At a glance. One lesson of about 60 minutes. It assumes the counting rule for sets and the complement rule used informally. Students state the three axioms, prove the complement and addition rules with a reason on each line, deduce P(∅) = 0 and P(A) ≤ 1, and use the rules to bound probabilities.
Calculator: no calculator. No calculator is needed. The numbers are short decimals, and the focus is on logical steps and reasons.
What students will learn
Know
- The axioms: P(A) ≥ 0, P(U) = 1, and P(A ∪ B) = P(A) + P(B) for mutually exclusive events.
- The complement rule P(A′) = 1 − P(A), P(∅) = 0 and 0 ≤ P(A) ≤ 1 are theorems or corollaries.
- The addition rule P(A ∪ B) = P(A) + P(B) − P(A ∩ B) is proved by splitting into disjoint parts.
Understand
- Probability is an axiomatic system: a few accepted statements give every rule by logic.
- A proof is a chain in which each line is an axiom, a definition or an earlier result.
- The axioms also restrict which collections of numbers can be probabilities.
Do
- Name the axiom or theorem that justifies each line of a proof.
- Prove a simple probability result from the axioms.
- Decide whether given probabilities can exist, with a reason.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: using the complement rule and the addition rule to find probabilities (questions 1 to 5).
- Criterion B. Investigating patterns: the figure and the what-if. Students find the limits on P(A ∩ B) from the axioms.
- Criterion C. Communicating: a proof with a reason for each line, in correct notation (worked examples 1 to 3 and the faded example).
- Criterion D. Applying mathematics in real-life contexts: the weather hook, where a quoted chance is more than 100%.
The one thing to do with the figure
Ask for a prediction before every move. Set P(A ∩ B) to 0% first and ask students to raise P(A) and P(B) until a model fails. Ask which region goes negative and which axiom that breaks.
Then ask: “What is the smallest P(A ∩ B) can be when P(A) = 80% and P(B) = 50%?” and have students find it by sliding.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 5 min | Hook: the forecast. Collect predictions of the rule being used. |
| 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 B or C, ask what line 2 says about the size of P(A ∩ B). |
| 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 | 5/6; 19; x = 0.4. |
|---|---|
| Hook | The forecaster added the probabilities of two events that can both happen, so weekends with rain on both days were counted twice. The missing information is P(rain on both days). The least chance of some rain is 70%, when it rains on Sunday only in weekends when it also rains on Saturday (both days 50%); the greatest is 100%, when both days is 20%. |
| Worked examples | P(A′) = 1 − P(A) and P(A) ≤ 1; P(∅) = 0; the addition rule. |
| Faded example | P(B) = P(A) + P(A′ ∩ B) ≥ P(A) by axioms 3 and 1. |
| 1. Naming the axiom used | A. B fixes the whole; C assumes the events cannot overlap; D gives no sign on its own. |
| 2. Complement rule with an unknown | 0.18. 5x + 0.1 = 1. |
| 3. Probability of neither | 0.34. 1 − 0.66. |
| 4. A part of an event from the addition rule | 0.3. 0.5 − 0.2. |
| 5. Mutually exclusive with an unknown | 0.25. 3p = 0.75. |
| 6. The least possible intersection | 0.5. 0.7 + 0.8 − 1. |
| 7. A quadratic from the axioms | 0.2. 2x² + x − 0.28 = 0. |
| 8. Which probabilities cannot exist | B. A, C and D all have a union of at most 1 and an intersection no larger than either event. |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| B, C or D in question 1 | Named the axiom about the whole, assumed the events cannot overlap, or reached for a theorem. | Ask what line 2 says about the size of P(A ∩ B), and which axiom says that. |
| 0.2 or 0.22 in question 2 | Left out the 0.1, or added it instead of taking it from 1. | Write P(A) + P(A′) = 1 with every term. |
| 0.17 or 0.66 in question 3 | Subtracted both probabilities from 1, or gave the union. | Find the union first, then say the answer is its complement. |
| 0.2, 0.5 or 0.4 in question 4 | Gave the overlap, the whole of B, or P(A′). | Draw A and B and shade B but not A. |
| 0.6 or 0.28 in question 5 | Gave P(A), or left out the 0.1. | Write 2p + 0.1 + p = 0.85. |
| 0, 0.7 or 0.1 in question 6 | Used mutually exclusive, the greatest value, or the gap. | Use P(A ∪ B) ≤ 1 and rearrange. |
| −0.7 or 0.4 in question 7 | Kept the negative root, or divided by 2 in the formula. | Check the root against axiom 1, and use 2a on the bottom. |
| A, C or D in question 8 | Thought a pair of equal events, a disjoint pair, or a small intersection was impossible. | Compute the union and compare the intersection with each event. |
Questions worth asking
- Why is P(∅) = 0 not an axiom? It follows from the axioms: ∅ = U′, so P(∅) = 1 − P(U) = 0.
