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At least one success means one or more successes. In two trials, it includes success on the first trial only, success on the second trial only, and success on both trials. Leaving out the both-successes case turns the event into exactly one, which is a different question.
For G3 Mathematics, name the event before calculating its probability. With two independent fair coin tosses, at least one head includes HH, HT and TH. Its probability is three quarters. Exactly one head includes only HT and TH, giving one half.
If your child multiplies or adds correctly but answers the wrong event, more arithmetic practice may not fix the difficulty. The useful repair is to translate at least, exactly and none into a clear set of allowed outcomes before choosing the calculation.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–25
Sections 26–29
For any clearly defined collection of trials, either no success occurs or at least one occurs. Those events do not overlap and together cover every possible outcome. Therefore the probability of at least one success is one minus the probability of no successes.
For two independent fair coin tosses, no heads means TT, with probability one quarter. Subtracting from one gives three quarters for at least one head. This agrees with directly counting HH, HT and TH.
The complement relationship does not require independence. Independence may be needed for a particular multiplication used to calculate the no-success probability. Keep those two ideas separate: what event is the complement, and how is that event’s probability obtained?
All chance experiments in this guide are invented classroom models. Fairness, replacement and independence are stated where used. Do not transfer the numerical results to real events whose probabilities or relationships have not been established. The purpose is to read probability conditions accurately and justify the calculation.
The SEAB G3 Mathematics syllabus for 2027 includes probabilities of simple combined events, possible-outcome representations and appropriate addition and multiplication rules. Reading the event correctly is part of using those techniques meaningfully.
This guide concentrates on the phrase at least one. The examples use small sample spaces that can be inspected directly before a shorter complement calculation is introduced. No advanced probability distribution is needed for the core reasoning.
The broader probability mastery guide connects outcome reading with uncertainty, independence and changing information. This page develops one decision in detail: which outcomes belong to at least one and which belong to exactly one.
G3 is a subject level, not a judgement about a learner’s value or future. A student can know every basic probability operation yet still need help with a short phrase. Clarifying that phrase often makes the surrounding calculations more reliable without requiring the learner to start the whole topic again.
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In two coin tosses, HH is one ordered outcome: head first, then head again. The event at least one head contains several outcomes. Its probability is found by combining the probabilities of all outcomes that meet the condition.
HT and TH are different ordered outcomes because the head occurs on a different toss. They have the same number of heads but different sequences. Whether order matters depends on the sample space used to represent the experiment.
Do not compress the possibilities to “zero heads, one head, two heads” and assume those three categories are equally likely. For two independent fair tosses, the one-head category contains two equally likely sequences, while each of the other categories contains one.
This distinction explains why a quick count can go wrong even when the event wording is understood. Before dividing favourable cases by total cases, check that the elementary outcomes being counted are equally likely. A list of category labels is not automatically an equally likely sample space.
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Let the count of successes be the number of trials that meet the stated success condition. In two trials, the possible counts are zero, one and two. At least one means the count is greater than or equal to one, so counts one and two qualify.
Exactly one means the count equals one. At most one means the count is zero or one. These phrases are close in everyday language but define different events. Write the permitted counts first if the wording feels slippery.
For two fair coin tosses, success can be defined as a head. At least one head includes three sequences. Exactly one includes two. At most one includes TT, HT and TH, also three sequences, but a different set from at least one.
The last comparison is useful: two different events can have the same probability. A numerical answer alone does not prove that the event was read correctly. Ask the learner to name which outcomes were included, rather than accepting a familiar fraction without its interpretation.
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For two independent fair coin tosses, the four ordered outcomes are equally likely, each with probability one quarter. A table keeps event membership separate from the arithmetic.
| Outcome | Number of heads | At least one head? | Exactly one head? |
|---|---|---|---|
| HH | 2 | Yes | No |
| HT | 1 | Yes | Yes |
| TH | 1 | Yes | Yes |
| TT | 0 | No | No |
The at-least-one column has three qualifying outcomes, giving three quarters. The exactly-one column has two, giving one half. The difference is the HH outcome, whose probability is one quarter.
