OCR A Level Mathematics A: Probability: Venn and tree diagrams, conditional probability and modelling (H240) – Revision Notes

These notes condense section 2.03 Probability of the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018. For full explanations and worked examples, read the study guide. When you are ready, try the practice questions.

Stage 1 content (2.03a, 2.03b) is shared with AS. Content marked A Level only is Stage 2 (2.03c to 2.03e). Course hub: /boards/ocr/a-level/mathematics/. Printable checklist: /checklists/ocr/a-level/mathematics/. Find weak spots with the free diagnostics.

Notation

Symbol	Meaning
P(A)	probability of event A
A’	A does not happen, P(A’) = 1 − P(A)
P(X = 2), P(X = x)	probability that the random variable X takes that value
A ∩ B	A and B both happen
A ∪ B	A or B or both happen
A | B (A Level only)	A given that B has happened

Formulae

Result	Stage	When to use
Probabilities of all outcomes add to 1	1	finding an unknown constant in a distribution
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)	A Level only	any two events, with an overlap
Mutually exclusive: P(A ∩ B) = 0	1	events that cannot occur together
Independent: P(A ∩ B) = P(A) × P(B)	1	events that do not affect each other
P(A ∩ B) = P(A) × P(B | A)	A Level only	multiplying along a tree
P(A | B) = P(A ∩ B) / P(B)	A Level only	restricting to B

Independence also means P(B | A) = P(B). The specification excludes using the conditional formula to find P(A | B) from P(B | A).

Must-know distinctions

• Mutually exclusive means no overlap. Independent means no influence. Two events with non-zero probabilities cannot be both.
• P(A | B) is a share of B. P(B | A) is a share of A. They are usually different.
• “At least one” is nearly always easiest as 1 − P(none).
• Without replacement means the second-stage branches depend on the first outcome. With replacement, or from a very large population, they are the same.

Method boxes

Test for independence

1. Get P(A), P(B) and P(A ∩ B). If P(A ∪ B) is given, rearrange the union formula for the intersection.
2. Multiply P(A) by P(B).
3. Equal values mean independent. State the conclusion in a sentence.

Venn diagram

1. Start with the overlap.
2. Deduct the overlap from each group’s count to get what lies in that group alone.
3. Neither = total − (all three other regions).
4. Probabilities are region counts divided by the total.

Tree diagram

1. Draw one stage at a time. Label each branch with its probability.
2. Check that the branches from each node add to 1.
3. Multiply along the branches to get a combined outcome.
4. Add the outcomes you want. Use 1 − (unwanted) when that is shorter.

Conditional probability (A Level only)

1. Identify the “given” event. It is the denominator.
2. Find P(both) from the table, Venn diagram or tree.
3. Divide. Simplify only at the end.

Modelling comment

1. Name the assumption (fair, independent, constant probability, replacement).
2. Say why it may fail in the context given.
3. State which way the answer would move.

Choosing a diagram (Stage 1)

Situation	Best tool	Why
Two stages, outcomes you can list	sample space table	each cell is one outcome, easy to count
Ordered stages, possibly dependent	tree diagram	branches carry the changing probabilities
Overlapping groups with a total	Venn diagram	the overlap and the “neither” region are visible
Counts in two categories	two-way table	rows and columns give conditional probabilities directly

Small worked reminders

Union and independence. P(A) = 0.35, P(B) = 0.4, P(A ∪ B) = 0.6.

P(A ∩ B) = 0.35 + 0.4 − 0.6 = 0.15
P(A) × P(B) = 0.14

0.15 ≠ 0.14, so A and B are not independent.

Without replacement. A bag has 5 blue and 3 pink beads. Two are taken without replacement.

P(same colour) = 5/8 × 4/7 + 3/8 × 2/7 = 20/56 + 6/56 = 26/56 = 13/28

Conditional from a table. Of 100 students, 45 play chess and 18 play both chess and a second game.

P(second game | chess) = 18/45 = 2/5

Venn diagram with a total. Of 64 volunteers, 30 help with setup (S), 25 with catering (C) and 9 do both.

