At a glance. One lesson of about 60 minutes. It follows the axioms and the addition rule. Students test for independence by calculation, solve for an unknown probability, contrast independent with mutually exclusive, and prove a result about complements.
Calculator: no calculator. No calculator is needed. Products of one-decimal and two-decimal numbers are within reach, and the lesson is about the structure of the test.
What students will learn
Know
- A and B are independent exactly when P(A ∩ B) = P(A) × P(B).
- Mutually exclusive means P(A ∩ B) = 0.
- If A and B are independent then so are A and B′, A′ and B, and A′ and B′.
Understand
- Independence is a statement about numbers that can be checked, not a feeling that events are unrelated.
- Mutually exclusive events are the opposite of unrelated: one happening rules the other out.
- Independence is a symmetric property of the two events and of their complements.
Do
- Test two events for independence from given probabilities.
- Find an unknown probability for independent events.
- Prove that complements of independent events are independent.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: using the product test and the addition rule (questions 1 to 6).
- Criterion B. Investigating patterns: the figure and the what-if. Students find when the overlap equals the product.
- Criterion C. Communicating: a proof with each step justified (worked example 3 and the faded example).
- Criterion D. Applying mathematics in real-life contexts: the sleep and screens hook, which is a question about whether two facts about students are related.
The one thing to do with the figure
Ask for a prediction before every move. Before anyone reads the readout, ask: “How big would the overlap have to be for these to be independent?” and let students slide to it.
Then set P(A) to 50%, P(B) to 30% and the overlap to 0%, and ask: “Are mutually exclusive events independent?” Students should see that the product is not 0 and say why.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 5 min | Hook: sleep and screens. Collect predictions for the overlap. |
| 12 min | The figure with predictions; the key idea; settle the hook. |
| 10 min | Warning card and worked examples; pairs finish the faded example. |
| 5 min | Question 1, the check. If many choose C or B, return to the 0% state in the figure. |
| 18 min | Your turn, Levels 1 to 3, at each student's own pace. |
| 5 min | Something to think about, then the exit ticket, question 8. |
The answers
| Warm-up | 0.9; 1/4; p − pq. |
|---|---|
| Hook | 12%, because 0.3 × 0.4 = 0.12. An overlap larger than 12% means short sleep and late screens go together more than chance gives. |
| Worked examples | Independent (0.06 = 0.06); not independent (0 ≠ 0.06); P(A ∩ B′) = P(A) × P(B′). |
| Faded example | P(A′ ∩ B) = P(B) × (1 − P(A)) = P(A′) × P(B). |
| 1. Testing for independence | A. B confuses small overlap with mutually exclusive; C says overlap means dependent; D uses the sum. |
| 2. Intersection of independent events | 0.27. 0.45 × 0.6. |
| 3. A missing probability for independent events | 0.7. 0.6p = 0.42. |
| 4. How far from independent | 0.05. 0.3 − 0.25. |
| 5. A complement with independent events | 0.56. 0.7 × 0.8. |
| 6. Working back from neither | 0.6. 1 − 0.4. |
| 7. Both mutually exclusive and independent | 0. 0 = 0.5 × P(B). |
| 8. Independent pair on a die | C. A: 1/3 against 1/4; B: 1/6 against 1/12; D: mutually exclusive. |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| B, C or D in question 1 | Said small overlap means mutually exclusive, said overlap means dependent, or compared with the sum. | Write the product and the overlap on separate lines and compare. |
| 1.05, 0.15 or 0.78 in question 2 | Added, subtracted, or gave the union. | Say that independence multiplies. |
| 0.42 or 1.05 in question 3 | Ignored the overlap, or treated the extra as the overlap. | Write the addition rule with the product in it. |
| 0.25 or 0.3 in question 4 | Gave the overlap or the product, not the gap. | Find both numbers, then subtract. |
| 0.24, 0.5 or 0.14 in question 5 | Used A instead of A′, subtracted, or complemented the wrong event. | Underline which events are complemented. |
| 0.4 or 0.42 in question 6 | Gave P(B′), or assumed no overlap. | Write P(A′ ∩ B′) = P(A′) × P(B′). |
| 0.5 or 0.25 in question 7 | Assumed P(B) is positive, or computed a product. | Write both properties as P(A ∩ B) and equate. |
| A, B or D in question 8 | Compared the wrong numbers, or used a mutually exclusive pair. | List each event as a set and count. |
Questions worth asking
- If two events are independent, are their complements? Yes: the proof in worked example 3 gives A and B′, and the same argument gives the others.
