AQA GCSE Mathematics 8300: Probability – Study Guide

This study guide teaches Topic 5, Probability (specification references P1 to P9), of the AQA GCSE Mathematics (8300) specification, for teaching from September 2015 with exams from May/June 2017 (version 1.0). Probability can be tested on any of the three papers at Foundation or Higher tier. P1 to P8 are for both tiers; P9, conditional probability, is Higher tier only and is labelled where it appears. Exams run every May/June and November.

Use it with the Probability revision notes and the Probability practice questions. The course hub is AQA GCSE Mathematics, the printable checklist lists every statement, and the free 10-minute diagnostics show where to start.

What this topic covers

Spec	What you must be able to do	Tier
P1	Record, describe and analyse outcomes of experiments using tables and frequency trees	Both
P2	Use randomness, fairness and equally likely events to calculate expected outcomes	Both
P3	Link relative expected frequency to theoretical probability; use the 0 to 1 scale	Both
P4	Use the fact that probabilities of an exhaustive set of (mutually exclusive) outcomes sum to 1	Both
P5	Know that bigger unbiased samples give estimates closer to the theoretical probability	Both
P6	List sets and combinations systematically with tables, grids, Venn diagrams and tree diagrams	Both
P7	Build sample spaces for single and combined experiments and use them	Both
P8	Probabilities of independent and dependent combined events, including tree diagrams; know when to add and when to multiply	Both
P9	Calculate and interpret conditional probabilities using two-way tables, tree diagrams and Venn diagrams	Higher tier only

The whole of probability and statistics together makes up about 15% of the marks at both tiers. Probabilities can be written as fractions, decimals or percentages. Never write them as ratios or “1 in 4”.

The probability scale and the “sum to 1” rule (P3, P4)

A probability is a number from 0 (impossible) to 1 (certain). An even chance is 0.5. Words such as unlikely and likely sit either side of 0.5.

If a set of outcomes is exhaustive (covers everything that can happen) and mutually exclusive (no two can happen together), their probabilities add up to 1. So:

P(not A) = 1 − P(A)

Worked example 1. A four-colour spinner has P(red) = 0.3 and P(blue) = 0.25. Yellow is twice as likely as green. Find P(yellow).

P(green) + P(yellow) = 1 − 0.3 − 0.25 = 0.45
Let P(green) = x, so P(yellow) = 2x
x + 2x = 0.45  →  3x = 0.45  →  x = 0.15
P(yellow) = 2 × 0.15 = 0.3

Experiments, frequency trees and relative frequency (P1, P3, P5)

When you cannot work out a probability from equally likely outcomes, you run an experiment and use the relative frequency:

relative frequency = number of times the outcome happened ÷ number of trials

Relative frequency is an estimate of the probability. P5 is the key idea: an unbiased experiment with more trials tends to give an estimate closer to the theoretical probability. So always use the result from the largest number of trials.

Worked example 2. Mia rolls a dice 50 times and gets 14 sixes. Later she rolls it 500 times and gets 86 sixes.

• After 50 rolls the relative frequency is 14/50 = 0.28.
• After 500 rolls it is 86/500 = 0.172.
• A fair dice gives 1/6 ≈ 0.167. The 500-roll estimate is better because it uses more trials. It is close to 1/6, so there is no strong evidence that the dice is biased.

A frequency tree records how a group splits in two stages. Numbers, not probabilities, go on the branches.

Worked example 3. 240 people visit a museum. 3/8 are children and the rest are adults. 2/3 of the children and 40% of the adults buy something in the shop.

                    ┌── buy        60
      children 90 ──┤
                    └── no buy     30
240 ──┤
                    ┌── buy        60
      adults 150 ───┤
                    └── no buy     90

Children: 3/8 × 240 = 90, and 2/3 × 90 = 60 buy. Adults: 240 − 90 = 150, and 0.4 × 150 = 60 buy. So P(a visitor chosen at random buys something) = 120/240 = 1/2.

Expected outcomes and fairness (P2)

If an event has probability p and you repeat the experiment n times:

expected number of times = p × n

This is what you expect on average. It is not a guarantee, and it does not have to be a whole number.

Worked example 4. The probability of winning a fairground game is 0.15. Ali plays 60 times. The expected number of wins is 0.15 × 60 = 9.

A dice, coin or spinner is fair (unbiased) if every outcome is equally likely. To judge fairness, compare the relative frequency from many trials with the theoretical probability.

Sample spaces and systematic listing (P6, P7)

A sample space lists every possible outcome. For two experiments together, a grid is the clearest way.

Worked example 5. A fair four-sided spinner (1 to 4) and a fair three-sided spinner (1 to 3) are spun. The two scores are multiplied.

