Probability

This lesson covers probability as required by the Edexcel A-Level Mathematics specification (9MA0), Paper 3 Section A -- Statistics. You need to understand Venn diagrams, tree diagrams, conditional probability using the formula P(A|B) = P(A and B) / P(B), and the concepts of mutually exclusive and independent events.

Basic Probability Rules

The probability scale

All probabilities lie between 0 and 1 (inclusive): 0 ≤ P(A) ≤ 1.
P(A) = 0 means event A is impossible.
P(A) = 1 means event A is certain.
P(A') = 1 - P(A), where A' is the complement of A (i.e. A does not occur).

The addition rule

For any two events A and B:

P(A or B) = P(A) + P(B) - P(A and B)

Exam Tip: The subtraction of P(A and B) avoids double-counting the outcomes where both A and B occur.

Venn Diagrams

A Venn diagram uses circles inside a rectangle to represent events and their probabilities.

Setting up a Venn diagram

Draw a rectangle to represent the sample space.
Draw overlapping circles for events A and B.
Fill in the probabilities:
- The overlap region = P(A and B)
- The part of circle A that does not overlap = P(A) - P(A and B)
- The part of circle B that does not overlap = P(B) - P(A and B)
- The region outside both circles = 1 - P(A or B)

Example

Given: P(A) = 0.5, P(B) = 0.4, P(A and B) = 0.15

Only A: 0.5 - 0.15 = 0.35
Only B: 0.4 - 0.15 = 0.25
Both: 0.15
Neither: 1 - (0.35 + 0.15 + 0.25) = 0.25

Check: 0.35 + 0.15 + 0.25 + 0.25 = 1.00.

Tree Diagrams

A tree diagram shows the outcomes of successive events and their probabilities.

Rules for tree diagrams

Each branch represents a possible outcome and is labelled with its probability.
The probabilities on branches from the same node must sum to 1.
To find the probability of a particular path, multiply along the branches.
To find the probability of one outcome or another, add the relevant paths.

Example

A bag contains 3 red and 2 blue balls. You draw one ball, do not replace it, then draw another.

First draw:

P(Red 1st) = 3/5
P(Blue 1st) = 2/5

Second draw (given Red 1st):

P(Red 2nd) = 2/4 = 1/2
P(Blue 2nd) = 2/4 = 1/2

Second draw (given Blue 1st):

P(Red 2nd) = 3/4
P(Blue 2nd) = 1/4

P(Both Red) = 3/5 x 1/2 = 3/10 P(Red then Blue) = 3/5 x 1/2 = 3/10 P(Blue then Red) = 2/5 x 3/4 = 6/20 = 3/10 P(Both Blue) = 2/5 x 1/4 = 2/20 = 1/10

Check: 3/10 + 3/10 + 3/10 + 1/10 = 1.

P(one of each colour) = 3/10 + 3/10 = 6/10 = 3/5.

Conditional Probability

Conditional probability is the probability of event A occurring given that event B has already occurred.

The formula

P(A|B) = P(A and B) / P(B)

Rearranging the formula

P(A and B) = P(A|B) x P(B)

This is the multiplication rule for dependent events.

Example

In a class of 30 students, 18 study Maths, 15 study Physics, and 10 study both.

P(Maths) = 18/30 = 3/5
P(Physics) = 15/30 = 1/2
P(Maths and Physics) = 10/30 = 1/3

P(Maths | Physics) = (1/3) / (1/2) = 2/3

P(Physics | Maths) = (1/3) / (3/5) = 5/9

Exam Tip: P(A|B) and P(B|A) are generally different. Always read the question carefully to determine which conditional probability is being asked for.

Mutually Exclusive Events

Two events A and B are mutually exclusive if they cannot both occur at the same time.

Properties

P(A and B) = 0
P(A or B) = P(A) + P(B)

Testing for mutual exclusivity

If P(A and B) = 0, the events are mutually exclusive.

Independent Events

Two events A and B are independent if the occurrence of one does not affect the probability of the other.

Properties

P(A|B) = P(A) and P(B|A) = P(B)
P(A and B) = P(A) x P(B)

Testing for independence

Check whether P(A and B) = P(A) x P(B). If it holds, the events are independent.

