Two identical-looking classroom boxes contain different mixes of coloured counters. A pupil chooses one box at random, takes a counter and sees that it is red. Does that observation tell us which box was probably chosen? It changes what can reasonably be concluded, but it does not identify the box with certainty. The Mathematics of revising a probability after an observation is called conditional probability.
This Mathematics in Bukit Timah guide uses entirely fictional counters, boxes, scanner flags and classroom trials to teach probability trees, intersection and union, independence, replacement, total probability, Bayes’ rule and the role of starting frequencies. The examples do not describe actual Bukit Timah students, examination questions, products or automated identification systems. All counts and probabilities are deliberately defined as mathematical teaching data.
The quick example: Box A contains three red and two blue counters, while Box B contains one red and four blue. If either box is chosen with probability one-half and one counter is drawn uniformly, the probability of red is 2/5, while the probability the chosen box was A given red is 3/4. The conditional probability differs from the original one-half because red is more likely to come from A.
Article contents
- Quick answer: conditional probability changes the denominator
- Why this Bukit Timah classroom setting is fictional
- Define the first-stage choices
- Box contents set the conditional colour probabilities
- Worked example 1: draw the four tree leaves
- Why tree paths multiply
- Why different complete paths add
- Worked example 2: red from either box
- Updating the probability after seeing red
- Worked example 3: prove the reverse conditional is different
- Updating after blue gives another result
- Worked example 4: check the blue conditional probabilities
- Bayes’ rule expresses the updating step
- What is the prior probability?
- The starting selection rule can change the answer
- Worked example 5: recalculate with a new prior
- Why a conditional probability is not an intrinsic label
- A 100-record classroom scanner illustrates base rates
- Worked example 6: put the scanner counts in a table
- Why 90% is not the answer to the reversed question
- Worked example 7: a rarer symbol changes the flagged share
- What a base rate contributes
- Conditional probabilities can be written with ordinary counts
- Two-stage trees also handle repeated counter draws
- Worked example 8: two red with replacement
- Without replacement changes the branch probabilities
- Worked example 9: two red without replacement
- Probability of one red and one blue needs both orders
- Worked example 10: compare one-of-each with replacement
- Independence is a mathematical test
- Worked example 11: test independence numerically
- Mutually exclusive does not mean independent
- Worked example 12: one draw, two colours
- Conditional independence requires additional conditions
- Worked example 13: two reds with one hidden box
- A new box each time creates a different experiment
- A tree diagram should label conditions on branches
- Worked example 14: three red without replacement
- Why multiplying every probability is not always enough
- Worked example 15: at least one red in two draws
- Probability after evidence is not proof of certainty
- The denominator must match the new sample space
- Worked example 16: distinguish three scanner fractions
- Bayes’ rule is not a licence to invent prior probabilities
- Worked example 17: same likelihood, unknown prior
- The value of a two-way table
- Worked example 18: total probability with three boxes
- Exact fractions often make conditional checks easier
- One event can be both common and weak evidence
- Worked example 19: an uninformative colour
- Primary foundations: coloured counters and simple fractions
- Lower Secondary: sample spaces and two-stage choices
- Advanced extension: Bayes and evidence updating
- Three errors to diagnose separately
- A twenty-minute home diagnostic
- A seven-day retrieval plan
- Practice questions and checked answers
- Frequently asked questions
- Sources and connected Mathematics in Bukit Timah
Quick answer: conditional probability changes the denominator
For events A and R with P(R)>0, the conditional probability of A given R is P(A|R)=P(A∩R)/P(R).
The numerator counts or measures outcomes where both A and R occur. The denominator limits attention to outcomes where R occurs at all.
The vertical bar means given, not division of two event names. P(A|R) and P(R|A) ask different questions and generally have different values.
Why this Bukit Timah classroom setting is fictional
Families near Beauty World, Sixth Avenue and King Albert Park may recognise coloured counters and small containers from mathematical games.
But the contents of the boxes here were chosen to make calculations transparent. No claims are made about the frequency of any real event or the accuracy of a scanner used by a school.
