IB Maths AA: Probability and Expectation – Tutor Lesson Plan (independence, conditional probability, variance)
- 2026-11-05
Students know that independent events multiply, so when a question says given B occurred they multiply anyway. One value of 0.21 carries three different questions here.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Multiplying decimals | What is 0.4 times 0.7? | 0.28 |
| Fraction to decimal | What is 21 divided by 70? | 0.3 |
| Averages | The mean of 0, 1, 2 and 3 | 1.5 |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-8 min | Three question types | Both, given, and expected value | State how they differ |
| 8-20 min | Independent events | P(A) 0.4, P(B) 0.7 | Get 0.28 |
| 20-34 min | Conditional probability | P(A and B) 0.21, P(B) 0.7 | Get 0.3 |
| 34-48 min | Expectation | probabilities 0.2, 0.3, 0.4, 0.1 over 0 to 3 | Get 1.4 |
| 48-56 min | Variance | E(X squared) minus E(X) squared | Get 0.84 |
| 56-60 min | Exit check | Three questions marked | Finish alone |
03 Teaching script (read it out)
Scene 1: Multiply only after independence is established
Worked the other way: a question that gives P(A and B) = 0.21 while P(A)P(B) = 0.28 has already shown the events are not independent.
Student judges what the gap between 0.21 and 0.28 means (the events are linked).
Scene 2: Conditional probability changes the denominator
The usual slip is dividing by 0.4. The denominator is always the event that is known to have happened, not the one being asked about.
Student explains why the denominator is 0.7.
Scene 3: Expectation and variance from one table
E(X squared) = 0 + 0.3 + 1.6 + 0.9 = 2.8, so Var = 2.8 – 1.96 = 0.84.
Note: variance is not E(X squared); the square of the mean must be subtracted.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Independent events | P(A) 0.4, P(B) 0.7 | Multiply directly | 0.28 |
| Conditional probability | P(A and B) 0.21, P(B) 0.7 | Intersection over the given event | 0.3 |
| Expectation | table over 0 to 3 | Sum of x times P(x) | 1.4 |
| Variance | E(X squared) 2.8, E(X) 1.4 | E(X squared) minus E(X) squared | 0.84 |
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Probability of both, independent | 0.28 | Independence checked first |
| Value of P(A given B) | 0.3 | Denominator is 0.7, not 0.4 |
| Mean and variance | mean 1.4, variance 0.84 | The mean squared was subtracted |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Wrong denominator | answer of 0.525 | The denominator is the event known to have happened. |
| Multiplying dependent events | answer of 0.28 | The question already gives 0.21, so 0.28 contradicts it. |
| Variance left as E(X squared) | variance of 2.8 | Variance equals E(X squared) minus the mean squared. |
| Probabilities not summing to 1 | table totals 0.9 | Check the total before computing anything. |
07 Homework (paper, section and question numbers)
- IB maths chapter index — Use the three questions on this page
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| P(A and B) when independent | 0.28 | All three right: move to the normal distribution |
| P(A given B) | 0.3 | One wrong: re-teach conditional probability |
| Mean and variance | 1.4 and 0.84 | Two wrong: repeat scene 3 |
09 Tutor FAQ
Are probability formulas in the booklet?
Some are, but the definition of conditional probability must be reproduced.
Variance or standard deviation?
The standard deviation is the square root, in the same units as the data.
Memorise the table method?
No, but always check that the probabilities sum to 1.
Find a tutor — freeTutor Registration →
A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.


