It tells us that the selected item is not returned before the next selection. The next draw therefore takes place from a different collection. Start by recording that collection rather than changing a fraction mechanically.

Consider a bag containing three red counters and five blue counters. There are eight counters initially. Assume each individual counter is equally likely to be selected on each draw from the counters then available.

If the first counter is red and stays outside the bag, seven counters remain: two red and five blue. If the first counter is blue, the remaining seven are three red and four blue.

Parents can ask, “What is in the bag now?” That question supplies both the numerator and denominator for the next probability. It is more informative than simply reminding the child to subtract one.

The denominator describes the total number of equally likely individual counters available at that stage. The numerator describes how many of those counters satisfy the event being considered.

Without replacement does not mean that every numerator decreases. It depends on which item was removed and which colour the next branch asks for. Removing a red reduces the red count; it leaves the blue count unchanged.

A tutor can use a small sketch of the remaining counters beside each first-stage branch. The student then writes the next fractions from the sketch, explaining the counts before multiplying anything.

Read the complete selection rule. Some questions replace the item, add items or use a different collection for a later draw. The changing total must follow the actual process supplied.

For the simple bag used here, exactly one counter is removed after the first draw. That is why the second-stage total is seven. The mathematical habit is to reconstruct the current collection, so the same reasoning remains useful when a question changes its numbers or rules.

After a first red has been drawn without replacement, the bag contains two red and five blue counters. The next red probability is therefore 2/7, and the next blue probability is 5/7.

The first red probability was 3/8 because the original collection contained three red among eight counters. That first fraction describes a different stage from the second fractions.

Ask your child to write a small state note: “after red: 2 red, 5 blue, total 7”. From that note, both second branches can be read directly.

A child may correctly change the denominator to seven while leaving the red numerator at three. Their second-red fraction 3/7 counts the red counter that has already left the bag.

Another child may write 2/8. They have removed the red from the favourable count while keeping it inside the total. The fraction mixes two different collections.

Parents can point to the numerator and denominator separately. “Which counters are these two?” and “Which counters make these seven?” help the child see that both counts must refer to the same remaining bag.

The two second-stage possibilities are red and blue. Their probabilities add to 2/7 + 5/7 = 1. This checks that all remaining counters have been accounted for under the two-colour model.

A tutor can ask for the blue branch even when the question ultimately asks about two reds. The student should understand the full current collection, not merely adjust whichever fraction appears in the final calculation.

The first-red branch provides information about what happened. The next fraction is conditional on that information. It answers “red next, after a red first”, rather than “red on an unspecified draw”.

Keeping that description beside 2/7 gives the fraction a clear meaning. The next step is then to combine it with the first-red probability when calculating the chance of the complete two-red path.

A first blue removes one blue counter and leaves the three red counters in place. The second-stage bag contains three red and four blue, with seven counters altogether.

The next red probability is 3/7. The next blue probability is 4/7. Both fractions belong to the branch where blue happened first.

Compare this with the first-red branch, where the next probabilities were 2/7 for red and 5/7 for blue. The total is seven on both branches, but the colour counts differ.

Parents can ask the child to describe the two bags side by side. After red, the bag has fewer reds. After blue, it has fewer blues. This makes the branching structure meaningful.

A common error is copying the same pair of second-stage fractions onto both first-stage outcomes. That treats the different histories as if they left identical collections.

A tutor can draw two short remaining-bag notes before the tree: one for red first and one for blue first. The student should then label each outgoing branch from its matching note.

For example, “blue then red” uses 5/8 for the first blue and 3/7 for the later red. “red then blue” uses 3/8 followed by 5/7. The fractions differ in their stage positions, although the complete path products happen to be equal here.

Do not use that equality as permission to copy branch probabilities. The matching products come from the specific counting process; each individual fraction still has its own conditional meaning.

Checking the second-stage sum after blue gives 3/7 + 4/7 = 1. If a copied pair includes three reds and five blues over seven, its sum is eight-sevenths and reveals an impossible total.