- Can independent events be mutually exclusive? Only if one of them is impossible, because P(A) × P(B) = 0 forces one of them to be 0.
- How would you show events are dependent in a real survey? Compare the overlap share with the product of the two shares; a gap suggests a relationship, which still needs more than the data to call a cause.
- Why can't you decide independence by looking at the diagram alone? The areas must be compared with a calculation; sets can look unrelated and still not satisfy the test.
Differentiation
A way in for anyone stuck
- Provide the test as a frame: P(A ∩ B) = …, P(A) × P(B) = …, equal or not equal, so ….
- Draw a square split into a strip for A and a column for B, and ask students to shade the overlap that chance would give.
- Use whole-number counts out of 100 before moving to decimals.
Stretch
- Show that if A and B are independent, then P(A ∪ B) = 1 − P(A′) × P(B′).
- Three events A, B and C are pairwise independent but P(A ∩ B ∩ C) ≠ P(A) × P(B) × P(C). Find an example with two coin flips, where C is “the two coins match”.
Students learning in English as an additional language
“Independent” means that one event does not change the chance of the other. “Mutually exclusive” means that the events cannot both happen. These are different. Say “P of A and B” for P(A ∩ B), and “the product” for the result of multiplying.
| Word | Plain meaning | Example |
|---|---|---|
| independent events | events where one happening does not change the chance of the other | a head on one coin and a head on a second coin |
| mutually exclusive | events that cannot both happen | a die showing 1 and the same die showing 6 |
| dependent events | events that are not independent | drawing two cards from one pack without putting any back |
| product | the result of multiplying | 0.8 × 0.4 = 0.32 |
| overlap | the outcomes that are in both events | A ∩ B |
| test for independence | comparing P(A ∩ B) with P(A) × P(B) | 0.08 against 0.2 × 0.3 |
Sentence starters for explaining aloud or in writing:
- P(A ∩ B) = … and P(A) × P(B) = …, so the events are …
- They cannot be independent because …
- If A has happened, the chance of B …
Two things not to say
“Independent means they never happen together.” That is mutually exclusive, the opposite idea.
“They overlap, so they are dependent.” Independent events with positive probabilities always overlap.
Questions to set
Extra practice in three tiers, including one question from earlier work for spacing. Every answer has been checked.
1Fluency
The idea on its own, with friendly numbers. Set these first.
- TestP(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.25. Independent?
No: 0.4 × 0.5 = 0.2, not 0.25. - Earlier work (lesson 3)P(A) = 0.35. Find P(A′).
0.65.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- ContextIn a class, P(has a bicycle) = 0.7 and P(walks to school) = 0.4. Two students are chosen independently. What is the probability that the first has a bicycle and the second walks?
0.7 × 0.4 = 0.28. - Missing valueA and B are independent, P(B) = 0.25 and P(A ∩ B) = 0.1. Find P(A).
0.1 ÷ 0.25 = 0.4.
3Challenge
Reasoning, generalizing and working backwards.
- ReasoningProve that if A and B are independent then A′ and B′ are independent.
P(A′ ∩ B′) = 1 − P(A ∪ B) = 1 − P(A) − P(B) + P(A)P(B) = (1 − P(A))(1 − P(B)) = P(A′)P(B′). - CheckTwo coins are flipped. A: the first is heads. B: the two coins match. Are A and B independent?
P(A) = 1/2, P(B) = 1/2, P(A ∩ B) = 1/4 = 1/2 × 1/2. They are independent, even though B depends on both coins.
Practicalities
Works on a phone or a shared screen. Nothing a student types is saved or sent anywhere.
The figure works with the keyboard (tab to a slider, then the arrow keys) and by touch. “Show me the steps” gives the same states in words.
No calculator is needed. Use class data only if it is gathered anonymously.