Find your next learning step
Choose the question closest to what you need today, or read the guide from the beginning.
The probability that A and B both occur equals P(A) × P(B) when the events are independent. The word and alone does not establish that condition. If learning that A occurred changes the probability of B, the second factor must reflect that updated situation.
Start by identifying the two events and the chance model. Then ask what information the first event supplies about the second. That question explains which probabilities belong in a multiplication, rather than treating multiplication as a keyword response.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–25
Sections 26–30
Sections 31–32
An invented bag contains three red counters and two blue counters. Draw two without replacement. The probability of two reds is 3/5 × 2/4 = 3/10. After a red is removed, two red counters remain among four counters.
Using 3/5 × 3/5 would give 9/25, a different answer. The second original red probability cannot simply be reused along the first-red path. The first result changed the composition available for the second draw.
Multiplication along a path is still valid. What changes is the second factor’s meaning: it is the probability of the second event given the first event. For independent events, that conditional probability equals the original probability, which produces the familiar shorter formula.
The SEAB 2027 G3 Mathematics syllabus includes simple combined events, possibility or tree diagrams where appropriate, and probability operations involving independent and mutually exclusive events. Understanding the conditions behind a calculation supports that work.
This guide uses small equally likely outcome models and clearly stated selection procedures. Formal conditional notation is introduced only to explain the meaning of a path factor. A learner can understand the school-level examples by counting remaining outcomes without needing advanced probability methods.
The broader eduKate explanation of statistical independence develops the larger mathematical concept. Here the focused task is deciding whether the two original event probabilities may be multiplied unchanged.
All bags, cards and devices below are invented teaching models. A claim of fairness or independence belongs to the stated model. It should not be inferred merely from the fact that a physical object or a random-looking process is mentioned.
Let A mean the first counter is red and B mean the second counter is red. The combined event A and B means both specified conditions occur in the same trial. In the two-draw model, that is the ordered path red then red.
The probability of A describes the first event by itself. The probability of B describes the second event by itself before conditioning on the first outcome. The probability of B after learning A may be different.
Those distinctions are easy to hide if every number is labelled only red probability. Write first red, second red given first red, or second red overall as appropriate. The label tells you which outcome set supplies the denominator.
When a question involves a single draw with two properties, the same care applies. A may mean the selected number is even and B may mean it exceeds three. Both events concern the same selected outcome, so their overlap must be counted from that model.
SECTION 4 OF 32
3. Independence means the probability stays unchanged after the information
For events of positive probability, independence means learning one occurred does not change the probability of the other. If P(A) is positive, P(B given A) equals P(B) under independence.
This is a mathematical relationship, not simply a description such as separate, unrelated or random. A mechanism may support an independence assumption, or a complete outcome table may establish it. The assumption needs a basis in the problem.
For example, a fair coin result and a separate spinner result may be modelled as independent when the question states or justifies that arrangement. Then a coin head does not change the stated spinner probabilities.
By contrast, two drawings from one bag without replacement generally affect the available composition. The first outcome supplies information about the second. The chance process is random, but randomness alone does not make its events independent.
When P(A) > 0, the general multiplication relationship is P(A and B) = P(A) × P(B given A). The second factor is calculated within the outcomes where A has occurred.
If A and B are independent, P(B given A) equals P(B). Substituting gives P(A and B) = P(A) × P(B). Independence is the condition that allows the conditional factor to be replaced by the unchanged original probability.
For the bag example, P(A) = 3/5 and P(B given A) = 2/4. The product is three tenths. Using another second factor changes the event model being calculated.
If P(A) = 0, the conditional quotient given A is not defined in this elementary formulation. The combined event still has probability zero. Use the factorisation definition carefully in such boundary cases rather than dividing by zero to obtain a conditional probability.
Suppose a fair coin has head probability one half, and a spinner has five equally likely sectors, two red and three blue. The model states that the coin and spinner outcomes are independent.
The probability of a head and red is 1/2 × 2/5 = 1/5. Learning that the coin landed heads does not change the spinner’s red probability under the stated model.
