120i - Applying Complementary Events in Context

Lesson 120 — Applying Complementary Events in Context

Strand: Probability | Descriptor: AC9M8P01 | Duration: 45 minutes

Block note. Continuing from Lesson 119. Today extends the complement rule into richer, two-step applied contexts, and introduces “at least one” problems for two independent trials — the exact skill Lesson 121 will consolidate with harder, multi-trial problem solving.

Learning Intentions

• To apply the complement rule flexibly across varied, more complex applied contexts.
• To recognise “at least one” as the complement of “none”, and use this to solve two-trial probability problems.

Success Criteria

I can:

1. Apply to solve two-step applied problems, including estimating a count from a probability.
2. Recognise that, for two independent trials, “at least one” is the complement of “none”.
3. Calculate using for two independent trials.
4. Explain why the complement method is more efficient than listing outcomes directly.

Warmup

(5 minutes — rapid-fire, then a teaser)

1. . Find .
2. A coin is tossed twice. List all possible outcomes.
3. From your list, find by directly counting outcomes.
4. What is the complement of “at least one head” in this experiment? Find its probability. Do your answers to Q3 and Q4 sum to ?

Answers: 1. ; 2. ; 3. ( all qualify); 4. “No heads” , ; and ✓.

Today’s question: can we find without listing every outcome?

Activities

Activity 1 — Guided Practice: the Complement Rule in Richer Contexts (11 min)

I do / We do / You do.

I do — a two-step problem. A factory’s inspection finds . Out of items produced, estimate how many are expected to fail.

We do — a three-outcome weather problem. , . Find two ways.

You do:

1. A store finds . Of weekly customers, estimate how many do not prefer home delivery.
2. . Of customers, estimate how many do not buy online.
3. . Of a batch of , estimate how many are not defective.

(Answers: 1. ; . 2. ; . 3. ; .)

Activity 2 — Guided Practice: “at Least one” with Two Independent Trials (16 min)

Pairs.

I do: Two fair dice are rolled. Find .

We do: A biased coin has . It is tossed twice. Find .

You do:

1. Two spinners each have . Find .
2. Two independent matches each have . Find .
3. Two dice are rolled once each. Find .
4. (Harder.) A quality control test checks independent components, each with a chance of being faulty. Find . Then interpret your answer: out of every such pairs tested, about how many would be expected to have at least one faulty component?

Socratic scaffolding for Q4 (Polya cycle):

Prompt	Purpose
Understand: what exactly counts as “at least one faulty”?	Either component faulty, or both — everything except “neither faulty”.
What is the complement of “at least one faulty”?	”Neither component is faulty” — the single easiest outcome to calculate.
Devise a plan	Find , then subtract from .
What is ?	.
Carry it out	.
So the final answer?	.
Looking back — interpret in context	Out of pairs, about – would be expected to have at least one faulty component.

Answers:

Activity 3 — Inquiry: Direct Listing versus the Complement — Which Scales Better? (7 min)

Pairs.

1. For coin tosses, find : (a) by listing all outcomes and counting; (b) using the complement method.
2. How many outcomes did you need to check directly for (a)? How many for (b)?
3. Without listing them, predict how many outcomes “at least one head in tosses” would need checked directly, versus using the complement.
4. Which method scales better as the number of trials increases? Explain why.

Answers: 1. (a) outcomes, contain at least one head, . (b) , so — same answer, far less work. 2. Direct: all (to be sure of counting correctly); complement: just . 3. Direct would need to check up to of possible outcomes; the complement only ever needs the single “all tails” case. 4. The complement method scales far better — it always needs just one outcome (“none”), no matter how many trials there are, while direct listing grows exponentially.

Checks for Understanding

(6 minutes — exit ticket, collected)

1. . Find .
2. Two independent fair coins are tossed. Find , showing your working using the complement method.
3. A factory tests independent items, each with a fault rate. Find .
4. Of a school’s students, . Estimate how many do not pack lunch from home.
5. Reasoning. Explain why the complement method becomes more useful, relative to direct listing, as the number of trials increases.

Answers: 1. ; 2. , so ; 3. , so ; 4. ; ; 5. Direct listing requires checking a number of outcomes that grows very quickly (doubling with each extra trial), while the complement method only ever needs the single “none” outcome, regardless of how many trials there are.

Common Misconceptions

Misconception	How to pre-empt it
Treating “at least one” as meaning “exactly one”.	Contrast directly: “at least one head in 2 tosses” includes , and — three outcomes, not one.
Adding across trials instead of multiplying.	Reinforce with the dice/coin examples: , not .
Confusing “not sunny” with “rainy” when a third outcome (cloudy) exists.	Activity 1’s “We do” three-outcome weather example, checked two ways.
Applying the “at least one” complement method to trials that are not independent.	Flag explicitly: today’s method assumes each trial does not affect the others — this will be revisited in Lesson 121.
Forgetting the final "" step after correctly calculating .	Model the two-line working every time: find , then subtract from .

Enrichment — Competition-Style Problems

E1 (Kangaroo style). Two spinners each have equal sectors, one of which is gold. Both are spun. Find .

Answer

E2 (AMC Junior style). Two independent items each have a chance of being faulty. Find , as a percentage.

Answer

E3 (Challenge). Two independent, identical trials give . Find the probability of success on a single trial.

Answer

E4 (Investigation). A pair of independent trials each have . Compare for trials with for trials. Describe, without further calculation, what you expect to happen as the number of trials keeps increasing, and why.

Answer

As the number of trials keeps increasing, keeps shrinking (since it is raised to an ever-higher power), so keeps climbing towards, but never quite reaching, .

Homework

1. . Find .
2. Two fair coins are tossed. Find using the complement method.
3. Two independent students each have a chance of forgetting their homework. Find .
4. Of a company’s staff, . Estimate how many do not drive to work.
5. A three-outcome spinner has and . Find and .
6. Explain, in one sentence, why for two independent trials is found using , not .
7. Reasoning. A student calculates as . Explain what is wrong with this method and give the correct answer.
8. Reasoning. Explain why the complement method for “at least one” only works cleanly when the trials are independent, using an example of two events that are not independent (e.g. drawing two cards from a deck without replacement).
9. Challenge. Two independent machines each have the same unknown probability of producing a faulty item. If , find correct to decimal places.

Answers: Q1 — . Q2 — , so . Q3 — , so . Q4 — ; . Q5 — ; . Q6 — because “at least one” covers every outcome except “none at all”, so its probability is most directly found as minus , rather than trying to separately count every way to get exactly one, exactly two, and so on. Q7 — this method wrongly adds probabilities for events that are not mutually exclusive (both dice could show a at once) and does not represent “at least one” correctly; the correct method is , so . Q8 — if two cards are drawn from a deck without replacement, the probability of the second event depends on the outcome of the first (e.g. one fewer card of a given suit remains), so the probabilities cannot simply be multiplied as without adjusting for this dependency; with independent trials (like separate dice or separate machines), each trial’s probability is unaffected by the other, which is what allows simple multiplication. Q9 — .