The useful habit is to let the history determine the current bag. Every next probability should be read from the collection that exists on that branch.

Each branch represents a possible next outcome under the history that leads to its starting point. A probability tree is therefore a record of changing conditions, not only a pattern of lines to fill.

For the three-red, five-blue bag, the first branches are red with probability 3/8 and blue with probability 5/8. Those branches share the original collection.

After the first red, write second branches red 2/7 and blue 5/7. After the first blue, write second branches red 3/7 and blue 4/7.

Parents can ask the child to read a branch label as a full sentence. “Two-sevenths” should become “the probability of red next after one red was removed”. The sentence makes the conditioning visible.

The second fractions are not the probabilities of the whole two-draw outcomes. The full red-red outcome requires both the first red and the second red, so its probability uses both branches.

A tutor can place the stage number above the tree and a remaining-count note near each split. This helps the student connect the diagram with time and the changing sample collection.

At each split, the outgoing probabilities should add to one when the listed branches cover all possible next outcomes and do not overlap. Here red and blue cover the two-colour bag.

Across the complete two-draw tree, the four paths are red-red, red-blue, blue-red and blue-blue. Each ordered colour sequence has one place in the tree.

The diagram is useful because it keeps histories separate. It should not encourage a student to memorise “multiply along, add across” without deciding what each path describes.

Begin with the bag contents, then label the branches, then identify the event the question asks for. That sequence gives the tree its mathematical purpose and makes later calculations easier to check.

The event “two reds” requires a red on the first draw and a red on the second draw. The second probability must be evaluated under the condition that the first red has already been removed.

For the example bag, the first-red probability is 3/8. Given that first red, the second-red probability is 2/7. The complete path probability is 3/8 × 2/7 = 6/56 = 3/28.

Multiplication here does not mean that the two colour events are independent. The second probability explicitly depends on the first outcome. The product uses that conditional probability.

Parents can ask, “Which second fraction belongs to the first-red branch?” If the child selects 3/8 again, they are treating the next draw as if the original collection were still available.

Using 3/8 × 3/8 gives 9/64, which describes the corresponding two-red calculation under replacement with restored counts and the same uniform selection rule. It models a different experiment.

A tutor can compare the two processes with physical counters or written counts. Without replacement, the first red is unavailable for the second draw. With replacement, it returns before the next selection.

The probability 3/28 can be left as an exact fraction when that is the requested form. Converting to a decimal is optional unless the task asks for it.

Check the event as well as the arithmetic. A single first-red probability of 3/8 is not enough to represent two reds. A second conditional probability of 2/7 is also not the full answer.

The two fractions work together because they describe successive requirements along one path. Reading them as “first red, then red given first red” keeps that relationship clear and avoids the mistaken rule that multiplication is allowed only when independence has been established.

For “red then blue”, follow the first-red branch and then the second-blue branch. The required order is part of the event.

The first-red probability is 3/8. After that red is removed, all five blue counters remain among seven counters. The second-blue probability is 5/7.

The path probability is 3/8 × 5/7 = 15/56. The numerator for blue does not decrease because a red, rather than a blue, left the bag.

Parents can ask the child to explain why the second numerator is five. This tests whether they are updating the colour counts, rather than subtracting one from every visible number.

A tutor can ask the paired question “blue then red”. Its probability is 5/8 × 3/7 = 15/56. The same final product does not make the two ordered outcomes identical; they are separate paths.

If the question asks specifically for red followed by blue, include only the red-blue path. Adding blue-red would answer the broader event “one of each colour in either order”.

The phrase “one red and one blue” can require careful reading. If no order is specified in a two-draw problem, both orders may be included. If the wording says first red and second blue, the order is fixed.

Do not decide from the word “and” alone. In both descriptions, two requirements must be satisfied. The difference is whether the event permits more than one sequence.

A helpful written label is “RB: red first, blue second”. That keeps the path meaning beside the product and reduces the chance of combining an unwanted route later.

The arithmetic is short, but the selection of the path is the important reasoning step. First identify the full event, then use the branch fractions that match its sequence and the bag remaining at each stage.

