OxfordAQA International AS and A-level Mathematics: Probability laws, conditional probability and independence (9660) – Revision Notes

These revision notes follow the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, section S1.1 (Further probability). The spec states this content is part of International AS, Unit PSM1. For full explanations and worked examples, read the study guide first. Then test yourself with the practice questions, and tick off progress on the course checklist. The whole course sits on the OxfordAQA Mathematics hub, and the free 10-minute diagnostics show where to focus.

Key definitions

• Event: a set of outcomes. P(A) lies between 0 and 1.
• Equally likely outcomes: P(A) = (outcomes in A) / (all outcomes).
• Relative frequency: P(A) is estimated by (times A occurred) / (number of trials).
• Complement: A’ means “not A”.
• Mutually exclusive: A and B cannot both happen, so P(A ∩ B) = 0.
• Independent: knowing A has happened does not change the probability of B, so P(B | A) = P(B).

Laws to know

Law	Result	Notes
Complement	P(A’) = 1 − P(A)	Use for “at least one”
Addition, any two events	P(A ∪ B) = P(A) + P(B) − P(A ∩ B)	Spec: two events only
Addition, mutually exclusive	P(A ∪ B) = P(A) + P(B)	Spec: two or more events
Multiplication	P(A ∩ B) = P(A) × P(B	A) = P(B) × P(A
Conditional	P(B	A) = P(A ∩ B) / P(A)
Independent	P(A ∩ B) = P(A) × P(B)	Spec: two or more events

For three events in sequence: P(A ∩ B ∩ C) = P(A) × P(B | A) × P(C | A ∩ B).

Method boxes

Method: probability from a frequency table

1. Find the row, column and grand totals.
2. For a plain probability, divide the cell or total you want by the grand total.
3. For “given”, divide by the total of the condition only.
4. Simplify, or give a decimal if asked.

Method: test for independence

1. Find P(A), P(B) and P(A ∩ B) from the data.
2. Multiply P(A) × P(B).
3. Compare with P(A ∩ B).
4. Write a conclusion: “equal, so independent” or “not equal, so not independent”.

Method: selection without replacement

1. Write each pick as a fraction.
2. Reduce the numerator and denominator by one for the items removed.
3. Multiply along the sequence.
4. For “at least one”, find “none” and subtract from 1.

Method: unknown probability with independence

1. Let P(B) = p.
2. Write P(A ∩ B) = P(A) × p.
3. Substitute into the addition law.
4. Solve the linear equation and check that p lies between 0 and 1.

Must-know distinctions

	Mutually exclusive	Independent
Test	P(A ∩ B) = 0	P(A ∩ B) = P(A) × P(B)
Union	P(A) + P(B)	P(A) + P(B) − P(A)P(B)
Picture	Events do not overlap	Overlap is exactly the product

Two events with non-zero probabilities cannot be both. If they are exclusive, one happening rules out the other, which is a strong dependence.

Also remember that P(A | B) and P(B | A) are normally different. They share the numerator P(A ∩ B) but divide by different totals.

Small worked reminders

Reminder 1: addition law. For P(A) = 0.5, P(B) = 0.3 and P(A ∩ B) = 0.1:

P(A ∪ B) = 0.5 + 0.3 − 0.1 = 0.7

Reminder 2: conditional probability from a table. Of 120 pupils, 45 play chess, and 27 of those also play an instrument.

P(instrument | chess) = 27/45 = 3/5

Reminder 3: without replacement. A bag has 5 red and 3 blue counters. Two are taken without replacement.

P(both red) = 5/8 × 4/7 = 20/56 = 5/14

Useful consequences

These follow from the laws above. Each can be derived in a line or two, so you do not need to memorise them separately.

Situation	Result
Split B by whether A happens	P(B) = P(A ∩ B) + P(A’ ∩ B)
Neither event happens	P(A’ ∩ B’) = 1 − P(A ∪ B)
A happens but B does not	P(A ∩ B’) = P(A) − P(A ∩ B)
Exactly one of A, B happens	P(A) + P(B) − 2 × P(A ∩ B)
Independence and complements	If A and B are independent, so are A and B’

The first row is the key step in many “find P(B)” questions. A Venn diagram or a table is permitted, so draw one whenever you are unsure which region you are counting.