A possibility diagram would contain ten equally likely ordered combinations: two coin outcomes multiplied by five labelled spinner sectors. Two combinations have a head and a red sector, giving two tenths, or one fifth.
The diagram and multiplication agree because they describe the same joint model. The product is not justified merely by seeing two denominators. It is justified by the outcome structure and the stated independence.
Return to three red and two blue counters without replacement. The first red probability is three fifths. After that event, the bag contains two reds and two blues, so the next red probability is one half.
The red-red path therefore has probability 3/5 × 1/2 = 3/10. The original probability of a red on a draw from the full bag was three fifths, so learning the first red has changed the second-stage probability.
This is exactly the condition that blocks multiplication of the unchanged original probabilities. The operation along the path remains multiplication, but the factors must correspond to the states actually reached.
For a focused companion, read why a second draw needs a new denominator without replacement. This guide adds the wider independence question: when is an unchanged second probability justified at all?
| Quantity in the three-red, two-blue model | Value | Meaning |
|---|---|---|
| First draw red | 3/5 | Before drawing |
| Second draw red given first red | 2/4 | Two reds among four remaining |
| Second draw red given first blue | 3/4 | Three reds among four remaining |
| Both draws red | 3/10 | Product along the red-red path |
| Second draw red overall | 3/5 | Combine all first-draw paths |
The final row is important. The second draw’s overall red probability can equal the first draw’s red probability even though the two events are dependent. Equal individual probabilities do not establish independence.
Independence concerns what happens after information is supplied, or equivalently whether the joint probability equals the product of the individual probabilities. Do not compare only the first and second marginal probabilities and stop there.
The second counter can be red after a first red or after a first blue. The first route has probability 3/5 × 2/4 = 3/10. The second has probability 2/5 × 3/4 = 3/10.
Those first-draw routes are mutually exclusive and exhaust the possibilities leading to a second red. Add them to obtain 3/10 + 3/10 = 3/5. That is the overall second-red probability before learning the first colour.
Yet after a first red, the probability is one half; after a first blue, it is three quarters. The overall three fifths averages over those different branches using their probabilities.
This example separates two ideas that are often confused. A marginal probability describes an event without a specified conditioning result. A branch probability describes the event within a stated earlier outcome. They can be different even when the stage labels sound similar.
In the same bag model, the red-red path is three tenths. Red-blue is 3/5 × 2/4 = 3/10. Blue-red is 2/5 × 3/4 = 3/10. Blue-blue is 2/5 × 1/4 = 1/10.
The four path probabilities add to one. Each represents a different ordered colour sequence, and together they exhaust the possible colour outcomes of two draws.
If your complete path list totals more than one or less than one, check for omitted paths, overlapping events or branch probabilities taken from the wrong bag state. The sum test can expose a structural mistake before the final requested event is assembled.
A total of one alone does not prove every branch is correct. Some errors can compensate. Use it alongside direct checks of the remaining counts at each branch, so both local probabilities and the full outcome structure are inspected.
Suppose a model chooses one fair coin result, then displays that same result in two windows. Let A mean the first window shows heads and B mean the second shows heads. Each event has probability one half.
But the probability both show heads is one half, not one quarter. Knowing A tells you B occurs with probability one. The windows have equal marginal probabilities and complete dependence under this copying mechanism.
This is a simple counterexample to the claim that equal probabilities imply independence. The relevant comparison is the joint event or the probability after learning the other event, not numerical equality of the marginals.
It also shows why two separate-looking displays are not enough evidence. A shared mechanism can connect their outcomes. Read how results are generated rather than inferring independence from the number of objects pictured.
Choose one integer uniformly from one through twelve. Let A mean the number is even and B mean it is divisible by three. There are six even numbers and four multiples of three, so P(A) = 1/2 and P(B) = 1/3.
The numbers satisfying both are six and twelve. Thus P(A and B) = 2/12 = 1/6, which equals 1/2 × 1/3. The events are independent in this particular finite model.
This prevents the opposite shortcut: events must come from separate trials to be independent. Independence is a property of the event probabilities in the joint model, and one draw can contain independent properties.