In two draws, one red and one blue can occur as red then blue or blue then red. If the question allows either order, both paths belong to the requested event.

The red-blue probability is 3/8 × 5/7 = 15/56. The blue-red probability is 5/8 × 3/7 = 15/56. Adding them gives 30/56 = 15/28.

These two paths cannot both occur in the same two-draw trial. The first draw cannot be both red and blue. They are disjoint alternatives, so their probabilities can be added without an overlap correction.

Parents can ask the child to list the allowed sequences before calculating. Writing RB and BR makes it harder to overlook one route or add a route that the question does not permit.

A tutor can contrast three prompts using the same bag: “red then blue”, “one of each colour” and “at least one red”. Each prompt uses the same tree but selects a different collection of paths.

Multiplying 3/8 and 5/8 would ignore the changing bag. Multiplying only one valid path would omit the other allowed order. Both errors can lead to plausible-looking fractions.

The completed one-of-each probability 15/28 is greater than either single path probability 15/56 because it includes both distinct ways of meeting the condition.

This is a useful reasonableness check. Adding a second allowed route should not leave the result smaller than the first route alone.

Do not multiply by two automatically in every problem. First verify that exactly two disjoint paths are allowed and, if using a doubled product, that their probabilities are equal. More stages or different selection rules may require a different calculation.

The reliable method is to list the permitted ordered routes, calculate each from its own remaining counts and add the disjoint route probabilities. The tree then serves the wording instead of replacing it.

The event “same colour” in two draws includes red-red and blue-blue. It excludes the two mixed-colour paths.

The two-red probability is 3/8 × 2/7 = 3/28. For two blues, the first-blue probability is 5/8 and the second-blue probability after a blue is removed is 4/7. Their product is 20/56 = 5/14.

Add the disjoint same-colour paths: 3/28 + 5/14 = 3/28 + 10/28 = 13/28.

Parents can ask, “Which complete colour sequences count as same colour?” That question separates choosing the event from calculating its probability.

A child may write only the two-red route because they focused on the red counters mentioned first. Another may multiply the first red and first blue probabilities, which describes neither same-colour route under this process.

A tutor can colour-code the four completed paths and mark the two selected ones. The student should explain the selection before using any fraction.

The complementary event is different colours, which is the one-of-each event from the previous chapter. Its probability is 15/28. The two results add to 13/28 + 15/28 = 1.

That check works because every two-draw outcome is either same colour or different colours, and no outcome can be both. The two events form a complete nonoverlapping split.

Keep the changing numerator on the same-colour branches. A red first leaves only two red available; a blue first leaves only four blue available. The decrease in the matching-colour count is essential.

The final answer 13/28 describes all same-colour outcomes. It does not describe one named colour alone. Clear event labels help the child preserve that distinction when a word problem uses a short phrase rather than spelling out each route.

“At least one red” in two draws includes red-red, red-blue and blue-red. Its only excluded colour sequence is blue-blue.

The blue-blue probability is 5/8 × 4/7 = 5/14. Therefore the probability of at least one red is 1 − 5/14 = 9/14.

The direct method gives the same result. Add 3/28 for red-red, 15/56 for red-blue and 15/56 for blue-red. Using denominator fifty-six, the sum is 6/56 + 15/56 + 15/56 = 36/56 = 9/14.

Parents can ask the child to name the excluded event in words before subtracting. “No red in either draw” is the complement of “at least one red”. It is not merely “blue on the first draw”.

Subtracting the first-blue probability 5/8 would ignore what happens on the second draw. A blue first can still be followed by a red, so that first-stage event is not the required complement.

A tutor can compare “exactly one red”, “at least one red” and “two reds”. Exactly one includes the two mixed-colour routes; at least one also includes two reds.

This helps the child see why the result for at least one red must be at least as large as the result for exactly one red. Here 9/14 exceeds 15/28 because it includes an additional allowed path.

The complement method is useful when the excluded event has fewer routes to calculate. It is not a rule to apply without identifying what is left out.

