Mini-Lesson

Probability

This mini-lesson covers sample spaces and Venn diagrams, the addition rule, mutually exclusive and independent events, tree diagrams (with and without replacement) and conditional probability P(A|B).

Where this sits: Unit AS 2: Applied Mathematics — probability: mutually exclusive and independent events, Venn diagrams and tree diagrams. Unit A2 2 extends this to conditional probability and set notation.

Work through each screen, answer the questions as you go (some are reasoning, some are calculations) and collect ⭐ stars. Press Start when you are ready.

Foundations

Sample space, Venn diagrams and the addition rule

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)subtract the overlap, or you count it twice · P(A′) = 1 − P(A)

A ∪ B (union) = "A or B or both".
A ∩ B (intersection) = "A and B".
A′ (complement) = "not A".

Draw the Venn diagram and fill it from the middle outwards — put P(A ∩ B) in first, then subtract to get the parts of A and B that do not overlap.

Special cases

Mutually exclusive and independent

mutually exclusive: P(A ∩ B) = 0 ⇒ P(A ∪ B) = P(A) + P(B)independent: P(A ∩ B) = P(A) × P(B) — one event does not affect the other

These are different ideas and are often confused. Mutually exclusive events cannot happen together (rolling a 2 and a 5 on one die). Independent events can happen together — knowing one occurred simply does not change the chance of the other (two separate coin tosses). In fact, two events with non-zero probability that are mutually exclusive are never independent.

Worked example

P(A) = 0.4, P(B) = 0.5, independent.

P(A ∩ B) = 0.4 × 0.5 = 0.2

P(A ∪ B) = 0.4 + 0.5 − 0.2 = 0.7

Calculate

Independent events

1A 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

Addition rule

2With P(A) = 0.4, P(B) = 0.5 and 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.

A2 skill

Conditional probability

P(A | B) = P(A ∩ B) / P(B)"the probability of A GIVEN that B has happened" — B becomes the new sample space

Worked example — a two-way table

Of 60 students, 25 study Maths, 20 study Physics and 10 study both.

P(Maths | Physics) = P(M ∩ P)/P(P) = (10/60) ÷ (20/60) = 10/20 = 0.5

In words: of the 20 physicists, 10 also do Maths.

Rearranged: P(A ∩ B) = P(A|B) × P(B). And A and B are independent exactly when P(A|B) = P(A) — knowing B tells you nothing about A.

Calculate

Conditional probability

3Of 60 students, 25 study Maths, 20 study Physics and 10 study both. Find P(Maths | Physics).

Hint: Restrict to the 20 physicists: 10 of them also do Maths, so the probability is 10 ÷ 20.

Tree diagrams

With and without replacement

On a tree diagram you multiply along branches and add between paths. The crucial question is whether the item is replaced:

With replacement — the probabilities on the second set of branches are unchanged (independent).
Without replacement — the second set of probabilities changes, because both the numerator and the total have gone down (conditional).

Worked example — 5 red, 3 blue, two drawn without replacement

P(both red) = (5/8) × (4/7) = 20/56 = 0.357 (3 d.p.)

P(at least one blue) = 1 − P(both red) = 1 − 0.357 = 0.643

Calculate

Without replacement

4A bag holds 5 red and 3 blue counters. Two are drawn without replacement. Find P(both red) to 3 decimal places.

Hint: First red: 5/8. Then only 4 red of 7 remain: 4/7. Multiply: (5/8) × (4/7) = 20/56.

Calculate

At least one

5From the same bag (5 red, 3 blue, two drawn without replacement), find P(at least one blue) to 3 decimal places.

Hint: The complement of 'at least one blue' is 'both red', so P = 1 − 20/56 = 36/56.

Sort it

Mutually exclusive, independent or neither?

Tap a pair of events, then tap what they are.

🚫 Mutually exclusive

🔗 Independent

❓ Neither

Quick check

Not the same thing

?Events A and B both have non-zero probability and are mutually exclusive. Are they independent?

Quick check

Reading a tree

?Two counters are drawn without replacement from 5 red and 3 blue. What is the probability on the second branch for 'red' after a first red?

Match it

Match the probability rule

Tap the rule on the left, then its formula on the right.

Statement

Answer

Quick check

Complement trick

?A fair coin is tossed 4 times. What is P(at least one head)?

Quick check

Conditional in context

?In a class, P(late) = 0.2 and P(late and cycles) = 0.08. Find P(cycles | late).

Examiner traps

Probability pitfalls

Mutually exclusive vs independent — these are different and (for non-zero probabilities) incompatible.
Forgetting to subtract the overlap: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
Without replacement: BOTH the numerator and the denominator drop on the second draw.
Conditional probability: the given event goes on the bottom.
"At least one" is nearly always 1 − P(none).

Recap

The big ideas to know

Addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) · P(A′) = 1 − P(A)

Mutually exclusive: P(A ∩ B) = 0 — they cannot happen together

Independent: P(A ∩ B) = P(A)P(B) — one does not affect the other

Conditional: P(A|B) = P(A ∩ B)/P(B) — the condition goes on the bottom

Trees: multiply along, add between; without replacement the second probabilities change

'At least one' = 1 − P(none)

Probability underpins the binomial and normal distributions and every hypothesis test. Press Finish to see your score.

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Mini-lesson complete!

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