger root would invent information. A complete solution reports all admissible answers, including genuine ambiguity.
Worked Example 8: Use the Complement for At Least One
A bag contains three red and five blue counters. Two are drawn without replacement. To find the probability of at least one red, consider the opposite event: no red, meaning BB.
P(BB) = (5/8)(4/7) = 5/14. Therefore P(at least one red) = 1 − 5/14 = 9/14. This route has one unfavourable path instead of three favourable paths.
The complement must cover exactly the outcomes not wanted. “Not two red” is not the same as “at least one red”: it includes BB and excludes RR. Write the event and its complement in words before subtracting from one. The arithmetic is simple only after the event boundary is correct.
Worked Example 9: Three Draws, Exactly Two Red
A bag contains four red and three blue counters. Three are drawn without replacement. Exactly two red counters can occur through RRB, RBR or BRR.
P(RRB) = (4/7)(3/6)(3/5) = 6/35. P(RBR) = (4/7)(3/6)(3/5) = 6/35, but the middle numerator now counts blue counters and the last numerator counts remaining red counters. P(BRR) = (3/7)(4/6)(3/5) = 6/35.
The total is 18/35. The three products happen to be equal because the same colour counts are selected without replacement in different orders. It is still useful to write them once and inspect the changing collection. Do not assume that all colour strings in a different experiment have equal probabilities.
A student who finds only RRB has solved “red, red, then blue”, not “exactly two red”. A student who also includes RRR has solved a different event again. Enumerating the permitted strings before calculating is a small step that protects the whole answer.
Worked Example 10: Choosing a Bag Is Part of the Experiment
Bag A contains one red and one blue counter. Bag B contains one red and three blue counters. A bag is selected with probability 1/2 each, then one counter is selected uniformly from that bag. Find the probability of red.
Red can occur by A then red, with probability (1/2)(1/2) = 1/4, or B then red, with probability (1/2)(1/4) = 1/8. Add these disjoint paths to get 3/8.
Combining all counters physically into one bag would give two red among six counters, or 1/3. That answer is not correct for the stated two-stage selection process. Selecting the bags equally does not make all six counters equally likely: each A counter has probability 1/4, while each B counter has probability 1/8.
This is a useful warning against flattening a multi-stage experiment into one favourable-over-total fraction. The fraction method is valid for equally likely elementary outcomes. A tree preserves the unequal weights introduced by the first selection stage.
Worked Example 11: Replacement Followed by Another Change
A bag starts with two red and two blue counters. A counter is drawn and replaced. Before the next draw, one counter of the opposite colour is added. Find the probability that both draws are red.
After an initial red, replacing it restores two red and two blue. Adding the opposite colour then produces two red and three blue. Therefore P(RR) = (2/4)(2/5) = 1/5, not 1/4.
The remaining path probabilities are P(RB) = 3/10, P(BR) = 3/10 and P(BB) = 1/5. They total one. The first counter was replaced, but the bag did not remain unchanged. Reading only the familiar keyword “replacement” would select the wrong model.
Worked Example 12: The Unknown Can Be a Probability Rather Than a Count
Bag A contains three red and one blue counter. Bag B contains one red and three blue. Bag A is selected with probability p and Bag B with probability 1 − p. After selecting a bag, one counter is drawn uniformly. The overall probability of red is 11/20. Find p.
The weighted paths give (3/4)p + (1/4)(1 − p) = 11/20. Simplify to p/2 + 1/4 = 11/20. Then p/2 = 3/10, so p = 3/5.
The domain is now 0 ≤ p ≤ 1, not an integer-count condition. A fraction is exactly the kind of answer required. The same algebraic discipline applies, but the admissibility test depends on what the variable represents. Here the overall red probability also lies between 1/4 and 3/4, the two bag-specific probabilities, which supplies a useful check.
When the Given Probability Is Impossible
Suppose a bag has n red and four blue counters, and its exact theoretical red probability is stated as 2/5. The equation n/(n + 4) = 2/5 gives 5n = 2n + 8, so n = 8/3. That is not a valid count. There is no whole-number collection satisfying all the stated conditions.
Do not round 8/3 to three counters. Three red and four blue give probability 3/7, not 2/5. Exact probability conditions cannot be repaired by rounding an impossible collection into existence. A classroom question may deliberately test inconsistency; in other circumstances the data or wording would need checking.
A different statement would be “red occurred in 40 of 100 experimental draws”. That is an observed proportion, not an exact description of the bag’s composition. Random experimental outcomes can differ from a model’s theoretical probability. The reverse equations in this guide use probabilities specified by the model, not noisy observations silently treated as exact.
One Probability May Determine a Ratio but Not a Total
An initial red probability of 2/5 fits two red and three blue counters, four red and six blue counters, and many larger collections. It determines a proportion but not the total number of counters.
Without replacement, those collections need not give the same two-red probability. The five-counter bag gives (2/5)(1/4) = 1/10. The ten-counter bag gives (4/10)(3/9) = 2/15. Knowing the first-draw proportion is therefore insufficient for every later calculation. A total count or another independent condition may be necessary.
This is why a good solution distinguishes “I have not solved it yet” from “the information permits several answers”. The latter can be a mathematical conclusion supported by two explicit models that satisfy the same given information.