Write “1 − P(no red)” before substituting the numbers. That line makes the chosen complement visible and gives the tutor or parent a precise place to inspect if the final probability is wrong.

The changing denominator still matters inside the excluded path. Taking a complement does not restore the original bag; blue-blue remains a without-replacement sequence.

With one counter removed after each draw and none returned, the successive totals are eight, seven and six. The numerators depend on the colours already removed.

For three reds, the path is 3/8 × 2/7 × 1/6. After the first red, two reds remain; after the second, only one remains. The product is 6/336 = 1/56.

A child who writes 3/8 × 2/7 × 1/7 has updated the red count but not the total after the second removal. The third collection has six counters, not seven.

Parents can ask the child to list the number of counters outside the bag at each stage. Before the first draw, none have left. Before the second, one has left. Before the third, two have left.

A tutor can then ask for a mixed route, such as red-blue-red. Its factors are 3/8, 5/7 and 2/6. The third red numerator is two because only one red has been removed so far.

The two different routes use the same third-stage denominator but different numerators. The total depends on how many counters were removed; the colour count depends on which colours were removed.

After red-blue, the remaining bag contains two red and four blue. After red-red, it contains one red and five blue. Both bags contain six counters.

Keep the stage counts visible when first extending a tree. A longer diagram should not become a reason to copy fractions without reconstructing the collection.

For the simple fixed bag here, one can say “total minus the number already drawn”. More complicated questions may insert replacement or additions, so the total must still follow their actual rules.

The next useful check is that outgoing branches at the third split add to one. After red-blue, 2/6 + 4/6 = 1; after red-red, 1/6 + 5/6 = 1.

Exactly one red in three draws means the other two counters are blue. The allowed colour sequences are red-blue-blue, blue-red-blue and blue-blue-red.

For red-blue-blue, the probability is 3/8 × 5/7 × 4/6 = 60/336 = 5/28. After the red, all five blues remain; after a blue is then removed, four blues remain.

For blue-red-blue, use 5/8 × 3/7 × 4/6. The first blue leaves three reds and four blues. Removing a red next leaves four blues among six counters for the third draw.

For blue-blue-red, use 5/8 × 4/7 × 3/6. Two blues have left, so all three reds remain among the final six available counters.

Each path has probability 5/28 under this model. Their sum is 15/28. The paths are distinct and cannot occur together in one trial.

Parents can ask the child to locate the red in each permitted position. This gives a concrete way to list all routes without guessing how many products are needed.

A tutor should still have the student explain each branch fraction. The equality of the completed products does not mean that the same second or third fraction belongs everywhere.

The phrase “exactly one” excludes two reds and three reds. A calculation of “at least one red” would include those extra outcomes and give a different answer.

For comparison, the probability of no red in three draws is 5/8 × 4/7 × 3/6 = 5/28. The probability of at least one red is therefore 23/28.

Exactly one red and at least one red are not interchangeable. Write the event description beside the selected routes so the final result remains attached to the question actually asked.

A small table can show the remaining bag after each first draw. It is especially useful when a child’s tree looks tidy but its second-stage labels are copied from the original collection.

For red first, the remaining counts are two red and five blue, total seven. The next probabilities are 2/7 for red and 5/7 for blue.

For blue first, the remaining counts are three red and four blue, total seven. The next probabilities are 3/7 for red and 4/7 for blue.

The table below includes these states and the corresponding complete two-draw path products. Read the remaining counts before looking at the multiplication.

Parents can cover the probability columns and ask the child to reconstruct them from the counts. This checks whether the fraction is being interpreted as favourable counters out of all currently available counters.

A tutor can reverse the task too. If a branch is labelled red 2/7 and blue 5/7, ask which first colour must have been removed from the original bag.

This makes the representation work in both directions. The student reads fractions from a collection and recognises the collection represented by those fractions.

Do not mix a first-stage numerator with a second-stage total. The fraction 3/7 after a first red would claim three reds remain when only two do.

