Use complements for at-least and not events
Identify the exact excluded event, then subtract its probability from one without losing boundary cases.
Section 5.3 · Probability
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.3. Define the sample space and state its probability assumptions before counting. Fair independent dice give equiprobable ordered pairs; their sums are not equiprobable. Random selection means each specified individual is equally likely under the stated sampling model. Without replacement, later branch probabilities change; with replacement and independent mixing, they remain the same. A′ is a complement, A ∩ B means both, A ∪ B means either or both, and P(A|B) means A given B with P(B) > 0. Mutually exclusive and independent are different properties. The explicit general addition formula is not required by this section; explain overlap through disjoint regions. Conditional probabilities may use P(A|B) = P(A ∩ B)/P(B). Numerical models describe the stipulated experiment, not automatically a real population. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- List finite outcomes and use combinations.
- Read inclusive and exclusive count conditions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which event complements at least one blue ball?
What you will learn
- Use P(A′) = 1 − P(A).
- Translate at-least and neither statements into exact complements.
A complement partitions the whole experiment
An event A and its complement A′ cannot both happen and together include every outcome. Therefore P(A′) = 1 − P(A). For three independent fair coin tosses, at least one head is the complement of no heads, which is the single TTT string. The probability is 1 − 1/8 = 7/8.
For three balls selected uniformly without replacement from five red and three blue, at least one blue is the complement of all red. Its probability is 1 − 5C3/8C3 = 1 − 10/56 = 23/28. ‘At least one’ includes two or three as well; ‘exactly one’ does not.
Translate the boundary before subtracting
For an integer count X, the complement of X ≥ 2 is X ≤ 1, not X < 1. The complement of exactly two successes includes zero, one, three and any other possible counts. Write the entire excluded condition in words or symbols before calculating.
For two draws without replacement from three red and two blue balls, at least one blue has probability 1 − (3/5)(2/4) = 7/10. A separate direct count gives six mixed pairs and one all-blue pair out of ten subsets, also 7/10. A complement can be easier, but only after establishing the correct sample space and sampling rule.
Pause and explain
Three fair independent coins: what is P(at least one head)?
Worked example
Select three balls from five red and three blue uniformly without replacement. Find P(at least one blue).
Show the worked solution
- The excluded event is all three red, not exactly one red.
- Its probability is 5C3/8C3 = 10/56 = 5/28.
- Subtract from one: 1 − 5/28 = 23/28.
Answer 23/28.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Three fair independent coins are tossed. Find P(at least one head) using a complement.
Give me a hint
The only excluded string is TTT.
Compare my reasoning
- There are eight equiprobable strings.
- No heads has probability 1/8.
- Subtract: 1 − 1/8 = 7/8.
7/8.
Look for these in your work
- Identify the exact complement.
- Include all positive head counts.
Select three balls uniformly without replacement from five red and three blue. Find P(at least one blue).
Give me a hint
Subtract the all-red subset probability.
Compare my reasoning
- All subsets number 8C3 = 56.
- All-red subsets number 5C3 = 10.
- The probability is 1 − 10/56 = 23/28.
23/28.
Look for these in your work
- Use every permitted positive blue count.
- Keep the same subset model in both counts.
Draw two balls at random without replacement from three red and two blue. Find P(at least one blue) and check by counting unordered pairs.
Give me a hint
All red is the excluded event; mixed and all-blue pairs are favourable.
Compare my reasoning
- P(all red) = (3/5)(2/4) = 3/10.
- The complement gives 7/10.
- There are 3 × 2 = 6 mixed pairs and 2C2 = 1 blue pair out of 5C2 = 10, also giving 7/10.
7/10 by complement and pair counting.
Look for these in your work
- Change the second draw denominator without replacement.
- Make the direct and complement counts agree.
Common mistakes
- At least one is not exactly one.
- Include the equality boundary when taking a complement.
- Use the stipulated replacement rule in the excluded probability.
What can you explain now?
What is the complement of at least two successes?
Compare with the explanation
Zero or one success.
Every possible success count must belong to exactly one of the event and its complement.
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Make it stick
Write the exact complement, calculate it under the same model, and subtract from one.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.