A mathematical prediction is only as reliable as its supplied probabilities and selection rules. Those rules belong near the beginning of every solution.
Define the first-stage choices
Our first experiment chooses Box A or Box B uniformly, so P(A)=1/2 and P(B)=1/2.
These two events are mutually exclusive and exhaustive for the chosen box. Exactly one is selected per trial under the stated procedure.
The second stage chooses one counter uniformly from the selected box. The resulting colour depends on which box was chosen, so the tree must preserve the two branches.
Box contents set the conditional colour probabilities
Box A has three red among five counters, so P(R|A)=3/5 and P(blue|A)=2/5.
Box B has one red among five, so P(R|B)=1/5 and P(blue|B)=4/5.
Each colour pair sums to one inside its box. The red probability is not a universal 3/5 until a particular box has been specified.
Worked example 1: draw the four tree leaves
Begin with Box A or B, each with branch probability 1/2. From A, draw red with probability 3/5 or blue with 2/5. From B, draw red with 1/5 or blue with 4/5.
Multiply branch probabilities to obtain the four complete outcomes: A and red = (1/2)(3/5)=3/10. A and blue = (1/2)(2/5)=2/10. B and red = (1/2)(1/5)=1/10. B and blue = (1/2)(4/5)=4/10.
Their sum is one, a useful check of the complete sample space.
Why tree paths multiply
A full path represents a sequence of events. The probability of reaching its second stage equals the chance of its first stage multiplied by the conditional chance of the second event.
Formally, P(A∩R)=P(A)P(R|A). This identity follows from the conditional-probability definition when P(A)>0.
Students should not automatically multiply probabilities as though all stages were independent. The second factor can depend on the earlier outcome, as it does in this box experiment.
Why different complete paths add
The outcomes “A and red” and “B and red” cannot happen together in one trial because only one box was selected.
Therefore P(R)=P(A∩R)+P(B∩R)=3/10+1/10=4/10=2/5.
Addition is appropriate because the paths are disjoint and together exhaust the ways to observe red. The multiplication within a path and addition across suitable paths have distinct justifications.
Worked example 2: red from either box
Using the complete tree, add both red leaves: 3/10+1/10=4/10=0.4.
Thus the chance of a red counter is 40%. The chance of blue is the remaining 60%, or 2/10+4/10=6/10.
The red result lies between the box-specific red chances 1/5 and 3/5, as expected from averaging the two equal-prior alternatives.
Updating the probability after seeing red
Once red is observed, restrict attention to the two red leaves. Their combined probability is 4/10.
The A-red path contributes 3/10 of the full sample space, so its share among red outcomes is (3/10)/(4/10)=3/4.
Therefore P(A|R)=75%. This does not say that 75% of counters in Box A are red. It says that among trials producing red under the stated selection procedure, three quarters originate from A.
Worked example 3: prove the reverse conditional is different
P(R|A)=3/5=60% because Box A contains three red out of five. P(A|R)=3/4=75% because three quarters of red outcomes come from A under the complete two-stage experiment.
The first conditions on A and asks about colour. The second conditions on red and asks about the box.
A pupil who writes P(A|R)=3/5 has reversed the condition. Label the restricted denominator in full words before substituting values.
Updating after blue gives another result
The two blue leaves have probabilities 2/10 from A and 4/10 from B, for total 6/10.
The chance the box was A given blue is (2/10)/(6/10)=1/3. The chance it was B given blue is 2/3.
Blue makes B more plausible under these assumptions because blue counters are relatively common in B. This is inference about a fictional random selection, not certainty about an unseen object.
Worked example 4: check the blue conditional probabilities
Calculate P(A|blue)=1/3 and P(B|blue)=2/3. They sum to one.
Each is conditioned on the same observed colour, so they partition the remaining box possibilities.
The analogous red conditional probabilities P(A|red)=3/4 and P(B|red)=1/4 also sum to one. This is a useful consistency check when only A and B are possible.
