Mini-Lesson

Probability

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.

Hint: P(RB) + P(BR) = (5/8)(3/7) + (3/8)(5/7) = 15/56 + 15/56 = 30/56.

Sort it

When is it true?

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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