Scope and prerequisites
AQA A-level Mathematics 7357, section M (Probability), version 1.3 (first assessment 2018). Original teaching notes; objective-level mapping lives in content/teaching/aqa_a_level_mathematics/objective-review.yaml.
Prerequisites: GCSE probability trees and Venn diagrams. Each lesson develops its own method before independent transfer.
These are original teaching examples, not official AQA material.
Probability 概率
What would explain this observation?
- Two counters leave a bag of 3 red and 2 blue, one after another. The second draw's probabilities are different from the first — is that an error or the point?
- Start with a prediction. State what "without replacement" changes, and how a tree shows it.
Build the model
- Multiply along a tree branch and add disjoint branches for one outcome. Conditional probability 条件概率 restricts the sample space: P(A given B) = P(A ∩ B)/P(B) for P(B) > 0. Without replacement, the composition changes for the next draw.
- Mutually exclusive 互斥 means no overlap; independent 独立 means knowing one event does not change the other's probability. Two disjoint events with positive probability are never independent: if A happens, P(B) drops to 0.
- conditional probability: The probability of an event within the group selected by given information; mutually exclusive: Events that cannot happen together; independent: Events where knowing one leaves the other\u2019s probability unchanged.
Choose evidence that can test it
- Fill a Venn diagram from the OVERLAP outward: with 18 cycling, 12 swimming and 7 doing both, the only-regions are 11 and 5, so at least one is 23 and neither is 7 of 30. Adding 18 + 12 double-counts the 7.
- A conditional probability uses the conditioned group's total as its denominator: P(swim given cycle) = 7/18, not 7/30. Reversing the condition usually changes the value.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: 3 red and 2 blue, two draws without replacement. P(two red) = 3/5 × 2/4 = 3/10. P(one of each) = 3/5 × 2/4 + 2/5 × 3/4 = 3/10 + 3/10 = 6/10. The branches multiply and the disjoint routes add.
Example:
Given P(A ∩ B) = 0.12 and P(B) = 0.3, find P(A given B) using the same sequence: identify the conditioned group → divide → interpret. (Answer: P(A given B) = 0.12/0.3 = 0.4. The denominator is B's probability, not the whole space.)
Check the conclusion and its limits
- Independence and exclusivity answer different questions: disjoint events with positive probability are always dependent. Expected counts (np) are long-run averages, not guarantees for a single run.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Warn:
Mutually exclusive events with positive probabilities must be independent. This claim is false: if A and B cannot happen together, then P(B given A) = 0 ≠ P(B) — knowing A happened changes B's probability completely, the opposite of independence.
Key:
Probability: multiply along branches, add disjoint routes; conditional probability divides by the conditioned group; without replacement the probabilities evolve; disjoint with positive probability means dependent, never independent.