More worked reminders

Reminder 4: four regions from three numbers. P(A) = 0.45, P(B) = 0.35 and P(A ∩ B) = 0.2.

P(A ∩ B')  = 0.45 − 0.2 = 0.25
P(A' ∩ B)  = 0.35 − 0.2 = 0.15
P(A' ∩ B') = 1 − (0.25 + 0.2 + 0.15) = 0.4
Check: 0.25 + 0.2 + 0.15 + 0.4 = 1

Reminder 5: testing independence. In the example above, P(A) × P(B) = 0.45 × 0.35 = 0.1575, while P(A ∩ B) = 0.2. These differ, so A and B are not independent.

Reminder 6: relative frequency table. A survey of 200 customers gives these relative frequencies.

	Online	In store
Under 30	0.24	0.16
30 or over	0.22	0.38

P(under 30)              = 0.24 + 0.16 = 0.4
P(online)                = 0.24 + 0.22 = 0.46
P(online | under 30)     = 0.24 / 0.4 = 0.6
P(under 30) x P(online)  = 0.4 x 0.46 = 0.184, not 0.24

The probabilities in the table add to 1. The two events are not independent, because 0.184 is not equal to 0.24.

Reminder 7: three independent events. Three independent switches work with probabilities 0.9, 0.8 and 0.7.

All work:       0.9 x 0.8 x 0.7 = 0.504
At least one fails: 1 - 0.504 = 0.496

Checking an answer

Before you move on, run through these checks.

1. Is every probability between 0 and 1?
2. Do the probabilities of all regions or outcomes add to 1?
3. Does the answer make sense? A union cannot be smaller than either event, and an intersection cannot be larger than either.
4. For a “given” question, did you divide by the probability or total of the condition?
5. For independence, did you compare numbers rather than just state a conclusion?

Quick self-test

1. P(A) = 0.35. Find P(A’).
2. A, B and C are mutually exclusive with probabilities 0.2, 0.15 and 0.4. Find P(A ∪ B ∪ C).
3. P(A) = 0.5, P(B) = 0.3 and P(A ∩ B) = 0.1. Find P(A ∪ B).
4. P(A ∩ B) = 0.12 and P(B) = 0.4. Find P(A | B).
5. A and B are independent with P(A) = 0.25 and P(B) = 0.8. Find P(A ∩ B).
6. P(A) = 0.5, P(B) = 0.4 and P(A ∪ B) = 0.9. Are A and B mutually exclusive?
7. A and B are independent, P(A) = 0.4 and P(A ∪ B) = 0.64. Find P(B).
8. A fair six-sided die is rolled twice. Find P(sum is 8) and P(sum is 8 | first roll is 3).
9. In a group of 120 pupils, 45 play chess. Find P(chess) as a fraction in its simplest form.
10. A bag has 5 red and 3 blue counters. Two are drawn without replacement. Find P(both red).

Answers

1. 1 − 0.35 = 0.65
2. 0.2 + 0.15 + 0.4 = 0.75
3. 0.5 + 0.3 − 0.1 = 0.7
4. 0.12 / 0.4 = 0.3
5. 0.25 × 0.8 = 0.2
6. P(A ∩ B) = 0.5 + 0.4 − 0.9 = 0, so yes, mutually exclusive.
7. 0.64 = 0.4 + p − 0.4p, so 0.24 = 0.6p and p = 0.4.
8. There are 5 ways to make 8 out of 36, so 5/36. If the first roll is 3, only a second roll of 5 works: 1/6.
9. 45/120 = 3/8
10. 5/8 × 4/7 = 20/56 = 5/14

Where marks are usually lost

• Dividing by the grand total instead of the condition total in a “given” question.
• Adding P(A) and P(B) without subtracting the overlap when the events can occur together.
• Saying events are independent because they are “unrelated”, with no numerical test.
• Confusing mutually exclusive with independent.
• Keeping the denominator unchanged on the second pick when there is no replacement.
• Finding P(A | B) when the question asked for P(B | A).
• Rounding intermediate decimals, which can spoil a final answer.
• Giving a conclusion with no working on a question that says “show that” or “determine whether”.
• Setting up the equation for an unknown probability correctly but not checking that the result is between 0 and 1.

Exam-technique advice for the whole course is in exam preparation.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA. Section S1.1 (Further probability), Unit PSM1, International AS exams from May/June 2018 and International A-level exams from May/June 2019.