Check the specified range. If the same event descriptions are used on a different set of numbers, the counts may change and independence may fail. The words even and divisible by three do not carry a universal independence guarantee detached from the sample space.
Choose one integer uniformly from one through six. Let A mean even and B mean greater than three. Each event contains three outcomes, so both probabilities are one half.
Their overlap is {4, 6}, containing two of the six outcomes. Therefore P(A and B) = 1/3, while P(A)P(B) = 1/4. The events are not independent in this model.
After learning the number is even, the remaining possible values are two, four and six. Two of those three exceed three, so P(B given A) = 2/3. The path rule gives 1/2 × 2/3 = 1/3, matching the direct count.
The word and still identifies an intersection. It does not decide whether to multiply unchanged marginal probabilities. The outcome list supplies the relationship needed for that decision.
Mutually exclusive events cannot occur together. Independent events leave each other’s probabilities unchanged. For events with positive probabilities, mutual exclusivity means learning one occurred makes the other impossible, which changes its probability to zero.
For one fair six-outcome draw, odd and even are mutually exclusive. Each has probability one half, but their joint probability is zero rather than one quarter. They are not independent.
The positive-probability condition matters. A probability-zero event can satisfy the formal product definition of independence with another event. For introductory examples, use positive-probability events so the distinction is visible without a boundary-case distraction.
The existing guide to mutually exclusive and independent events develops this comparison. Keep the two questions separate: can both happen, and does learning one change the other’s probability?
Question: twenty equally likely cards have two possible symbols, a star and a triangle. Six have both symbols, four have only a star, two have only a triangle and eight have neither. Are star and triangle independent events on one draw?
There are ten star cards, so P(star) = 10/20 = 1/2. There are eight triangle cards, so P(triangle) = 8/20 = 2/5. Their product is 1/5.
The actual joint probability is six twentieths, or three tenths. Since 3/10 differs from 1/5, the events are not independent. Given a star, six of the ten possible star cards also have a triangle, so the triangle probability becomes three fifths.
Both checks agree: the joint does not factor into unchanged marginals, and learning star changes the triangle probability from two fifths to three fifths. The conclusion comes from the complete card model, not an assumption about whether the two symbols look related.
Use exact fractions where possible. For each multiplication, label whether the second factor is an unchanged probability or a probability after a stated first result. Assume the fairness and selection rules explicitly given.
Question A. An independent fair coin and five-sector spinner have two red sectors. Find head and red. Question B. From three red and two blue counters without replacement, find red then red. Question C. Find blue then red in the same bag model.
Question D. Find the second-red probability overall in that model. Question E. One fair coin result is copied into two windows. Find the probability both show heads and decide whether the window events are independent.
Question F. Select uniformly from one through twelve. Check independence of even and divisible by three. Question G. Select uniformly from one through six. Check independence of even and greater than three.
Question H. For one fair six-outcome draw, compare odd and even as mutually exclusive and independent events. Question I. In the twenty-card worked example, find P(triangle given star). Question J. Explain why equal first- and second-red probabilities overall do not prove the two draws are independent.
Answer A. The probability is 1/2 × 2/5 = 1/5, using the stated independence. Answer B. It is 3/5 × 2/4 = 3/10. Answer C. It is 2/5 × 3/4 = 3/10.
Answer D. Add the red-red and blue-red paths: three tenths plus three tenths equals three fifths. Answer E. Both windows show heads with probability one half. They are dependent because learning the first result fixes the second, rather than leaving its probability at one half.
Answer F. The marginals are one half and one third; the overlap {6, 12} has probability one sixth. The product matches, so the events are independent in that range. Answer G. The overlap {4, 6} has probability one third, different from one quarter, so the events are dependent.
Answer H. Odd and even are mutually exclusive and each has positive probability one half. Their joint is zero, so they are not independent. Answer I. Six of the ten star cards have a triangle, giving three fifths.
Answer J. Independence concerns the probability after learning the first outcome, or the joint factorisation. In the bag model, the second-red probability is one half after red and three quarters after blue, despite an overall marginal of three fifths. Equal marginals do not remove that dependence.