Bayes’ rule expresses the updating step
For events A and R with suitable nonzero probabilities, Bayes’ rule is P(A|R)=P(R|A)P(A)/P(R).
For our boxes, substitute (3/5)(1/2)/(2/5). The numerator is 3/10 and the denominator 4/10, giving 3/4.
Bayes’ rule does not magically create evidence; it reorganises probability relationships that already depend on a well-defined model.
What is the prior probability?
Before seeing a counter, the probability that Box A was selected is P(A)=1/2. This is the prior for the hidden box event.
After observing red, P(A|red)=3/4 is the posterior under the stated conditional colour rates.
A prior is not automatically a personal opinion. In our experiment it comes from an explicit fair box-selection mechanism. In other problems it may depend on measured frequencies or additional assumptions.
The starting selection rule can change the answer
Now imagine Box A is selected only one quarter of the time, and B three quarters of the time. Keep the counter contents unchanged.
Then P(A∩R)=(1/4)(3/5)=3/20, and P(B∩R)=(3/4)(1/5)=3/20.
Red outcomes are equally likely to have come from either box, so P(A|red)=1/2, despite red remaining more common *within* A.
Worked example 5: recalculate with a new prior
Use P(A)=1/4, P(B)=3/4, P(R|A)=3/5 and P(R|B)=1/5.
The overall red chance is 3/20+3/20=6/20=3/10. Bayes’ rule gives P(A|R)=(3/20)/(6/20)=1/2.
The posterior changed from 3/4 to 1/2 because the prior box-selection frequencies changed. The likelihoods of red inside the two boxes were held constant.
Why a conditional probability is not an intrinsic label
“Red means A” is not a universal rule. The probability of A after seeing red depends both on the tendency of A to produce red and on how often A was selected beforehand.
A student who inspects only P(R|A) while ignoring P(A) can be misled. The initial population or selection mechanism belongs in the calculation.
This matters far beyond coloured counters, but an educational example is preferable to inventing high-stakes factual probabilities for real people.
A 100-record classroom scanner illustrates base rates
Suppose 100 purely imaginary printed labels contain a special symbol on twenty labels and lack it on eighty. A fictional scanner flags 90% of symbol labels and 10% of plain labels.
Under this constructed count model, it flags 18 of the 20 symbol labels and eight of the 80 plain ones. Thus 26 labels are flagged, but only 18 flagged labels have the symbol.
The chance a uniformly selected flagged label truly contains the symbol is 18/26=9/13≈69.2%, not 90%. The 90% was the flagging rate *given a symbol*, not the symbol rate *given a flag*.
Worked example 6: put the scanner counts in a table
Imagine rows labelled actual symbol and no symbol, and columns labelled flagged and not flagged.
Symbol row: 18 flagged, 2 not flagged; total 20. Plain row: 8 flagged, 72 not flagged; total 80.
Column totals are 26 flagged and 74 not flagged, for overall total 100. This two-way table verifies all inputs before the conditional calculation.
Why 90% is not the answer to the reversed question
The model states P(flag|symbol)=18/20=90%. It does not state P(symbol|flag).
Once a flag has been observed, the relevant denominator is 26 flagged labels, not 20 symbol labels. Therefore P(symbol|flag)=18/26≈69.2%.
The direction of the conditioning event controls what the denominator includes. A reliable solution says that restricted group aloud.
Worked example 7: a rarer symbol changes the flagged share
Use another imaginary collection of 200 labels, with ten containing the special symbol and 190 plain. Keep the same ideal scanner rates: 90% of symbol labels flagged, 10% of plain labels flagged.
The flagged counts are nine symbol and nineteen plain, totalling 28. Therefore P(symbol|flag)=9/28≈32.1%.
Although the scanning percentages did not change, the positive share among flagged labels is much lower because the special symbol is now rare in the larger set.
What a base rate contributes
A base rate is the frequency or prior probability of the event being investigated before the new indicator is observed.
In the two scanner examples, the symbol occurs among 20% and 5% of the respective fictional collections. Those starting proportions change the meaning of the same flagging performance.