S only = 30 − 9 = 21     C only = 25 − 9 = 16     both = 9
Neither = 64 − (21 + 9 + 16) = 18
P(S ∪ C) = 46/64 = 23/32

Distribution link. P(X = x) = k/x for x = 1, 2, 3, 6, and no other values.

k(1 + 1/2 + 1/3 + 1/6) = 2k = 1, so k = 1/2
P(X = 3) = 1/6

Dependent chain. If P(late today) = 0.1, P(late tomorrow | late today) = 0.5 and P(late tomorrow | on time today) = 0.05:

P(late both days) = 0.1 × 0.5 = 0.05
P(late tomorrow) = 0.05 + 0.9 × 0.05 = 0.095

An independence model with 0.095 each day would give 0.1 × 0.095 = 0.0095 for both days, much lower than 0.05. This shows how ignoring dependence can understate a joint probability.

Reading a two-way table. Of 80 members, 50 are adults. Of the 50 adults, 20 own a bike. Of the 30 children, 24 own one.

P(bike) = 44/80 = 11/20
P(adult | bike) = 20/44 = 5/11
P(bike | child) = 24/30 = 4/5

Modelling check. Sampling two students from 1200, of whom 540 are girls.

Exact: 540/1200 × 539/1199 = 0.2023
Independent model: 0.45 × 0.45 = 0.2025

The gap is tiny because the population is large. With 12 students the same idea would be much less accurate.

Stage map

Reference	Topic	Stage
2.03a	mutually exclusive, independent, notation, link to distributions	1
2.03b	tree, sample space and Venn diagrams	1
2.03c	conditional probability from diagrams and tables; two formulae	A Level only
2.03d	conditional probability from first principles	A Level only
2.03e	modelling and critique of assumptions	A Level only

Quick self-test

1. Events C and D cannot happen together, with P(C) = 0.35 and P(D) = 0.2. Find P(C ∪ D).
2. P(A) = 0.35, P(B) = 0.4, P(A ∪ B) = 0.6. Find P(A ∩ B).
3. P(A) = 0.4 and P(A ∩ B) = 0.12. Find P(B | A).
4. Two beads are taken without replacement from 5 blue and 3 pink. Find P(one of each colour).
5. P(B) = 0.45 and P(A ∩ B) = 0.18. Find P(A | B).
6. P(A) = 0.7, P(B | A) = 0.2 and P(B | A’) = 0.5. Find P(B).
7. Each of four items independently fails a check with probability 0.1. Find P(at least one fails).
8. P(X = x) = k(x + 2) for x = 0, 1, 2, 3. Find k and P(X ≤ 1).
9. P(A) = 0.3, P(B) = 0.6 and P(A ∩ B) = 0.18. Are A and B independent? Find P(A’ | B).
10. A model treats ten days of weather as independent with the same chance of rain each day. State one reason that this may be unrealistic.

Answers

1. 0.35 + 0.2 = 0.55.
2. 0.35 + 0.4 − 0.6 = 0.15.
3. 0.12 / 0.4 = 0.3.
4. 2 × 5/8 × 3/7 = 30/56 = 15/28.
5. 0.18 / 0.45 = 0.4.
6. 0.7 × 0.2 + 0.3 × 0.5 = 0.14 + 0.15 = 0.29.
7. 1 − 0.9⁴ = 1 − 0.6561 = 0.3439.
8. 14k = 1 so k = 1/14; P(X ≤ 1) = 2/14 + 3/14 = 5/14.
9. 0.3 × 0.6 = 0.18 = P(A ∩ B), so independent. P(A’ | B) = P(A’ ∩ B)/P(B) = (0.6 − 0.18)/0.6 = 0.7.
10. For example, wet days tend to come in spells, so rain on one day makes rain the next day more likely. The days are not independent.

Where marks are usually lost

• Writing “independent” because the numbers look similar, with no product test shown.
• Adding probabilities for events that overlap, or forgetting to subtract P(A ∩ B).
• Using the first-stage denominator again on the second stage of a without-replacement tree.
• Putting the wrong event in the denominator of a conditional probability.
• Stopping at an unsimplified or mis-copied fraction when the question asks for an exact value.
• Giving a decimal that has been rounded early, so the independence test fails by a small amount.
• Missing the “neither” region in a Venn diagram, so the regions do not total the full sample.
• Answering a modelling question with “not realistic” and no named assumption or direction of effect.
• Leaving out the final conclusion in words when the question asks you to decide or comment.

Official syllabus

OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR. Section 2.03 Probability.