For the twenty-card model, arrange the counts by star status and triangle status. The four cells are both six, star only four, triangle only two and neither eight. The row totals are ten and ten; the triangle column totals eight.
Within the star row, six out of ten have a triangle. Within the no-star row, two out of ten have a triangle. Those conditional proportions differ, and neither must be replaced by the overall triangle proportion eight out of twenty.
The table makes the information change visible without a story about time. Learning star restricts attention to one row. The denominator changes because the relevant possible cards have changed.
This is a useful representation when a student associates conditional probability only with drawing objects out of a bag. Dependence can concern two properties of one selected outcome as well as two stages of a sequential process.
Build another twenty-card model with four cards having both symbols, six star only, four triangle only and six neither. There are still ten star cards and eight triangle cards, giving the same marginals one half and two fifths.
Now the joint probability is four twentieths, or one fifth, matching the product. Within either star-status row, the triangle proportion is two fifths. The events are independent in this second model.
Comparing the two models shows why marginal probabilities alone are insufficient. Both have the same individual star and triangle frequencies, but the arrangement of overlap differs. Independence concerns that joint arrangement.
If a question gives only P(A) = 1/2 and P(B) = 2/5 without a relationship, do not infer a unique joint probability. Additional information such as independence, overlap counts or a conditional probability is needed.
If a drawn counter is replaced and the next draw is modelled as a fresh uniform draw from the restored bag, the second red probability remains three fifths regardless of the first colour. Under that procedure, the two colour events are independent.
The red-red probability is then 3/5 × 3/5 = 9/25. That is the value that was inappropriate for the earlier without-replacement model. The arithmetic is valid for a different procedure.
Include the full procedure rather than treating the word replacement as magical. The counter must be returned and the next selection must follow the stated random mechanism. A model that deliberately chooses the same counter again would have another relationship.
This reinforces a useful review question: is the answer wrong arithmetic, or correct arithmetic for the wrong model? The distinction helps identify whether the repair concerns multiplication, counting or reading the selection rule.
SECTION 21 OF 32
20. Complements inherit independence when the original events are independent
If A and B are independent, A and not B are independent too. The joint probability is P(A) − P(A and B) = P(A)[1 − P(B)], which is P(A)P(not B).
In the independent coin-and-spinner model, head and blue therefore has probability 1/2 × 3/5 = 3/10. Together with head and red, the total is one half, as expected for all head outcomes.
This result is useful for completing a simple table or checking a missing branch. It follows from the probability relationships rather than from a new unsupported assumption for every complement.
Do not use it backwards without the required information. Knowing one particular numerical joint result may or may not establish the full intended event relationship unless the marginals and conditions have been checked. State which events the model says are independent.
Under a model of independent fair coin tosses, a run of heads does not change the next toss’s head probability from one half. The model’s independence concerns conditional probabilities after any specified previous results.
That does not mean a long specified sequence is as likely as one single result. A sequence of several heads has a probability obtained by multiplying several one-half factors. The complete sequence and the next toss are different events.
Keep the requested event visible. Before the sequence begins, the probability of a particular long run may be small. After a particular prefix has already occurred, the next-step probability is still one half under the independent fair model.
This example connects the arithmetic to interpretation. Surprise about a past sequence does not by itself establish a changed chance for the next independent trial. A different physical mechanism would require a different stated model and evidence.
Because A and B together is contained within A and within B, its probability cannot exceed either individual probability. For marginals one half and two fifths, the joint can be at most two fifths.
The dependent card joint three tenths fits those limits, as does the independent joint one fifth. A proposed joint of three fifths would exceed the triangle marginal two fifths and therefore cannot describe that model.
This is a useful quick check, but it does not establish independence. Many different joint probabilities can lie within valid bounds. Passing a size check only shows that this particular obvious contradiction has not occurred.
Use the bound alongside an outcome count, tree or factorisation test. A probability answer should fit the event containment and the actual joint structure. Neither condition alone replaces the full reasoning when independence is the question.
The star and triangle events in the dependent card model can occur together, because six cards have both. That shows they are not mutually exclusive. It does not show they are independent.