The term is useful when an observed signal seems compelling. Its mathematical lesson is to consider the whole starting population, not only the cases where a signal appears.
Conditional probabilities can be written with ordinary counts
For a finite sample space where every individual record is equally likely to be selected, conditional probability can be calculated as favourable records inside the condition divided by all records satisfying the condition.
Our flagged-label example uses 18 symbol-and-flagged records divided by 26 flagged records.
When individuals are selected with unequal probabilities, raw counts alone may not capture the correct probability. The sampling rule must be stated before using frequency ratios.
Two-stage trees also handle repeated counter draws
Return to Box A, with three red and two blue counters. Suppose a counter is drawn, recorded and replaced before drawing again.
Replacement restores the original contents. Each draw has red probability 3/5, and the successive draws are independent under the stipulated uniform procedure.
The probability of red followed by red is therefore (3/5)(3/5)=9/25. This differs from drawing without replacement.
Worked example 8: two red with replacement
In Box A, the first red chance is 3/5. After replacement, the second red chance remains 3/5.
Multiply the path: P(red then red)=3/5×3/5=9/25=36%.
The word replacement is not a minor detail. It determines whether the second-stage denominator and numerator remain the same.
Without replacement changes the branch probabilities
If a red counter is drawn from Box A and kept out, four counters remain, of which two are red.
Thus P(second red | first red)=2/4=1/2. The chance of two reds in order is (3/5)(2/4)=3/10.
The two events are not independent under this procedure because the first outcome changes the available contents.
Worked example 9: two red without replacement
Calculate 3/5×2/4=6/20=3/10=30%.
Compare with 9/25=36% under replacement. The result changes because returning a red counter makes it available again.
A student who uses 3/5 twice in the no-replacement question has ignored the altered composition after the first draw.
Probability of one red and one blue needs both orders
For two draws without replacement from Box A, “one red and one blue” can occur red–blue or blue–red.
Red then blue has probability (3/5)(2/4)=6/20. Blue then red has probability (2/5)(3/4)=6/20.
The routes are mutually exclusive, so add them: 12/20=3/5. Using only the first order would answer a narrower question than was asked.
Worked example 10: compare one-of-each with replacement
With replacement from Box A, P(red then blue)=(3/5)(2/5)=6/25.
P(blue then red)=(2/5)(3/5)=6/25. Therefore the probability of one of each in either order is 12/25=48%.
Without replacement it was 3/5=60%. Identical event wording can produce different answers when the physical sampling rule changes.
Independence is a mathematical test
Two events E and F are independent when P(E∩F)=P(E)P(F). Equivalently, when P(F)>0, P(E|F)=P(E).
This does not mean the events happen at different times or look unrelated. Independence describes an equality of probabilities under a particular experiment.
In the without-replacement Box A experiment, seeing red first changes the second-red probability from 3/5 to 1/2, so the red outcomes at the two positions are dependent.
Worked example 11: test independence numerically
Without replacement, P(first red)=3/5. By symmetry, P(second red)=3/5 as well, even though the conditional probability changes after the first draw.
If independent, the chance of two reds would be (3/5)²=9/25. Actual two-red probability is 3/10.
Since 3/10 differs from 9/25, the events are dependent. The calculation distinguishes unconditional probability at the second position from its conditional probability given first red.
Mutually exclusive does not mean independent
For one draw from a box, the events “red” and “blue” are mutually exclusive. They cannot both occur on the same draw.
If both have positive probabilities, their intersection probability is zero, while the product of their probabilities is positive. Therefore they are not independent.
This is another reason to use definitions rather than conversational synonyms. Disjoint and independent are different logical relationships.
Worked example 12: one draw, two colours
In Box A, P(red)=3/5 and P(blue)=2/5. The intersection of red and blue on a single one-colour draw has probability zero.
The product is (3/5)(2/5)=6/25, not zero. Thus the events are not independent.
Their mutual exclusivity follows from the stipulated one-counter, one-colour outcome. It does not imply statistical independence.
