Did you know that taking one counter out of a bag changes the next probability even when nobody changes the rules? If a student repeats the same fraction for the first and second draw, the arithmetic may look tidy while the model is wrong. The crucial question is whether the first counter goes back.
Drawing without replacement means that the selected object is kept out of the pool before the next draw. The second probability is therefore conditional on what happened first: the number of available objects decreases, and the number of objects of the chosen colour may decrease too. To solve the problem, update both counts, multiply probabilities along a specified path, and add the disjoint paths that meet the question.
Here is the useful order: identify what can be selected; decide whether the selection is replaced; label the first branch; update the remaining objects for each second branch; calculate each complete path; and verify that all complete paths total one.
Begin with a picture of the remaining objects, not a memorised formula
Imagine five equally likely counters: three green and two orange, in an opaque bag. Two are drawn at random, one after the other, and the first stays outside the bag. We will ask three different questions about the same experiment. These numbers are invented for instruction, not observations of a particular classroom.
On the first draw, the chance of orange is 2/5, and of green is 3/5. If orange is selected first, four remain: three green and one orange. If green is selected first, four remain: two green and two orange. This is why the second branch needs a separate numerator for each first-draw outcome.
Worked example: the probability of two orange counters
The orange–orange path has probability (2/5) × (1/4) = 2/20 = 1/10. Multiplying 2/5 × 2/5 would wrongly assume the first orange went back and the bag again held five objects. A fraction represents a particular situation, not a colour label that remains fixed throughout the calculation.
The result can be checked by counting ordered draws. There are 5 × 4 = 20 equally likely ordered selections of two distinct counters. The two physical orange counters can be selected in either order, giving two qualifying selections. Hence 2/20 = 1/10, agreeing with the branch method.
Worked example: exactly one orange versus at least one orange
Exactly one orange has two separate paths: orange then green, (2/5) × (3/4) = 6/20; or green then orange, (3/5) × (2/4) = 6/20. Adding gives 12/20 = 3/5. Those paths are distinct, so adding them is correct; simply using the first path would omit half the possibilities.
At least one orange includes those two mixed paths and orange–orange. It is often easier to calculate the opposite: neither orange means green–green, (3/5) × (2/4) = 6/20 = 3/10. Thus P(at least one orange) = 1 − 3/10 = 7/10. Direct addition confirms 3/5 + 1/10 = 7/10.
One final check completes the model. Green–green is 6/20, green–orange is 6/20, orange–green is 6/20, and orange–orange is 2/20. Together, 6 + 6 + 6 + 2 = 20 ordered possibilities out of 20. A complete tree with branches that do not overlap must sum to probability one.
Diagnose the learner before adding more questions
- Unchanged-denominator error: the learner writes fifths on both stages, suggesting that the first object was mentally replaced.
- Unchanged-numerator error: the denominator correctly becomes four but the orange numerator stays two even after an orange was removed.
- One-path error: the student solves orange–green but forgets green–orange when asked for exactly one orange.
- At-least/exactly confusion: the student leaves out orange–orange from ‘at least one’.
- Path-versus-branch confusion: the learner adds adjacent probabilities where a complete ordered path calls for multiplication.
Ask a learner to explain what the bag contains immediately after the first orange draw. If they can say ‘three green and one orange’ without calculating, they have the physical model. If not, return to five actual tokens or a simple sketch before teaching notation. This diagnosis is much more useful than asking the student to memorise ‘multiply then add’.
Progress from guided demonstration to independent transfer
Step 1: move five paper counters
Use three green and two orange paper squares. Remove an orange and count the remaining objects. Repeat by removing a green instead. Record the second orange probabilities 1/4 and 2/4. Say aloud why they differ; the physical difference should be visible.
Step 2: complete a half-drawn tree
Give the first branches 2/5 orange and 3/5 green. Ask the learner to complete all four second branches without looking at the worked example. The orange-first branches should be orange 1/4 and green 3/4; the green-first branches orange 2/4 and green 2/4. The second branches after each first outcome must sum to one.
Step 3: answer a changed question
Ask for the chance that the two colours differ. It is the orange–green path plus the green–orange path, giving 6/20 + 6/20 = 12/20 = 3/5. Challenge the learner to explain why orange–orange and green–green do not qualify. The sentence in the question, not the appearance of the tree, chooses the paths.
Step 4: try a fresh collection independently
A new bag holds four yellow counters and two purple counters. Two are taken without replacement. What is the probability of two yellow? The learner should obtain (4/6) × (3/5) = 12/30 = 2/5. What is the probability of at least one purple? Use the complement: 1 − 2/5 = 3/5. Explain that ‘at least one purple’ is everything except two yellow.
Step 5: compare replacement and non-replacement
Return to the same four-yellow, two-purple bag but put the first counter back and mix before drawing again. Two yellow now has probability (4/6) × (4/6) = 4/9, not 2/5. The learner should explain the difference in words before comparing the fractions. The first probability is unchanged; the second is different because the experiment changed.
What a parent or teacher can observe
Check whether the child labels the remaining objects at every branch, selects the correct paths for words such as ‘both’, ‘exactly one’, and ‘at least one’, and completes an unfamiliar example without copying the first. Three short checks a few days apart can be more informative than one page of repeated trees. Keep a small record of which part failed: reading the question, updating the bag, multiplying along a path, or adding qualifying paths.
A parent can make this playful with coloured slips of paper. A teacher can use a student’s incorrect ‘2/5 then 2/5’ to ask whether the second object could be the very same object that was removed. Avoid turning a diagnostic conversation into a race: accurate modelling should precede speed.
Limits and assumptions matter
Our calculation assumes each remaining physical counter is equally likely to be picked on each draw. If counters differ in size, the selection mechanism is biased, or the first is replaced, the model changes. If a question only says ‘draw twice’ and omits the replacement rule, do not pretend the rule was specified. Explain the ambiguity or state the assumption explicitly. Probability describes the stated random mechanism; it is not a guarantee of what two draws will produce.
Frequently asked questions
Why multiply along a path?
A complete path says that the first event happens and then the second event happens under that first outcome. The appropriate multiplication is P(first) × P(second given first). In our orange–orange example, the second factor is 1/4 precisely because orange happened first.
Why do we add some paths?
When the question can be satisfied by several mutually exclusive ordered paths, such as orange–green or green–orange, their probabilities add. Make sure paths cannot happen simultaneously in a single two-draw experiment; otherwise a simple addition may double-count.
Is without replacement always dependent?
For these colour-count examples, the first colour changes the composition of the bag and therefore the next colour probability. Dependence should be checked for the actual events being studied, rather than assumed solely from a familiar diagram.
Do students have to draw a tree?
No. A table of ordered outcomes, sequential fractions, or a correct complementary calculation can work. The tree is particularly helpful while learning because it makes the changing counts visible and reduces forgotten cases.
Continue learning
For the underlying probability language, read OpenStax on dependent events and sampling without replacement. Within eduKateSG, compare at least one versus exactly one, and the Secondary Mathematics probability-tree guide. The Education Hub connects subjects and learning practices; the Clementi Mathematics tutorial reference shows why clear explanation and independently checked practice belong together.
The core aim is to reason about what remains possible after an action, rather than move symbols mechanically. Once a learner can explain the changing bag, their probability calculations become easier to verify and transfer.