This mini-lesson covers AQA section M — Probability: mutually exclusive and independent events, the addition and multiplication rules, set notation, Venn diagrams, two-way tables, tree diagrams, and conditional probability with the formula P(A|B) = P(A ∩ B) ÷ P(B).
A ∩ B = both happen · A ∪ B = at least one happens · A′ = A does not happen.
Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. You will also be asked to critique the assumptions in a probability model. Press Start when you're ready.
Section M1 · set notation
The language of events
Events are sets of outcomes inside the sample space ξ.
A ∩ B — the intersection: A and B both occur.
A ∪ B — the union: A or B (or both) occurs.
A′ — the complement: A does not occur. P(A′) = 1 − P(A).
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)the addition formula — true for ANY two events (subtract the overlap once)
Mutually exclusive events cannot both happen, so P(A ∩ B) = 0 and the formula collapses to P(A ∪ B) = P(A) + P(B).
Independent events do not affect each other's probability, so P(A ∩ B) = P(A) × P(B), equivalently P(A | B) = P(A).
Classic trap: mutually exclusive and independent are not the same — in fact, if P(A) and P(B) are both non-zero, mutually exclusive events are necessarily dependent (if A happens, B definitely cannot).
Calculate
Independent events
1Events A and B are independent, with P(A) = 0.4 and P(B) = 0.5. Find P(A ∩ B).
Hint: For independent events P(A ∩ B) = P(A) × P(B) = 0.4 × 0.5.
Calculate
The addition formula
2Using the same events (P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2), find P(A ∪ B).
Hint: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.4 + 0.5 − 0.2.
Quick check
Mutually exclusive
?Two events A and B are mutually exclusive. Which statement must be true?
Quick check
Testing for independence
?P(A) = 0.3, P(B) = 0.4 and P(A ∩ B) = 0.15. Are A and B independent?
Section M2 · conditional
Conditional probability
P(A | B) is the probability of A given that B has already happened. Restricting attention to the B outcomes only:
P(A | B) = P(A ∩ B) ÷ P(B)rearranged: P(A ∩ B) = P(B) × P(A | B) — the multiplication rule
Without replacement the second branch probabilities change — that is conditional probability in action.
Tree rules: multiply along the branches, add between them. Check: 20 + 15 + 15 + 6 = 56 ✓
Calculate
Conditional probability
3Events A and B satisfy P(A ∩ B) = 0.12 and P(B) = 0.3. Find P(A | B).
Hint: P(A | B) = P(A ∩ B) ÷ P(B) = 0.12 ÷ 0.3.
Calculate
Tree diagram: both red
4A bag holds 5 red and 3 blue counters. Two are taken without replacement. Find P(both red), to 3 decimal places.
Hint: P(RR) = (5/8) × (4/7) = 20/56.
Calculate
Exactly one red
5Same bag (5 red, 3 blue, without replacement). Find P(exactly one red), to 3 decimal places.
Tap a probability statement, then tap the condition under which it holds.
🚫 Mutually exclusive only
🔗 Independent only
♾️ True for ANY two events
Quick check
Venn diagram counting
?In a class of 50 students, 22 study French, 18 study German and 7 study both. How many study neither?
Match it
Read the notation
Tap a symbol on the left, then its meaning in words on the right.
Notation
In words
Quick check
Critique the model
?A model assumes that whether it rains today and whether it rains tomorrow are independent events. What is the best critique?
Section M2 · two-way tables
Conditional probability from a two-way table
A two-way table is often the fastest route to a conditional probability: the condition just tells you which row or column to look in.
100 people surveyed. Every conditional probability here is just "one cell ÷ one total".
Worked example
P(bike | car) — restrict to the car row (60 people): 24 ÷ 60 = 0.4
P(car | bike) — restrict to the bike column (44 people): 24 ÷ 44 = 0.545 (3 s.f.)
These are different — P(A | B) ≠ P(B | A). Confusing the two is a classic error.
Check for independence: P(bike) = 44/100 = 0.44 but P(bike | car) = 0.4. They are not equal, so owning a car and owning a bike are not independent in this survey.
Calculate
Conditional from a table
6Using the table above (24 own both, 44 own a bike in total), find P(owns a car | owns a bike), to 3 decimal places.
Hint: Restrict to the bike column: 24 of the 44 bike-owners also own a car, so 24 ÷ 44.
Quick check
At least one
?A bag holds 5 red and 3 blue counters; two are taken without replacement. What is P(at least one red)?
Recap
The big ideas to know
Notation: ∩ = and · ∪ = or · A′ = not A · P(A′) = 1 − P(A)
Addition formula (any events): P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Mutually exclusive: P(A ∩ B) = 0 · Independent: P(A ∩ B) = P(A) × P(B)
Conditional: P(A | B) = P(A ∩ B) ÷ P(B); independence ⇔ P(A | B) = P(A)
Trees: multiply along branches, add between them; without replacement, the second branches change
Modelling: always state and critique your assumptions
That is the whole of section M — Probability for AQA A-level Mathematics (7357). Press Finish to see your score.
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Mini-lesson complete!
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