Conditional independence requires additional conditions
Even if two observations look independent after conditioning on one hidden cause, they may be associated when that cause is unknown.
Imagine each box has a fixed red probability and counters are drawn with replacement. Given the chosen box, the two draws are independent.
But when the box identity is hidden and shared across both draws, observing the first colour can update beliefs about which box was selected, changing the predicted distribution for the second colour.
Worked example 13: two reds with one hidden box
Choose A or B with equal probability, then draw twice with replacement from that same chosen box.
The probability of two reds is (1/2)(3/5)²+(1/2)(1/5)² = (1/2)(9/25+1/25)=5/25=1/5.
A single draw has red probability 2/5, so two reds would have probability (2/5)²=4/25 if the two unconditional red events were independent. Because 1/5≠4/25, they are associated through the hidden box.
A new box each time creates a different experiment
Suppose the box is independently reselected with equal probability before each draw. Then each draw has red chance 2/5.
Under that stated fresh-selection procedure, two red results have probability (2/5)²=4/25.
Compare with 1/5 when the hidden box is held fixed for both draws. The same box contents and individual draw process produce different joint probabilities because the first-stage selection mechanism differs.
A tree diagram should label conditions on branches
Do not write 3/5 on every red branch merely because it appeared in the first row of the story. The Box B branch uses 1/5.
Similarly, without-replacement second-stage probabilities depend on the first colour, because the composition changes.
The diagram’s branch labels must reflect the specific history leading to each point. That is the real purpose of a probability tree.
Worked example 14: three red without replacement
From Box A’s three red and two blue counters, draw three counters without replacement.
P(all three red)=(3/5)(2/4)(1/3)=6/60=1/10. The numerator of the red probability decreases after each successful red draw.
The event is ordered in execution but has only one colour path RRR. There is no need to multiply by 3! because rearranging three identical red outcomes does not create distinct colour sequences.
Why multiplying every probability is not always enough
A problem asking “at least one red” includes many possible colour orders. Rather than enumerate all red-containing paths, it may be easier to use the complement of “no red”.
If the drawing procedure and number of draws are clear, P(at least one red)=1−P(all blue).
The complement method works because the two events are exhaustive and mutually exclusive. It is not a generic instruction to subtract whichever probability is easiest from one.
Worked example 15: at least one red in two draws
From Box A without replacement, P(two blue)=(2/5)(1/4)=2/20=1/10.
Therefore P(at least one red)=1−1/10=9/10. An independent path enumeration would add RR, RB and BR and yield the same total.
This check shows the complement method’s efficiency while preserving the dependence of the second draw.
Probability after evidence is not proof of certainty
Even after a red observation, P(A|red)=3/4 leaves a one-quarter probability of B under the original fair box-selection model.
A student should not write “therefore the box was A.” The supported conclusion is “A is more likely than B under the given assumptions”.
Probability quantifies uncertainty; it does not usually identify one hidden event with certainty when multiple branches remain possible.
The denominator must match the new sample space
The most frequent conditional-probability mistake is retaining the original population total after a condition has restricted attention to a subset.
In the 100-label scanner example, the chance of a symbol among flagged labels is 18/26, not 18/100.
Both fractions are valid for different events: 18/100 is P(symbol and flagged), while 18/26 is P(symbol | flagged). Naming the denominator resolves the confusion.
Worked example 16: distinguish three scanner fractions
In the first fictitious scanner table, P(flagged)=26/100, P(symbol and flagged)=18/100, and P(symbol | flagged)=18/26.
These are approximately 26%,18% and 69.2%. Their difference follows from their event definitions.
A pupil who obtains one of these numbers in the wrong question has probably confused an intersection probability with a conditional probability.
Bayes’ rule is not a licence to invent prior probabilities
If a problem gives how often a scanner flags an item with a symbol but says nothing about how common symbols are, P(symbol | flag) generally cannot be determined uniquely.
Different underlying frequencies can give different posteriors, as our two fictional scanner populations demonstrated.
The correct mathematical response may be “insufficient information” or an expression in terms of the unknown prior. Guessing a convenient base rate is not a justified solution.