Independence requires the precise joint probability to equal the product of the marginals. The fact that the overlap is nonempty says only that the joint is positive in this equally likely finite model.
Likewise, separate-looking event names do not establish independence. Number even and number greater than three sound like different properties, yet their probabilities are dependent on the one-through-six sample space.
When a learner says both can happen, ask which question that answers. It addresses exclusivity. To address independence, compare conditional and original probability, or compare the actual joint with the product.
Suppose a student says and means multiply and writes three fifths times three fifths for two red draws without replacement. The useful correction keeps the idea of multiplying along the path while repairing the second factor.
After a first red, count the remaining bag: two reds, two blues, four total. The second-red branch is two fourths. The path now represents the situation actually reached rather than repeating the initial state.
If the student adds the two factors instead, the difficulty is different: they may be combining events as alternatives rather than following both stages of one path. Draw or list the ordered outcomes to clarify the requested event.
Precise diagnosis avoids replacing one blanket rule with another. Multiplication and addition both appear in probability, but their use depends on event relationships, path structure and whether alternatives overlap.
Begin with the same three-red, two-blue bag under two procedures: replacement with a fresh uniform draw, and no replacement. Keep the requested event red then red fixed. Explain why the second factors differ.
Next keep the marginals fixed using the two twenty-card symbol models. Compare their different overlaps and check factorisation. This shows that knowing the individual probabilities does not automatically determine their joint.
Then use one single-draw example with independent properties and another with dependent properties. This prevents a procedural association such as different trials always independent from taking the place of the mathematical definition.
Finally ask for a sentence naming the evidence: the mechanism restores the same distribution, the table’s conditional proportions match, or the joint equals the product. The learner should be able to justify the condition as well as calculate the result.
Define A and B precisely and identify the sample space or selection procedure. Ask whether the first event changes the probability of the second. For a path, label the second factor with the state after the first result.
Use unchanged marginal probabilities only when independence is given or established. Otherwise count the overlap directly or use an appropriate conditional probability. Keep mutually exclusive alternatives separate from events that must both occur.
Check fractions against the actual remaining outcomes, complete path totals where relevant, and the fact that an intersection cannot exceed either marginal. Explain what condition makes the chosen multiplication valid.
The mathematics learning hub connects probability models, trees, replacement, overlap and answer checking. The aim is to choose probabilities that describe the event, so a familiar operation does not quietly calculate a different question.
Label the three red counters R1, R2 and R3, and the two blue counters B1 and B2. For two uniform draws without replacement, there are five choices for the first counter and four for the second, giving twenty equally likely ordered individual outcomes.
Six outcomes are red then red: three choices of the first red multiplied by two remaining red choices. Thus the joint probability is six twentieths, or three tenths, matching the path calculation.
Red then blue has three times two, or six outcomes. Blue then red has two times three, also six. Blue then blue has two times one, or two. The counts total twenty and produce the earlier four path probabilities.
This direct enumeration checks the bag model without simply repeating the same branch fractions. It also explains why the four colour sequences are not equally likely: they summarise different numbers of equally likely individual outcomes.
A student who assigns one quarter to each colour sequence has changed the probability model. List or count the labelled individual selections first, then combine them into colour events. Equal-looking event names do not guarantee equally likely outcomes.
In the first twenty-card symbol model, P(triangle given star) is six out of ten, or three fifths. Reversing the condition gives P(star given triangle) equal to six out of eight, or three quarters.
The numerator counts the same six overlapping cards, but the reference group changes from star cards to triangle cards. The conditional probabilities therefore differ. Reversing a condition does not merely reverse the wording of an unchanged fraction.
Dependence itself is symmetric: if the joint fails to equal the product of the marginals, it fails in either multiplication order. That symmetry does not require the two directional conditional probabilities to have equal numerical values.
Even for independent events, the two conditional values may differ because they equal different marginals. With P(A) = one half and P(B) = two fifths, independence gives P(B given A) = two fifths and P(A given B) = one half.