If a student gives one half for at least one, ask them to point to HH and explain why it was excluded. This makes the missing case concrete. The issue is not a mysterious probability rule; two successes still satisfy the requirement of having one or more successes.
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The outcomes HH, HT and TH cannot occur simultaneously in a single two-toss experiment. Exactly one complete sequence occurs. Their probabilities can therefore be added directly to find the probability of the event containing them.
For fair independent tosses, each sequence has probability one quarter, so the sum is three quarters. With a biased coin under an independent repeated-toss model, the sequence probabilities can differ, but the disjoint-outcome addition principle still applies.
Do not confuse the disjoint complete sequences with the events “head on the first toss” and “head on the second toss.” Those larger events overlap at HH. Adding their probabilities without adjustment would count that outcome twice.
The level of description matters. Adding separate complete outcomes is not the same as adding two overlapping events. Before adding, ask whether a single experimental outcome could belong to more than one of the categories being added. If it could, the overlap must be handled rather than ignored.
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Let A mean a head on the first toss and B mean a head on the second. Each has probability one half. Adding gives one, but at least one head is not certain: TT remains possible.
The overcount occurs because HH belongs to both A and B. Its probability of one quarter has been included twice. Subtracting that overlap once gives one half plus one half minus one quarter, or three quarters.
In general, the probability of A or B is the probability of A plus the probability of B minus the probability of both. For this two-event union, the overlap correction ensures each qualifying outcome is counted once.
This formula is a compact description of the counting logic, not a replacement for defining the events. If A and B mean something different, their overlap changes. If they are mutually exclusive, the overlap probability is zero and direct addition is appropriate. Check that condition rather than automatically adding every pair of probability values in the question.
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No successes and at least one success are complementary events. They cannot occur together, and every possible outcome belongs to one of them. Their probabilities add to one because they cover the entire experiment without overlap.
This is why subtracting the no-success probability works. It is not a shortcut based on the word least alone. The event must be identified accurately, and its complement must include exactly the outcomes that fail it.
The complement of exactly one success is not simply no successes. In two trials, it includes zero successes and two successes. In more trials, it includes every permitted count other than one. Confusing these complements changes the question.
When using a complement, write a short sentence first: “At least one head fails only when both tosses are tails.” That sentence justifies the subtraction and identifies the event whose probability must be calculated. A correct complement statement often prevents a larger error later in the working.
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For two independent trials with success probability p on each, the failure probability on each is 1 − p. Independence allows the probability of two failures to be calculated as (1 − p)(1 − p). The probability of at least one success is then 1 − (1 − p)².
The complement relationship remains valid even if the trials are dependent, but the simple product of identical failure probabilities may not. In a dependent experiment, use the appropriate conditional branch probability or another valid description of the no-success event.
Students sometimes remember the final expression and apply it to any two opportunities. That silently assumes both equal trial probabilities and independence. Two stated chances do not automatically supply those conditions.
Ask what the model says about the second trial after the first outcome is known. If the available objects change, or the experiment explicitly links the trials, revisit the multiplication. A familiar complement method can still be used, but its internal probability calculation must match the actual experiment.
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Suppose an invented trial has success probability one quarter and failure probability three quarters. Repeat it twice independently under the same conditions. The probability of no successes is three quarters multiplied by three quarters, or nine sixteenths.
The probability of at least one success is therefore seven sixteenths. Exactly one success has two separate paths: success then failure, or failure then success. Each has probability three sixteenths, giving six sixteenths altogether.
Both successes has probability one sixteenth. Adding exactly one and both gives six sixteenths plus one sixteenth, agreeing with the at-least-one result. The direct and complement methods provide independent representations of the same event.
Simply doubling one quarter gives one half, which overcounts the both-successes outcome by one sixteenth. The error is small enough to look plausible, but its cause is clear. An additional opportunity can increase the probability of at least one success without making that probability exactly twice the single-trial value.