Worked example 17: same likelihood, unknown prior
Suppose P(flag|symbol)=0.9 and P(flag|no symbol)=0.1, but the symbol rate q is unknown.
Bayes’ rule gives P(symbol|flag)=0.9q/[0.9q+0.1(1−q)], provided the denominator is positive.
At q=0.2 the result is 0.18/(0.18+0.08)=9/13. At q=0.05 it is 0.045/(0.045+0.095)=9/28. The posterior depends on q.
The value of a two-way table
Tree diagrams organise sequential choices. A two-way table organises joint counts or probabilities of two categorical events.
For the scanner example, both are valid representations. The two-way table makes the restricted denominator for a flagged outcome easy to see.
Students should be able to move between a clearly labelled tree and table, but they should not alter totals when switching representations.
Worked example 18: total probability with three boxes
Imagine three fictional boxes chosen with probabilities 1/2,1/3 and 1/6, containing red probabilities 1/2,1/4 and 1/6 respectively.
The total red chance is (1/2)(1/2)+(1/3)(1/4)+(1/6)(1/6) =1/4+1/12+1/36 =9/36+3/36+1/36=13/36.
The three box choices are mutually exclusive and exhaustive. Each term represents one complete red path, and their sum gives total red probability.
Exact fractions often make conditional checks easier
Keeping probabilities as fractions avoids premature rounding. In the original box example, P(A∩R)=3/10 and P(R)=4/10.
Their ratio simplifies exactly to 3/4. Rounded decimals can be useful in the final answer, but repeated rounding mid-calculation can obscure simple relationships.
A tutor can ask for the exact fraction first, then a percentage to the requested precision.
One event can be both common and weak evidence
An observation can occur frequently in both boxes and therefore give only modest information about which was chosen.
The amount of updating depends on how different the conditional likelihoods are, as well as the prior odds. A red observation is more informative about A when P(red|A) greatly exceeds P(red|B).
This does not automatically make one outcome “good” or “bad”. Information is about how much the observation changes the probabilities under the model.
Worked example 19: an uninformative colour
Suppose two fictional boxes each contain two red and three blue counters and are chosen with equal probability.
The chance of red is 2/5 regardless of the box. Given red, P(A|red)=(1/2)(2/5)/(2/5)=1/2.
The observation does not change the prior probability because the colour is equally likely under both possibilities. A correct Bayes calculation can therefore leave the probability unchanged.
Primary foundations: coloured counters and simple fractions
Younger pupils can begin by counting favourable counters out of a known total and recording possibilities as fractions.
For one draw from a box of three red and two blue, ask why the red probability is 3/5 and the blue probability 2/5. Use replacement language only after the single-draw sample space is secure.
Formal Bayes’ rule is an advanced extension, not a prerequisite for learning elementary theoretical probability.
Lower Secondary: sample spaces and two-stage choices
Secondary learners can use tree diagrams, complementary events and ordered outcomes once they can multiply fractions and distinguish “and” from “or”.
Start with two independent draws with replacement. Then remove replacement and ask what changed in the second branch.
This structure turns a memorised formula into a consequence of the experiment’s stated choices.
Advanced extension: Bayes and evidence updating
Learners ready for algebra can use Bayes’ rule, unequal prior box choices and total probability to interpret an observation.
The SEAB 2027 SEC G3 Mathematics syllabus directory supplies official course context. Bayesian posterior reasoning here should be treated as enrichment unless the applicable syllabus explicitly includes the relevant technique.
The core mathematical goal is not terminology for its own sake. It is to identify which event has become known and how that changes the denominator.
Three errors to diagnose separately
One child may multiply events as if they are independent despite removing a counter. Another may add probabilities along one path rather than multiply. A third may reverse P(A|R) and P(R|A).
These are different conceptual mistakes. The correction should begin with the specific selection rule, tree structure or conditional sample space that was misunderstood.
A learner who can explain the first wrong step is better prepared for a changed problem than one who only copies the corrected calculation.