Similarly, using 5/8 after the first red would pair the unchanged blue count with a total that still includes the removed red. Both counts in a fraction need to describe the same state.

The table is a checking aid, not a separate selection rule. The original experiment determines the states; the table records them; the tree organises paths through them.

Once the updates are reliable, the child may need only a brief count note. The purpose is understanding that remains available in a fresh question, rather than a large diagram for every simple draw.

Imagine labelling the three red counters R1, R2 and R3, and the five blue counters B1 through B5. The labels distinguish individual counters even when their colours match.

There are eight possible first counters. For each first counter, seven different individual counters remain available as the second. There are therefore 8 × 7 = 56 ordered pairs of distinct counters.

Under uniform selection at each stage, each such ordered pair has probability 1/8 × 1/7 = 1/56. This gives a second way to check the colour-path calculations.

For two reds, choose one of three red counters first and one of the remaining two reds second. There are 3 × 2 = 6 favourable ordered pairs. The probability is 6/56 = 3/28.

Parents can ask why there are not nine red-red pairs. The missing pairs would select the same labelled red twice, such as R1 followed by R1. Without replacement, that second selection is impossible.

A tutor can list the six red-red pairs: R1R2, R1R3, R2R1, R2R3, R3R1 and R3R2. The concrete list explains the numerator before the count is compressed into multiplication.

For red then blue, there are 3 × 5 = 15 ordered labelled pairs, confirming 15/56. For blue then red, there are another fifteen.

The labelled-counter model explains why having two colour labels does not make the two colour categories equally likely. Here, red and blue represent different numbers of individual counters.

Do not count ordered favourable pairs over an unordered total, or the reverse. The numerator and denominator must use the same outcome convention.

The method is useful for checking a short problem, especially when the tree’s changing total seems mysterious. It shows that the denominator seven comes from excluding the one individual counter already selected, not from a rule attached only to the word probability.

For an event that ignores draw order, unordered pairs can provide a valid alternative count. Both favourable and total outcomes must then be counted without order.

The fifty-six ordered pairs of distinct counters form twenty-eight unordered pairs, because each pair has two orders. For example, R1 then B1 and B1 then R1 represent the same unordered pair of counters.

There are three unordered red-red pairs: R1 with R2, R1 with R3 and R2 with R3. Their probability is 3/28, matching the tree calculation for two reds.

For one red and one blue, there are 3 × 5 = 15 unordered mixed-colour pairs. The probability is 15/28, matching the sum of red-blue and blue-red paths.

Parents can ask the child to explain what counts as one outcome in their chosen method. Are they recording the sequence of draws, or only the final pair of distinct counters?

A tutor can compare the same event using both conventions. Six favourable ordered red-red outcomes over fifty-six total gives the same ratio as three favourable unordered pairs over twenty-eight.

Using six over twenty-eight would count favourable outcomes with order but total outcomes without order. That mismatch doubles the correct probability.

The unordered method here works because uniform drawing without replacement makes the distinct unordered pairs equally likely. If the experiment weights items differently or uses additional rules, equal likelihood needs to be reconsidered.

Order-specific questions should retain the order. “Red first, blue second” selects fifteen out of fifty-six ordered pairs, rather than all fifteen mixed unordered pairs out of twenty-eight.

The goal is consistency in the sample space. Trees are often convenient because they preserve order automatically. Unordered counting can be shorter for an order-free event, but it should be used only when the child can justify both the outcome convention and the equal-likelihood assumption.

A conditional second-draw probability changes according to the first outcome. The unconditional probability of red on the second draw, before we learn the first colour, is a different question.

In the example bag, red next after a first red has probability 2/7. Red next after a first blue has probability 3/7. Those fractions describe two different remaining bags.

If the first colour has not been specified, calculate the probability of second red through both possible histories. Red-red contributes 3/8 × 2/7 = 6/56. Blue-red contributes 5/8 × 3/7 = 15/56.

Their sum is 21/56 = 3/8. Under the uniform without-replacement process here, the unconditional chance that the second counter is red matches the original red proportion.