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Roll a fair six-sided die twice independently. There are thirty-six equally likely ordered outcomes. Define success as rolling a six. Six outcomes have a six on the first roll, and six have a six on the second.
The outcome (6, 6) is in both lists. Therefore the number with at least one six is 6 + 6 − 1 = 11, giving probability 11/36. Exactly one six has ten outcomes, giving 10/36 = 5/18.
The complement calculation counts outcomes with no six. Each roll has five non-six possibilities, producing twenty-five ordered outcomes. Subtracting from thirty-six leaves eleven, so the probability is again 11/36.
This example shows why two times one sixth is not the exact at-least-one probability. It would count twelve outcome slots, with (6, 6) occupying two of them. Count physical outcomes in the sample space, not duplicated appearances across overlapping event descriptions.
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A bag contains two red counters and three blue counters. Draw one counter, replace it, mix, and draw again. If the draws are modelled as independent random selections, each draw has red probability two fifths and blue probability three fifths.
At least one red has probability one minus the probability of two blues: 1 − (3/5)(3/5) = 16/25. Exactly one red has probability (2/5)(3/5) + (3/5)(2/5) = 12/25. Both red has probability 4/25.
The counts in the bag return to their original values before the second draw. That stated reset supports using the same branch probabilities again. Do not assume replacement merely because two draws are mentioned.
The event distinction remains the same as with coins. At least one red includes red-red as well as the two mixed-colour paths. Changing the physical experiment changes the probabilities, but it does not change the meaning of at least one. Keep the event definition and the experiment’s probability model separate.
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Use the same bag but do not replace the first counter. No red means blue then blue. The first blue probability is three fifths. After a blue has been removed, two blue counters remain among four counters, so the second blue probability is two fourths.
The no-red probability is (3/5)(2/4) = 3/10. Therefore at least one red has probability 7/10. The complement relationship has not changed; the second branch probability has.
For a direct check, both red has probability (2/5)(1/4) = 1/10. Exactly one red has probability (2/5)(3/4) + (3/5)(2/4) = 3/5. Their sum is 7/10, matching the complement result.
The without-replacement guide develops the changing-denominator decision. Here it demonstrates a broader lesson: a valid complement can be combined with dependent branch probabilities. Independence is not a prerequisite for every use of one minus.
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For two independent fair coin tosses, head on the first toss and head on the second can both occur. Their overlap is HH, with positive probability. Independence does not mean that only one event is allowed to happen.
Mutually exclusive events cannot happen together in the same experiment. For one die roll, rolling a one and rolling a two are mutually exclusive. Their overlap is empty, so their probabilities can be added without an overlap correction.
For nonzero-probability events, mutual exclusivity generally prevents independence: knowing that one occurred rules out the other. These ideas answer different questions and should not be used as synonyms simply because both appear in combined-event chapters.
In an at-least-one question, ask whether the relevant success events can overlap. If both successes are possible, the event includes that case. Then calculate its probability under the correct independence or dependence assumptions. Reading the overlap comes before choosing whether a multiplication or direct addition is justified.
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In two independent fair coin tosses, the numbers of heads across HH, HT, TH and TT are two, one, one and zero. Their mean is one head per two-toss experiment. The probability of at least one head is three quarters.
Those are different quantities. The average count gives HH a contribution of two heads. The at-least-one event treats HH as one qualifying experimental outcome, just as it treats HT and TH as qualifying outcomes.
This section is an optional interpretation extension, not a requirement to learn a new formal probability topic. It helps explain why an average count of one does not mean that at least one success is certain. TT remains a possible outcome despite the mean count being one.
Check units in the explanation: one head per experiment is a count average, while three quarters is a probability. A number that looks like a valid probability is not automatically the requested probability. Name the quantity being calculated before deciding whether its numerical value answers the question.
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For three independent fair coin tosses, at least one head includes counts one, two and three. Only TTT has no heads. Its probability is one eighth, so at least one head has probability seven eighths.