A twenty-minute home diagnostic
Minutes 1–5: State P(A)=P(B)=1/2 and the two boxes’ red probabilities 3/5 and 1/5. Draw the four-leaf tree.
Minutes 6–10: Calculate P(red)=2/5 and P(A|red)=3/4. Ask why the two denominators differ.
Minutes 11–15: Draw twice from Box A with and without replacement. Answers for two reds are 9/25 and 3/10.
Minutes 16–20: Use the invented 100-label scanner table to calculate P(symbol|flag)=18/26=9/13. Ask whether 90% was the answer to that same question; it was not.
A seven-day retrieval plan
Day one counts simple favourable outcomes. Day two labels a two-stage tree. Day three multiplies path probabilities and adds disjoint paths.
Day four changes replacement rules. Day five uses conditional fractions and a two-way table. Day six introduces unequal prior box selection. Day seven presents a fresh fictional signal with a new base rate and asks for an evidence-based conclusion.
Do not move to Bayes notation while fraction division and conditional denominators remain unreliable.
Practice questions and checked answers
A. In Box A, probability of red on one draw? Answer: 3/5.
B. With equal box selection, overall red probability? Answer: 2/5.
C. Given red under equal box selection, probability of Box A? Answer: 3/4.
D. Two reds from Box A with replacement? Answer: 9/25.
E. Two reds without replacement? Answer: 3/10.
F. Under the fictional scanner table, 18 of 26 flagged labels contain a symbol. Conditional symbol probability? Answer: 9/13.
Frequently asked questions
What does the vertical bar in P(A|B) mean?
It means “given B”. Only outcomes satisfying B belong to the conditional denominator.
Is P(A|B) always equal to P(B|A)?
No. They ask different questions and can have different values even when A and B are related.
When do probabilities multiply in a tree?
Multiply conditional branch probabilities along one full path to find the probability of that joint sequence.
When do probabilities add?
Add mutually exclusive paths that together make the event whose probability is requested.
Does replacement matter?
Yes. Returning a counter can keep the next-draw composition unchanged; without replacement, the next conditional probabilities generally change.
Is a 90% scanner flagging rate the same as a 90% probability of a symbol after a flag?
No. The posterior also depends on how common symbols were before the flag and how frequently plain items were flagged.
Does Bayesian updating establish certainty?
Not generally. It revises probabilities under a model and can leave several possible causes.
Are these probabilities from real Bukit Timah school activities?
No. All boxes, drawings, labels and scanner rates are fictional.
How can a parent see genuine understanding?
The student can name the conditioned event, justify the denominator, distinguish joint and conditional probabilities, and adapt correctly when the prior or replacement rule changes.
Sources and connected Mathematics in Bukit Timah
The MOE Primary Mathematics syllabus and SEAB 2027 SEC G3 Mathematics syllabus directory provide official level context; all numbers in the counter and scanner models are original illustrative assumptions.
Continue with Mathematics in Bukit Timah: probability, dependent events and fair games for introductory sample-space reasoning, Venn diagrams and overlapping groups for intersections and unions, and scatter plots, correlation and causation for a different warning against unsupported inference.
The immutable Clementi Secondary 1 Mathematics small-group tutorial reference anchors clear stage-by-stage diagrams, misconception diagnosis and independent checking.
The durable habit is to define how outcomes are selected, label every conditional branch, multiply within a full path, add disjoint paths and update only after identifying exactly which evidence has changed the sample space.
Continue with the companion Mathematics in Bukit Timah articles from this four-part batch: Direct and inverse proportion through team tasks · Differentiation, tangent gradients and rates · Prime factorisation, HCF and coprime integers. These original guides provide distinct lessons in proportional relationships, conditional probability, calculus and number theory, with worked checks and clearly stated assumptions.
More Mathematics in Bukit Timah from this four-article batch: Direct and inverse proportion through team-task rates · Differentiation, tangent gradients and rates · Prime factorisation, HCF and coprime numbers. Every guide includes original worked problems, clear assumptions, targeted misconceptions and links to suitable curriculum context.