Mini-Lesson

Probability

This mini-lesson covers Topic 3 of Edexcel A-level Mathematics (9MA0): set notation and Venn diagrams, mutually exclusive and independent events, the addition formula, conditional probability and its formula, tree diagrams and two-way tables, and modelling with probability — including criticising the assumptions a model makes.

Everything inside the rectangle is the sample space. Probabilities of all the distinct regions add to 1.

Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Throughout, take g = 9.8 m s⁻² unless a question says otherwise. Press Start when you are ready.

Probability · notation

Set notation & the basic rules

A ∩ B — "A and B": both events happen (the intersection).
A ∪ B — "A or B or both": at least one happens (the union).
A′ — "not A": the complement, so P(A′) = 1 − P(A).
A | B — "A given B": B is known to have happened.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)the addition formula — subtract the overlap so it is not counted twice

Why subtract? Adding P(A) and P(B) counts everything in the overlap A ∩ B twice. Subtracting P(A ∩ B) once puts it right. This formula is true for any two events.

Quick check

Mutually exclusive

?Two events A and B are mutually exclusive. Which statement must be true?

Probability · mutually exclusive & independent

Two ideas that are often confused

These two words sound similar but mean completely different things.

Mutually exclusive — the events cannot both happen. P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B). On a Venn diagram the circles do not overlap.
Independent — one event happening does not change the probability of the other. P(A ∩ B) = P(A) × P(B), and equivalently P(A | B) = P(A) and P(B | A) = P(B).

Worked example — is it independent?

P(A) = 0.5, P(B) = 0.4, P(A ∩ B) = 0.2.

Test: P(A) × P(B) = 0.5 × 0.4 = 0.2 = P(A ∩ B) ✔ — so A and B are independent.

They are not mutually exclusive, because P(A ∩ B) = 0.2 ≠ 0.

Key insight: two events with non-zero probabilities can never be both mutually exclusive and independent — if they cannot occur together, then knowing B happened tells you A definitely did not, which is a huge change in probability.

Calculate

Union, exclusive events

1Events A and B are mutually exclusive with P(A) = 0.4 and P(B) = 0.35. Find P(A ∪ B).

Hint: Mutually exclusive ⇒ P(A ∩ B) = 0, so P(A ∪ B) = 0.4 + 0.35.

Calculate

Union, independent events

2Events A and B are independent with P(A) = 0.6 and P(B) = 0.5. Find P(A ∪ B).

Hint: Independent ⇒ P(A ∩ B) = 0.6 × 0.5 = 0.3. Then P(A ∪ B) = 0.6 + 0.5 − 0.3.

Quick check

Test for independence

?P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2. Are A and B independent?

Sort it

Always, exclusive, or independent?

Tap a formula, then tap the column it belongs in.

♾️ True for ANY events

🚫 Only if mutually exclusive

🎲 Only if independent

Probability · conditional probability

Conditional probability

P(A | B) is the probability of A given that B has happened. Knowing B has occurred shrinks the sample space down to B alone, so we scale by P(B):

P(A | B) = P(A ∩ B) ÷ P(B)rearranged: P(A ∩ B) = P(B) × P(A | B) = P(A) × P(B | A)

Worked example

P(A ∩ B) = 0.18 and P(B) = 0.45.

P(A | B) = 0.18 ÷ 0.45 = 0.4.

Interpretation: among the outcomes where B happens, A happens 40% of the time.

Warning: P(A | B) and P(B | A) are not the same. Nearly all people who have a rare disease test positive; that does not mean nearly all people who test positive have the disease. Divide by the probability of the event you are given.

Calculate

Conditional probability

3P(A ∩ B) = 0.18 and P(B) = 0.45. Find P(A | B).

Hint: P(A | B) = P(A ∩ B) ÷ P(B) = 0.18 ÷ 0.45.

Match it

Set notation → words

Tap an item on the left, then its partner on the right.

Notation

In words

Probability · tree diagrams

Tree diagrams & sampling without replacement

A tree diagram is the fastest way to handle two-stage experiments — especially without replacement, where the second set of branches uses conditional probabilities.

Two counters drawn without replacement from 5 red and 3 blue. Multiply along the branches; add the outcomes you want. Check: 20 + 15 + 15 + 6 = 56. ✔

Multiply along the branches of a path (this is P(A ∩ B) = P(A) × P(B | A)).
Add the paths that give the outcome you want.
All the final probabilities must sum to 1 — always check.

With replacement the counter goes back, so the second draw has the same probabilities: the draws are independent. Without replacement they are not — there is one fewer counter and possibly one fewer red.

Calculate

Both red

4A bag holds 5 red and 3 blue counters. Two are drawn without replacement. Find the probability that both are red, to 3 decimal places.

Hint: P(RR) = 5/8 × 4/7 = 20/56 = 5/14. Now write that as a decimal.

Calculate

Exactly one red

5From the same bag (5 red, 3 blue, no replacement), find the probability that exactly one counter is red, to 3 decimal places.

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

Quick check

The formula

?Which expression correctly gives the conditional probability formula?

Probability · modelling

Modelling — and criticising the assumptions

Spec point 3.3 asks you to model with probability and to critique the assumptions made.

"The die is fair" — is it? A biased die would make every calculated probability wrong.
"The events are independent" — often false. Two children in the same family catching a cold are not independent.
"Each person is equally likely to be chosen" — only true if the sampling really is random.

Continuous variables: for a continuous distribution, probability is the area under the curve, so P(X = a) = 0 for any single value. That is why P(X < a) and P(X ≤ a) are equal for a continuous random variable — but they are different for a discrete one such as the binomial.

Quick check

Read the shading

?A Venn diagram is shaded to show the region that is inside B but outside A. Which probability does the shaded region represent?

Probability · two-way tables

Two-way tables

A two-way table is just a Venn diagram in a grid. Conditional probabilities become "divide by the row or column total".

	Passed	Failed	Total
Boys	35	15	50
Girls	40	10	50
Total	75	25	100

P(Girl and Passed) = 40 ÷ 100 = 0.4 — divide by the grand total.
P(Passed | Girl) = 40 ÷ 50 = 0.8 — given a girl, divide by the Girls total.
P(Girl | Passed) = 40 ÷ 75 = 0.533 (3 s.f.) — given a pass, divide by the Passed total.

Notice P(Passed | Girl) = 0.8 but P(Girl | Passed) = 0.533. Conditioning is not symmetric — the two are completely different questions.

Quick check

Conditional from a table

?Using the table above, a student is chosen at random from those who passed. Find the probability that the student is a girl.

Recap

The big ideas to know

Notation: A ∩ B (both) · A ∪ B (at least one) · A′ (not A) · A | B (A given B)

Any events: P(A ∪ B) = P(A) + P(B) − P(A ∩ B) and P(A′) = 1 − P(A)

Mutually exclusive: P(A ∩ B) = 0 ⇒ P(A ∪ B) = P(A) + P(B)

Independent: P(A ∩ B) = P(A) × P(B) ⇔ P(A | B) = P(A)

Conditional: P(A | B) = P(A ∩ B) ÷ P(B) — and P(A | B) ≠ P(B | A)

Trees: multiply along branches, add the paths, check they total 1

Modelling: state and question the assumptions (fairness, independence)

That is the whole of 9MA0 Topic 3 — Probability. Press Finish to see your score.

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