Exactly one head has the outcomes HTT, THT and TTH, giving three eighths. At least two heads includes HHT, HTH, THH and HHH, giving one half. The phrases select different collections even though the experiment is unchanged.
The complement of at least two heads is at most one head, not no heads alone. It contains the zero-head outcome and all three one-head outcomes. Read the count condition before choosing which outcomes to subtract.
A small complete list remains useful at this stage. It makes the language visible and provides a check on a shorter formula. Students should understand why the complement is the correct collection, rather than identifying every phrase beginning with at least with the same no-success calculation regardless of the number required.
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A. For two independent fair coin tosses, find the probability of at least one tail. B. In the same experiment, find exactly one tail. C. Find at most one tail and list its outcomes.
D. For two independent fair die rolls, find the probability of at least one six. E. Find exactly one six. F. An invented independent trial has success probability one third on each of two repetitions. Find the probability of at least one success.
G. A bag contains one red and two blue counters. Draw twice without replacement. Find the probability of at least one red. H. Repeat G with replacement and independent random draws. I. Explain why the two results differ.
J. For three independent fair coin tosses, find the probability of at least one head. K. Identify the complement of exactly one head in the three-toss experiment. L. Explain why adding the first-toss-head probability to the second-toss-head probability overcounts in the two-toss experiment.
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A gives three quarters, using TT, TH and HT. B gives one half, using TH and HT only. C gives three quarters, using HH, HT and TH. A and C have the same probability but different event memberships.
D is 11/36, while E is 10/36 = 5/18. For F, no successes has probability (2/3)² = 4/9, so at least one has probability 5/9. The independence and repeated success probability were stated in the question.
For G, no red means two blues, with probability (2/3)(1/2) = 1/3. At least one red therefore has probability 2/3. For H, no red has probability (2/3)² = 4/9, giving at least one red probability 5/9.
For I, the bag composition changes in G but resets in H. J gives seven eighths. K includes zero, two or three heads. For L, HH belongs to both single-toss-head events and is counted twice. The correction must remove one copy of that overlap or use a valid complement calculation.
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Suppose the first of two independent trials has success probability one half, while the second has success probability one quarter. The failure probabilities are one half and three quarters. No success has probability three eighths.
At least one success therefore has probability five eighths. The formula 1 − (1 − p)² would not fit unless one common p represented both trials. Here the correct product uses two different failure probabilities.
A direct check separates the outcomes by success pattern. First only has probability (1/2)(3/4) = 3/8. Second only has probability (1/2)(1/4) = 1/8. Both has probability (1/2)(1/4) = 1/8. Their sum is five eighths.
This is a useful extension because independence and equal probabilities are separate assumptions. Trials can be independent without having the same success chance. Read both features explicitly instead of treating the word independent as a guarantee that every branch uses an identical number.
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Imagine a mathematical model in which the second result is defined to copy the first fair coin result exactly. The only possible two-result outcomes are HH and TT, each with probability one half. This is not a pair of independent tosses.
At least one head has probability one half, not three quarters. A second reported result has added no independent opportunity for a head because it is completely determined by the first. No heads and at least one head remain complementary events.
This example shows why the phrase two chances is insufficient by itself to determine a combined probability. The relationship between the chances matters. Equal single-event probabilities can coexist with very different overlap structures.
It also prevents a casual claim that an additional event must always strictly increase the probability of at least one success. Adding an alternative cannot reduce that probability, but it need not increase it if the new event contributes no previously uncovered outcomes. State the experiment before making a numerical prediction about its gain.
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A probability must lie between zero and one. If adding several success probabilities produces a result above one, it cannot be the probability of at least one event without further adjustment. An overlap or a model assumption may have been ignored.
However, a result below one is not automatically correct. Doubling one sixth gives one third, which is within the allowed range but still overcounts the at-least-one-six event for two independent die rolls. A basic bound detects some errors, not every error.
For two events, the probability of at least one is at least as large as either individual event probability, because the union includes every outcome in each event. It is no larger than their probability sum. These checks can help assess a proposed result before detailed review.