- Why must the two parts in the addition proof be mutually exclusive? Axiom 3 only allows adding probabilities of events that cannot happen together.
- Can P(A ∩ B) exceed P(A)? No: A ∩ B ⊆ A, so by the faded example P(A ∩ B) ≤ P(A).
- What is the difference between an axiom and a theorem? An axiom is accepted; a theorem is proved from axioms and earlier results.
Differentiation
A way in for anyone stuck
- Give the three axioms on a card, numbered, and ask students to write the number beside each line of the proof.
- Draw the bar of the figure on paper and have students colour the four regions before writing any algebra.
- Provide proof frames with the reasons given and the lines missing, then the other way round.
Stretch
- Prove that P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C) using the two-event addition rule.
- Prove that P(A ∪ B ∪ C) ≤ P(A) + P(B) + P(C) for any three events, using the two-event result twice.
Students learning in English as an additional language
“Axiom” means a statement accepted without proof. “Theorem” means a statement that has been proved. “Mutually exclusive” means that the events cannot happen at the same time. “Deduce” means to get a new result from earlier ones by reasoning.
| Word | Plain meaning | Example |
|---|---|---|
| axiom | a statement accepted without proof | P(A) ≥ 0 |
| theorem | a statement proved from the axioms and earlier theorems | the addition rule for any two events |
| corollary | a result that follows from a theorem in a step or two | P(∅) = 0, from the complement rule |
| mutually exclusive | events that cannot happen at the same time | rolling a 1 and rolling a 6 on one throw |
| proof | a chain of steps, each justified, from the axioms to the result | the proof of the complement rule |
| sample space | the set of all possible outcomes, written U | {H, T} for a coin |
Sentence starters for explaining aloud or in writing:
- This line is true because … (axiom 1, 2 or 3).
- The two events are mutually exclusive because …
- These values cannot be probabilities because the … would be negative.
Two things not to say
“Just add the probabilities.” Only for mutually exclusive events.
“The rule is true because the diagram shows it.” A diagram suggests a rule; the axioms prove it.
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.
- ComplementP(rain) = 0.35. Find P(no rain).
0.65. - Earlier work (lesson 1)Simplify (A ∩ B′)′.
A′ ∪ B.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- ContextIn a class, P(plays football) = 0.44, P(plays hockey) = 0.31 and P(plays neither) = 0.37. Find P(plays both).
P(at least one) = 0.63, so P(both) = 0.44 + 0.31 − 0.63 = 0.12. - TestCan P(A) = 0.9, P(B) = 0.8 and P(A ∩ B) = 0.6 all be true? Give a reason.
No: the union would be 0.9 + 0.8 − 0.6 = 1.1, which is more than 1.
3Challenge
Reasoning, generalizing and working backwards.
- ProofProve that P(A ∩ B′) = P(A) − P(A ∩ B).
A = (A ∩ B) ∪ (A ∩ B′), and the two parts are mutually exclusive, so P(A) = P(A ∩ B) + P(A ∩ B′) by axiom 3. - BoundP(A) = 0.55 and P(B) = 0.65. State the greatest and least values of P(A ∪ B).
Greatest 1, when the union is everything; least 0.65, when A ⊆ B.
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. Index cards with the three axioms help in the proof section.