 ×  |  1   2   3
----+------------
 1  |  1   2   3
 2  |  2   4   6
 3  |  3   6   9
 4  |  4   8  12

There are 4 × 3 = 12 equally likely outcomes. 8 products are even, so P(even) = 8/12 = 2/3. Three products are greater than 6 (8, 9, 12), so P(product > 6) = 3/12 = 1/4.

When listing combinations without a grid, fix the first item and run through all choices of the second, then move on. This stops you missing or repeating outcomes.

Venn diagrams

A Venn diagram sorts items into overlapping sets inside a rectangle that holds everything.

Worked example 6. In a class of 40, 23 play football, 18 play tennis and 7 play both.

• Fill the overlap first: 7.
• Football only: 23 − 7 = 16. Tennis only: 18 − 7 = 11.
• Neither: 40 − (16 + 7 + 11) = 6. Write this outside both circles.
• P(neither) = 6/40 = 3/20.

Combined events and tree diagrams (P6, P8)

You must know these two results. They are not given in the exam.

Rule	Use it when
P(A or B) = P(A) + P(B) − P(A and B)	“or” questions; if A and B are mutually exclusive, P(A and B) = 0 so you just add
P(A and B) = P(A given B) × P(B)	“and” questions; if A and B are independent, P(A given B) = P(A), so P(A and B) = P(A) × P(B)

Check with the Venn example: P(football or tennis) = 23/40 + 18/40 − 7/40 = 34/40 = 17/20.

Multiply along branches, add between branches. On a tree diagram, each set of branches from one point must sum to 1.

Independent events. The first outcome does not change the second. The probability that a bus is late is 0.2 and the probability that it rains is 0.3, and you assume these are independent. P(late and rain) = 0.2 × 0.3 = 0.06. P(neither) = 0.8 × 0.7 = 0.56, so P(at least one) = 1 − 0.56 = 0.44.

Dependent events. The first outcome changes the second, as in picking “without replacement”.

Worked example 7. A bag holds 5 red and 3 blue counters. Two are taken at random without replacement.

First        Second        Outcome   Probability
             R  4/7   →    RR        5/8 × 4/7 = 20/56
R  5/8  ──<
             B  3/7   →    RB        5/8 × 3/7 = 15/56
             R  5/7   →    BR        3/8 × 5/7 = 15/56
B  3/8  ──<
             B  2/7   →    BB        3/8 × 2/7 =  6/56

The four outcomes total 56/56 = 1, which is a useful check.

• P(same colour) = 20/56 + 6/56 = 26/56 = 13/28.
• P(at least one blue) = 1 − P(RR) = 1 − 20/56 = 36/56 = 9/14.

“Know the underlying assumptions” (P8) means you should be able to say why you multiplied: the counters are picked at random, and for independent events one result does not affect the other.

Conditional probability (P9) – Higher tier only

“The probability of A given B” means you already know B has happened. The group you are choosing from shrinks to B only.

P(A given B) = P(A and B) ÷ P(B)

This is the second formula above, rearranged.

From a two-way table. 120 students: 70 in Year 10 (28 walk to school) and 50 in Year 11 (15 walk).

	Walk	Other	Total
Year 10	28	42	70
Year 11	15	35	50
Total	43	77	120

A student who walks is chosen at random. P(Year 11 given walks) = 15/43. The denominator is the walk total, 43, not 120.

From a Venn diagram. In worked example 6, a footballer is chosen. P(plays tennis given plays football) = 7/23.

From a tree diagram. In worked example 7, given that both counters are the same colour, P(both red) = (20/56) ÷ (26/56) = 10/13.

Expected frequencies make this easier to see. Imagine 56 repeats: about 20 give RR and 6 give BB, so 20 of the 26 “same colour” results are red.

Calculator and non-calculator

Probability can appear on Paper 1, so you must add, subtract and multiply fractions by hand. On Papers 2 and 3 a calculator helps with decimals, but keep exact fractions until the end and give a fraction answer unless the question asks for a decimal.

Common errors

• Adding probabilities along a tree branch instead of multiplying.
• Not changing the denominator on the second pick without replacement (writing 4/8 instead of 4/7).
• Adding “or” probabilities when the events overlap, so the overlap is counted twice.
• Using 120 (the grand total) instead of the row or column total for a “given” question.
• Putting the “both” number into each circle as well as the overlap in a Venn diagram.
• Writing “3 : 5” or “3 out of 8” as a probability. Write 3/8.
• Giving a probability greater than 1 and not noticing.

Next steps

Condense this into the revision notes, then test yourself with the practice questions. The AQA GCSE Mathematics hub links the other topics.

Official syllabus

AQA GCSE Mathematics (8300) specification, for teaching from September 2015, exams from May/June 2017, version 1.0, published by AQA – section 3.5 Probability (P1 to P9) and Appendix: mathematical formulae.