Use limits as evidence with a clear scope. A failed limit decisively rejects an answer. Passing a limit merely means the answer has survived that particular test. Return to the event definition and the correct probability model to establish the actual value.
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If success is impossible on both trials, at least one success is impossible too. Its probability is zero. If success is certain on the first trial, at least one success is already certain regardless of what the second trial does.
Under the equal independent-trial formula, p = 0 gives 1 − 1² = 0, while p = 1 gives 1 − 0² = 1. These endpoint checks agree with the meaning of the event.
Do not conclude that a probability of one means every trial must succeed. It means the stated event occurs with certainty in the model. At least one can be certain while some trials still fail, provided the model guarantees a success somewhere.
For example, a model could guarantee exactly one head across two reported results, allowing only HT and TH. Then at least one head is certain, but both heads is impossible. This dependent model again shows that the event wording and the relationship among trials determine what a probability statement actually guarantees.
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The mixed outcomes HT and TH can be combined into the event exactly one head, but they must not be treated as one equally likely elementary outcome alongside HH and TT. Their combined probability is one half, while each same-result outcome has probability one quarter in the fair independent model.
Grouping is useful when it simplifies a calculation without discarding weights. A count table with zero, one and two heads can carry probabilities one quarter, one half and one quarter. The categories then describe the experiment correctly because their different probabilities remain visible.
This resembles the grouped-data mean guide: a representation can compress information, but its meaning must be preserved. Here the critical retained information is the probability attached to each grouped category.
If a learner changes representation, ask what happened to the weights. A shorter list is not necessarily a list of equally likely outcomes. The denominator in a favourable-over-total count must correspond to elementary cases with justified equal likelihood, not merely the number of category names on the page.
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For two coin tosses, direct listing and the complement method are both short. For several independent trials, counting every outcome with at least one success can become cumbersome, while the no-success event may remain one simple repeated-failure path.
That makes the complement method efficient, but efficiency should come after interpretation. Write the event and its complement first. Then calculate the complement under the stated model and subtract once from one.
If the requested event is exactly one, direct mutually exclusive paths may be clearer than a complement with many cases. If the trials are dependent, a tree with changing branch probabilities may be the most transparent representation. There is no need to force every question into one favourite formula.
The worked-example revision guide supports this decision-focused practice. Before reading a solution’s arithmetic, predict which event is being counted and why the chosen method covers it exactly once.
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If a student excludes the both-successes outcome, repair the meaning of at least one. If they include it but add overlapping event probabilities without correction, repair the counting structure. If they use an unchanged second-draw probability without replacement, repair the experiment model.
If they list three count categories and assume equal likelihood, repair the sample-space weights. If their exact event and model are correct but a fraction product is wrong, repair the arithmetic. These are different difficulties that can produce similar-looking wrong answers.
Ask the learner to explain their event in words before showing a formula. “I want every outcome except no red” is informative. “I used one minus because it said at least” is less secure unless the student can identify the complement accurately.
Keep the feedback specific. The student may already understand most of the topic. A clear repair followed by one changed example provides better evidence of progress than asking for a whole chapter to be repeated without identifying which decision actually failed.
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Use one experiment and vary only the wording: at least one head, exactly one head, at most one head and no heads. Ask for the qualifying outcomes before calculating any fractions. This isolates language from arithmetic.
Next keep the event fixed and vary the model: fair independent tosses, unequal independent chances, or a defined dependent relationship. Then use a bag with and without replacement. The student can see which change affects the event membership and which affects the probabilities.
The Clementi Secondary 1 Mathematics small-group guide gives the wider teaching context. Effective support can begin with a small complete sample space and gradually remove the scaffold as the learner becomes able to justify the method independently.
A useful final explanation names the event, the complement if used, and the assumption supporting any multiplication. It can be short. The goal is not a long paragraph around every fraction, but enough reasoning to show that the calculation answers the actual probability question rather than a nearby one.
