121i - Problem Solving with Complementary Events

Lesson 121 — Problem Solving and Consolidation: Complementary Events

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

Learning Intentions

• To consolidate that complementary events have a combined probability of one.
• To apply efficiently to solve “at least one” and “none” problems in applied contexts.

Success Criteria

I can:

1. State the complement of an event in words and using .
2. Calculate given , and vice versa.
3. Recognise when a problem is solved more efficiently using the complement, especially “at least one” problems.
4. Solve multi-step applied problems using complementary events, showing full working.

Warmup

(6 minutes — rapid-fire, mini whiteboards)

1. If , what is ?
2. Two coins are tossed. List all four outcomes. What is the complement of “at least one head”?
3. A bag contains only red and blue counters. . What is ?
4. True or false: “The complement of rolling a on a die is rolling a .” Explain.

Answers: 1. ; 2. — the complement of “at least one head” is “no heads”, i.e. ; 3. ; 4. False — the complement of “rolling a ” is “not rolling a ”, i.e. rolling or . A complement must cover every remaining outcome, not just one of them.

Activities

Activity 1 — Explicit Consolidation: the “at Least one” Shortcut (14 min)

I do / We do / You do.

I do: A fair coin is tossed times. Find .

Direct method (inefficient): list all outcomes, count the containing at least one head, giving .

Complement method:

Say aloud: “The complement of at least one is always none. There is only ever one way to get ‘none’, so this is almost always the faster route.”

We do: A die is rolled twice. Find .

You do:

1. A -sector spinner (numbered –) is spun twice. Find .
2. A factory tests independent items; each has a chance of being faulty. Find .

Activity 2 — Applied Problems (19 min)

Pairs. Every answer must show full working.

Problem 1. A bag contains red, blue and green counters ( total). One counter is drawn at random. Find .

Problem 2. A football team wins each match with probability , independently of other matches. Find the probability the team wins at least one of its next matches.

Problem 3 (harder). A vaccine causes a mild side effect in of patients, independently of one another. In a group of patients, find the probability that at least one experiences the side effect, correct to decimal places. Then explain why the assumption of independence might fail in practice, and give an example of when it would.

Socratic scaffolding for Problem 3:

Prompt	Purpose
Understand: what is being asked?	The probability that at least one of the patients has the side effect, not exactly one.
What is the complement of “at least one”?	”None of the patients” experiences the side effect.
Devise a plan	Find , then subtract from .
What is for one patient?	.
Carry out the plan for	.
Compute it	.
So the final answer?	.
Looking back — is this sensible?	With independent trials of a not-so-rare event, a probability near one-half is believable — it is not , which would double-count patients who might have overlapping “no side effect” outcomes.
Looking back — independence?	If patients were family members sharing genetics or a shared exposure, one person’s reaction could make a relative’s reaction more likely, breaking independence and making the multiplication invalid.

Answers: P1: . P2: , so . P3: ; independence can fail for related or co-located individuals who share risk factors.

Checks for Understanding

(6 minutes — exit ticket, collected)

1. If , find .
2. A coin is tossed times. Find .
3. Three fuses each independently have a chance of blowing when a circuit is overloaded. Find , correct to decimal places.
4. Explain why computing by listing outcomes would be inefficient, and describe the complement method instead.

Answers: 1. ; 2. , so ; 3. , so ; 4. Listing directly needs checking of the outcomes; the complement needs only the single “all tails” outcome, giving instantly.

Common Misconceptions

Misconception	How to pre-empt it
”At least one” means “exactly one”.	Contrast directly: “at least one head in 3 tosses” includes , , etc. — only is excluded.
whenever there are two named events.	Insist events must be exhaustive and mutually exclusive to be true complements. Test with the warmup Q4 counterexample.
Complementing a compound event term-by-term, e.g. treating “not (A and B)” as “not A and not B”.	Anchor “at least one” firmly to its true complement, “none”, using the coin-toss diagram every time.
Adding individual probabilities instead of multiplying to find across independent trials.	Revisit the “we do” die example: , not .
Assuming independence always holds in real-world “at least one” contexts.	Problem 3’s reasoning question — require students to name a concrete reason independence could fail.

Enrichment — Competition-Style Problems

E1 (AMC Junior style). A fair die is rolled twice. Find the probability that the sum of the two rolls is not .

Answer

There are outcomes summing to out of : . So .

E2 (Kangaroo style). A biased coin lands heads with probability . It is tossed twice. Find .

Answer

E3 (Challenge). A -digit PIN uses digits –, chosen independently and uniformly (repeats allowed) by a lock’s owner. An attacker enters one random -digit guess. Find . If the attacker gets independent guesses (with replacement), find , correct to decimal places.

Answer

.

Homework

1. . Find .
2. A spinner has equal sectors: red, red, blue, green, yellow. Find .
3. A fair coin is tossed times. Find .
4. A factory has independent machines, each with an chance of a defect on a given day. Find today, correct to decimal places.
5. Reasoning. Explain why only works when "" and “not ” cover every possible outcome with no overlap. Give an example of two events that seem opposite but are not true complements.
6. Challenge. A single random digit guess (–) is made against a secret digit. An attacker gets independent guesses (repeats allowed). Find , correct to decimal places. Then, by trial, find the smallest number of guesses needed for this probability to first exceed .

Answers: 1. . 2. , so . 3. . 4. , so . 5. Complements require the two events to together cover all outcomes with no overlap — e.g. for a die, “greater than 4” () and “less than 3” () look opposite but leave out ; their probabilities sum to , so they are not complements. 6. ; . Checking successive powers, (probability , not yet over ) and (probability ) — so 7 guesses are needed.