This distinction is helpful when a learner correctly identifies independence but expects every probability in the explanation to match. Compare each conditional probability with its corresponding unconditioned event probability, rather than comparing the two conditional numbers indiscriminately.
Suppose twenty equally likely cards include ten with a star and eight with a triangle. Let k be the number with both. The remaining cell counts are ten minus k for star only, eight minus k for triangle only and two plus k for neither.
Every whole k from zero through eight gives nonnegative counts with the same marginals. Therefore the joint probability can take several values k/20 while the individual probabilities remain one half and two fifths.
Independence selects the particular joint one fifth, so k must equal four. The earlier independent model uses that overlap. The dependent model uses k = six, yielding three tenths instead.
This is a constructive explanation of insufficient information. The marginals alone do not identify one overlap because multiple complete valid tables satisfy them. Showing two such tables is enough to demonstrate that an extra relationship is needed.
The exercise stays within a finite counting model. There is no need to guess a hidden mechanism or introduce advanced formulas to see why multiplying unchanged probabilities is an additional structural claim rather than a consequence of knowing the two marginals.
SECTION 31 OF 32
30. Exact model counts and observed sample frequencies have different roles
In the invented twenty-card model, every card and its symbols are specified, and a uniform draw is part of the model. The cell counts therefore determine exact probabilities for that stated experiment.
If instead a table summarises twenty observations from a larger unknown process, its proportions are observed frequencies. Matching the product in that sample does not by itself prove exact independence throughout the larger process.
Likewise, a small sample difference does not automatically identify a particular causal mechanism. A real statistical investigation needs an appropriate design and analysis. For the school counting questions here, use the complete finite model actually supplied.
This boundary helps keep the calculation’s claim honest. A probability table can be an exact specification of an experiment or a sample record used to learn about one. The same arithmetic may appear, but the interpretation of the result differs.
When presenting an independence conclusion, say whether it follows exactly from a stated finite model or is an assumption in the question. That is enough for these examples and prevents a correct classroom calculation from claiming more evidence than its data provide.
A useful final explanation includes three pieces of information: the event being calculated, the probability assigned at each stage, and the reason the second factor is appropriate. For two reds without replacement, name the three initial reds among five and the two remaining reds among four before multiplying.
For the independent coin-and-spinner model, state that the spinner distribution remains unchanged after the coin result. The same multiplication sign appears, but its justification refers to a different relationship between the events. Writing that reason prevents a correct numerical answer from hiding an unsupported assumption.
If a problem supplies a table, use its actual overlap before deciding whether independence holds. If it supplies only marginals, look for an explicit relationship or recognise that the joint may remain undetermined. The ability to identify missing information is part of probability reasoning, not a failure to choose a formula quickly.
The questions below summarise those choices using the same clearly labelled models.
These answers distinguish a general path calculation from the shorter independent-events product. The event definitions and chance model decide which factors belong in the working.
Does the word and always mean multiply P(A) by P(B)?
No. The unchanged product applies when A and B are independent. In a general path with P(A) positive, use P(A) times the probability of B given A. Direct overlap counting may be simpler in a finite sample space.
Can I multiply along a dependent path?
Yes, using the correct conditional factor for that path. In the without-replacement example, three fifths times two fourths gives the two-red probability. Dependence changes the second probability, not the validity of the properly labelled path rule.
Do equal event probabilities prove independence?
No. Two copied displays can each show heads with probability one half while always matching. The joint is one half rather than one quarter. Independence concerns the information relationship or joint factorisation, not equality of the marginal numbers.
What should I remember first?
Identify the events and ask what changes after information is supplied. Justify independence before multiplying unchanged probabilities. If the situation changes, use the probability belonging to the actual branch and check the result against the complete model.
Continue the G1/G2/G3 Mathematics series
Browse every guide in the G1/G2/G3 Mathematics directory
What Is G1 Mathematics | Why a Lower Unit Price Can Still Mean a Higher Total Cost
What Is G2 Mathematics | Why You Can Split a Numerator Sum but Not a Denominator Sum
What Is G1/G2/G3 Mathematics | Why 0.999… Equals One but 0.999 Is Smaller