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Consider three independent repetitions of a fictional trial, each with success probability one quarter. Failure has probability three quarters. At least two successes means exactly two or exactly three successes, not merely anything except total failure.
There are three separate success patterns with exactly two successes: SSF, SFS and FSS. Independence and the common trial probabilities give each pattern probability (1/4)(1/4)(3/4) = 3/64. Their total is 9/64.
All three successes has probability (1/4)³ = 1/64. Therefore at least two successes has probability 10/64 = 5/32. The both-or-more interpretation from the opening has expanded to the relevant count threshold of two.
The complement includes zero successes and exactly one success. Zero successes has probability (3/4)³ = 27/64. Exactly one has three paths, each with probability (1/4)(3/4)(3/4) = 9/64, giving another 27/64. Subtracting their combined 54/64 from one leaves 10/64, agreeing with the direct calculation.
By contrast, one minus the zero-success probability alone gives 37/64. That is the probability of at least one success, a larger event. The arithmetic is valid for that event but does not answer at least two.
This optional extension is useful for checking whether the learner understands the complement or merely reacts to the words at least. Write the permitted success counts first. Then write the counts outside that event. The complement is determined by that partition of the complete experiment, not by a memorised instruction to subtract one particular path.
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Imagine one hundred equally likely elementary outcomes in a fictional experiment. Event A contains forty outcomes and event B contains thirty. Their individual probabilities are therefore 0.4 and 0.3, but those two numbers alone do not tell us how many outcomes they share.
If they share ten outcomes, their union contains 40 + 30 − 10 = 60, giving probability 0.6. If they share none, the union contains seventy and its probability is 0.7. If every B outcome also belongs to A, the union contains only forty and its probability is 0.4.
Each arrangement respects the same two individual probabilities. What changes is their overlap. Therefore a request for the probability of A or B needs enough information to determine that overlap, or a justified relationship such as independence or mutual exclusivity.
These examples also give bounds: the union cannot be smaller than the larger individual event and cannot exceed the sum of the two probabilities. Here its probability lies from 0.4 to 0.7 inclusive. The bounds are informative even when one exact value is not determined.
Do not choose the independence product simply because no relationship was stated. Independence is an assumption or a property to justify, not a default repair for missing information. If it is given in this model, the overlap probability is 0.4 × 0.3 = 0.12 and the union probability is 0.58.
This section concerns two named events, which may be success events associated with different trials or other features of one experiment. The same at-least-one logic applies: include every outcome in either event, including their shared outcomes, and count each only once. The unknown overlap is a genuine missing input, not an invitation to guess.
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Does at least one mean exactly one?
No. It includes one, two and every larger possible success count. In two trials, both successes belong to the event. Exactly one excludes the both-successes case and therefore usually has a different probability.
Can I always use one minus the probability of none?
For an at-least-one event, yes, once none is defined for the same complete experiment. But calculate the no-success probability correctly. Independence, changing compositions and unequal trial probabilities affect that calculation even though the complement relationship stays valid.
Does a second opportunity double the probability?
Not generally. The events can overlap, and dependence can change their relationship. For equal independent success probability p on two trials, the at-least-one probability is 2p − p², which accounts for the overlap rather than counting it twice.
Can rounding hide the overlap correction?
Yes. If each of two independent trials has success probability 0.01, the exact at-least-one probability is 1 − 0.99² = 0.0199. Simply adding the single-trial probabilities gives 0.02. Both may display as two percent after rounding to a whole percentage, but the exact calculations differ.
The matching display does not prove that the addition rule was valid. It only means that the reporting precision concealed a small difference. Keep the exact model and the final display separate. In a question requesting an approximation, report the requested accuracy after using the justified calculation. A correct rounded number obtained from an unjustified method is not evidence that the method will work on a changed example with a larger overlap.
Where should I continue?
Use the Mathematics explanation directory or the statistics and probability examination guide. The dependable habit is to define the event, account for every qualifying outcome exactly once, and justify the probabilities assigned to those outcomes.
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