Parents can ask, “Do we know the first colour, or are we considering all possible first colours?” That distinction prevents the child from selecting a branch condition that the question has not supplied.

The equality of the unconditional probability with 3/8 does not make the two draws independent. Knowing the first colour still changes the probability of red on the second draw.

A tutor can use the tree to show both facts together. The branch probabilities differ, but their appropriately weighted path contributions add to the original red proportion.

Do not label both second branches 3/8 merely because the unconditional second-red probability is 3/8. That would erase the different conditions represented by the branches.

The total remaining is seven in each particular second-draw bag. The calculation across all histories combines probabilities of those different bags; it does not put the removed counter back.

This is why updating the bag should be taught with a precise question attached. “Red next given red first” and “red second with first colour unspecified” are different events, and a reliable explanation keeps their meanings separate.

If a drawn counter is returned before the next draw, the original colour counts are restored. With the same uniform random selection rule on each draw, the next red probability is again 3/8 and the next blue probability is 5/8.

For two reds under that replacement model, the probability is 3/8 × 3/8 = 9/64. Without replacement, it was 3/8 × 2/7 = 3/28.

Parents can ask the child to describe the physical process between draws. Does the counter stay outside, return to the bag, or return with an additional counter? The answer controls the next collection.

A tutor should distinguish a process assumption from a remembered fraction pattern. Restoring the counts is relevant, but the selection rule must also support the stated probabilities. The worked replacement model assumes fresh uniform selection rather than an unexplained bias.

For the without-replacement model, the events “first red” and “second red” are dependent: the first colour affects the conditional probability of the second red.

They are not mutually exclusive, because both can happen in a two-draw trial. Red-red has positive probability. Dependence and mutual exclusion describe different relationships.

By contrast, “first red” and “first blue” are mutually exclusive under the two-colour model. A single first counter cannot have both colours.

This distinction matters when adding paths. Red-blue and blue-red are disjoint complete sequences, even though probabilities inside each sequence depend on its earlier draw.

Avoid the slogan that all multiplication means independence or all addition means mutual exclusion. Multiplication along a path uses the relevant conditional probabilities. Simple addition across complete paths works when those paths are disjoint.

The useful question is what relationship the events actually have. Read the draw rule, identify the stage and history, and use the probabilities that match that process. The labels then support the reasoning instead of replacing it.

Check the remaining counts, the outgoing branch sums and the total probability of the complete paths. These checks catch different kinds of errors.

After a first red, two red and five blue should total seven. The branch sum 2/7 + 5/7 is one. After a first blue, three red and four blue total seven, with branch sum 3/7 + 4/7 equal to one.

If a child writes 3/7 and 5/7 after a first red, the sum is eight-sevenths. It counts more favourable counters across the two colours than the total available.

Parents can ask, “Could the next counter be something outside these two colours?” In this bag, no. The two outgoing possibilities therefore need to cover all remaining counters exactly once.

The four complete two-draw path probabilities are 6/56, 15/56, 15/56 and 20/56. Their numerators sum to fifty-six, so the total is one.

A tutor can also check event size. The probability of exactly one red, 15/28, should not exceed the probability of at least one red, 9/14. Exactly-one outcomes are included inside at-least-one outcomes.

Every probability must lie between zero and one. A result outside that range is an immediate reason to revisit counts or arithmetic, although a result inside the range is not proof that the model is correct.

Check impossible branches too. If all red counters have already been removed, the next red probability must be zero. Keeping the original numerator would falsely permit a colour that is no longer available.

The changing denominator is one part of the check. A tree with totals eight, seven and six can still be wrong if its colour counts or selected routes are incorrect.

Use these checks after interpreting the event. A mathematically valid tree can answer the wrong question if the child adds paths that the wording excludes.

Bring the original probability question and the child’s complete tree or working. Include the sentence that explains how items are drawn and whether they are returned.

An enquiry can say, “My child keeps the original denominator on later draws without replacement.” That identifies a specific modelling step for a Secondary 4 Mathematics tutor to inspect.