Independent Practice: Record the Collection at Each Stage
All draws are uniform among the counters in the selected bag. State whether the unknown is an integer count or a probability, and check every algebraic candidate against that domain.
1. A bag contains n red and six blue counters. P(red) = 2/5. Find n.
2. Four red and six blue counters are in a bag. How many red counters must be added to make P(red) = 1/2?
3. A bag contains six red and nine blue counters. How many blue counters must be removed to make P(red) = 3/5?
4. Six red and four blue counters are in a bag. Find the probability of two reds with replacement, then without replacement.
5. Four red and three blue counters are sampled twice without replacement. Find P(exactly one red).
6. Two red and four blue counters are sampled twice without replacement. Find P(at least one red).
7. Bag A contains two red and one blue counter. Bag B contains one red and three blue. A bag is chosen with equal probability, followed by one counter from it. Find P(red).
8. Three red and two blue counters are sampled three times without replacement. Find P(exactly two red).
9. Five red and b blue counters are sampled twice without replacement. P(red then blue) = 15/56. Find all algebraic candidates for b and state the valid count.
10. Eight counters are red or blue. Two are drawn without replacement. The probability that their colours match is 13/28. Find all possible red counts.
11. A bag contains n red and four blue counters. Its exact red probability is 2/5. Explain why the conditions cannot all hold for a whole-number collection.
12. A student knows only that a bag’s red probability is 2/5. Explain why this is not enough to determine the probability of two reds without replacement.
Explained Answers
1. n/(n + 6) = 2/5 gives 5n = 2n + 12, so n = 4. Four red among ten gives the stated probability.
2. (4 + k)/(10 + k) = 1/2 gives 8 + 2k = 10 + k, so two red counters are added. The new bag has six red and six blue.
3. 6/(15 − k) = 3/5 gives 30 = 45 − 3k, so five blue counters are removed. Four blue remain, giving six red among ten.
4. With replacement, (6/10)² = 9/25. Without replacement, (6/10)(5/9) = 1/3. Only the latter reduces the red count and total after the first red.
5. Add RB and BR: (4/7)(3/6) + (3/7)(4/6) = 4/7. The two paths are mutually exclusive, so their probabilities are added.
6. Use the complement: 1 − (4/6)(3/5) = 1 − 2/5 = 3/5. The subtracted event is two blues.
7. P(red) = (1/2)(2/3) + (1/2)(1/4) = 1/3 + 1/8 = 11/24. Do not pool the counters because the bag-selection step changes individual selection weights.
8. RRB, RBR and BRR each have probability 1/5. For example, RRB gives (3/5)(2/4)(2/3) = 1/5. Their sum is 3/5.
9. The equation reduces to (b − 3)(3b − 20) = 0. Candidates are 3 and 20/3, but only b = 3 is an integer. The original ordered probability is then (5/8)(3/7).
10. Let red count be r. Matching colours mean RR or BB, so [r(r − 1) + (8 − r)(7 − r)]/56 = 13/28. Simplification gives r² − 8r + 15 = 0, so r = 3 or r = 5. Both counts are valid; swapping colour names preserves matching probability.
11. The equation gives n = 8/3, not a whole number. Rounding would change the exact probability, so there is no valid count under the stated conditions.
12. Both two-red/three-blue and four-red/six-blue bags have initial red probability 2/5, but their two-red probabilities without replacement are 1/10 and 2/15. The total collection size is additional necessary information.
A Repair Routine for Reverse Probability
First decide whether the event has been described correctly. Then write the collection before and after each action. Only after that should you inspect algebra. A wrong factorisation and a wrong sample space can both produce a wrong count, but they require different repairs.
For each corrected question, substitute the recovered count into the original experiment rather than only the final quadratic. Check every branch leaving a node totals one, and check a complete set of disjoint path probabilities totals one. When the answer is ambiguous, identify the extra information that would select one possibility instead of inventing it.
For a transfer task, keep the same numerical target but change the operation from adding red to removing blue or recolouring blue. The learner must alter the denominator appropriately. This tests whether the fraction represents a collection or is merely a familiar template.
Questions Students and Parents Often Ask
Do we multiply only when events are independent? No. A tree multiplies the first probability by the appropriate next-stage probability on that path. For dependent draws, that next-stage probability reflects the changed collection.
Why can two positive roots fail to be two answers? A counter count must be an integer and satisfy availability restrictions. A positive fraction such as 20/3 is still impossible as the number of counters.
Should every probability problem have one numerical answer? Not necessarily. Some conditions permit several collections, and some are inconsistent. Showing this through equations and explicit checks is a complete mathematical conclusion.
What should a parent ask when a child is stuck? Ask what the next bag contains after the stated action, which outcomes the event includes, and what the unknown counts. These questions reveal the modelling decision without supplying the equation.
Continue the Secondary 3 Learning Route
Continue with Distance-Time and Speed-Time Graphs for stage-specific formulas; Missing Values, Combined Means and Statistical Data Corrections for changing totals; and Geometric Algebra, Constraints and Maximum-Area Problems for admissible algebraic solutions.
A probability answer is secure when the event, selection process, changing collection and final count all agree. Return to the Secondary Mathematics Hub.