The tutor can ask the child to describe the bag after a first red and after a first blue. If those counts are clear but the fractions are wrong, the difficulty may lie in converting counts into probability.

If the remaining counts themselves are unclear, begin there. A small counter demonstration or written inventory can connect the selection rule with the next fraction.

Parents can ask how the lesson will check a fresh variation. Changing the number of counters, asking for a third draw or switching to replacement can show whether the child understands the process beyond one copied diagram.

In a three-pupil small group, students could compare their branch-state notes and explain why the second totals match while the colour counts differ. This is a possible activity for discussion, rather than a guarantee about a particular lesson or outcome.

Keep the work aligned with the child’s current school teaching and prerequisites. Two-draw colour paths may be enough for an initial repair; conditional notation or alternative counting methods can follow when appropriate.

The Secondary 4 Mathematics tuition page linked here provides the enquiry route. Confirm current arrangements directly and share a recent example rather than inferring a timetable or fee from the guide.

A useful home target is observable: before writing a second-stage fraction, the child states the remaining favourable count and total. That is more precise than asking them to be careful.

The first correction should preserve what the student already does well. A correctly drawn tree and accurate multiplication can remain useful once its branch probabilities represent the changing collection faithfully.

Use the same bag of three red and five blue counters, with uniform selection and no replacement between draws. Ask your child to name the remaining collection before reading each answer.

First, suppose the first counter is red. What is the chance of red next? Two red remain among seven counters, so the answer is 2/7. The first red has left both the red count and the total.

Second, suppose the first counter is blue. What is the chance of red next? Three red remain among seven, so the answer is 3/7. Removing a blue changes the total but leaves the red numerator unchanged.

Third, what is the probability of one red and one blue in two draws, in either order? Add red-blue and blue-red: 3/8 × 5/7 + 5/8 × 3/7 = 30/56 = 15/28.

Fourth, what is the probability of at least one red in two draws? The excluded event is blue-blue, with probability 5/8 × 4/7 = 5/14. Subtracting from one gives 9/14.

For a fresh variation, replace the drawn counter before the second selection and keep the uniform draw rule. The probability of two reds becomes 3/8 × 3/8 = 9/64. Ask the child to explain why the original total now reappears.

Parents can use the explanations to locate the first uncertain step. Incorrect remaining counts call for modelling; correct counts with wrong fractions call for probability interpretation; correct paths with wrong sums call for arithmetic.

A tutor can change the initial bag to four red and two blue counters. After a first red without replacement, the next red probability is 3/5. That checks whether the habit travels to new numbers.

The goal is an independent explanation of the current collection. Repeating seven as a remembered second denominator is not enough when the next question begins with a different total.

Choose one recent draw question and ask your child to explain what remains after the first selection. Keep the first discussion focused on the collection rather than the final probability.

If a red is removed from three red and five blue counters, write “two red, five blue, seven total”. Let the child use that note to produce both next fractions.

Then ask for the state after a blue instead. The answer becomes “three red, four blue, seven total”. The shared total and different colour counts make the branching history visible.

If those states are understood, move to one complete path. Ask for red then red and check that the second fraction belongs to the red-first branch. The calculation should be 3/8 × 2/7.

If the child can calculate a path but chooses the wrong collection of paths, inspect the wording: fixed order, either order, exactly one, same colour or at least one. Those phrases describe different events on the same tree.

Avoid rebuilding every probability topic at once. One accurate update and one fresh application can give the next lesson a clear starting point.

For related reading, the Secondary 4 probability guide connects events, Venn diagrams and trees. The conditional probability guide develops how information changes the relevant set of possibilities. The Mathematics article index offers other specific parent questions across the secondary levels.

If the difficulty persists, share the original draw rule and working through the Secondary 4 tuition enquiry page. A clear example helps the tutor distinguish a missing modelling step from arithmetic or event-selection errors.

The useful habit is to give every fraction a current collection and a precise question. What remains? Which items count as favourable? What event are we finding? With those answers in place, the tree becomes an explanation the child can trust, rather than a diagram whose numbers must be guessed.