Example

P(A) = 0.3, P(B) = 0.5, P(A and B) = 0.15

Test: 0.3 x 0.5 = 0.15 = P(A and B). So A and B are independent.

Example (not independent)

P(A) = 0.4, P(B) = 0.6, P(A and B) = 0.3

Test: 0.4 x 0.6 = 0.24 ≠ 0.3. So A and B are not independent.

Exam Tip: Independent events and mutually exclusive events are NOT the same thing. If two events are mutually exclusive (and both have non-zero probability), they cannot be independent, because knowing one occurred means the other definitely did not.

Combining Conditional Probability with Tree Diagrams

Tree diagrams naturally encode conditional probability. The second set of branches shows P(2nd event | 1st event). The probability of any path = P(1st event) x P(2nd event | 1st event).

Summary

P(A or B) = P(A) + P(B) - P(A and B) -- the addition rule.
Venn diagrams represent events as overlapping circles; ensure all regions sum to 1.
Tree diagrams multiply along branches and add between paths.
Conditional probability: P(A|B) = P(A and B) / P(B). Note P(A|B) ≠ P(B|A) in general.
Mutually exclusive: P(A and B) = 0.
Independent: P(A and B) = P(A) x P(B).
Mutually exclusive events (with non-zero probabilities) are never independent.

A-Level Deep Dive: Probability

Spec mapping

Edexcel 9MA0-03 specification, Paper 3 — Statistics and Mechanics, Section 5: Probability. The sub-strands require candidates to "understand and use mutually exclusive and independent events when calculating probabilities", "link to discrete and continuous distributions", and "understand and use conditional probability, including the use of tree diagrams, Venn diagrams, two-way tables". Although Section 5 is short on the specification page, probability ideas are tested throughout the statistics half of Paper 3 — they underlie Section 4 (Statistical distributions, where binomial probabilities require independent trials), Section 6 (Hypothesis testing, where p-values are tail probabilities) and Section 3 (Probability distributions and discrete random variables). The Edexcel formula booklet provides $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ and the conditional formula $P(A|B) = P(A \cap B)/P(B)$ , but candidates must recognise when to deploy each.

Worked example with full mark scheme

Question (8 marks): A factory has two machines, $M_1$ and $M_2$ . Machine $M_1$ produces 60% of all components and Machine $M_2$ produces the remaining 40%. The probability that a component from $M_1$ is defective is $0.03$ ; the probability that a component from $M_2$ is defective is $0.05$ .

(a) Draw a tree diagram representing this situation. (2)

(b) Find the probability that a randomly chosen component is defective. (2)

(c) Given that a randomly chosen component is defective, find the probability that it was produced by Machine $M_2$ . (4)

Solution with mark scheme:

(a) Step 1 — construct the tree. First branch: $P(M_1) = 0.6$ , $P(M_2) = 0.4$ . Second branch from each: $P(D|M_1) = 0.03$ , $P(D'|M_1) = 0.97$ , $P(D|M_2) = 0.05$ , $P(D'|M_2) = 0.95$ .

B1 — first-stage probabilities labelled correctly ( $0.6$ and $0.4$ ). B1 — second-stage conditional probabilities all four labelled correctly. Common error: writing $P(D) = 0.03$ on the second branch rather than $P(D|M_1) = 0.03$ — the branch represents a conditional event, and mislabelling here propagates into part (c).

(b) Step 1 — apply the law of total probability.

$P(D) = P(M_1)P(D|M_1) + P(M_2)P(D|M_2) = 0.6 \times 0.03 + 0.4 \times 0.05$

M1 — multiplying along each branch and summing the two "defective" routes.

$P(D) = 0.018 + 0.020 = 0.038$

A1 — correct numerical answer $0.038$ (or $\tfrac{19}{500}$ ).

$P(M_2 | D) = \dfrac{P(M_2 \cap D)}{P(D)}$

M1 — quoting the conditional formula with correct events. Many candidates flip the conditioning here, computing $P(D | M_2) = 0.05$ and writing it as the answer — that's the base-rate fallacy in miniature.

Step 2 — substitute.

$P(M_2 \cap D) = P(M_2) \cdot P(D|M_2) = 0.4 \times 0.05 = 0.020$

M1 — correct numerator from the tree.

$P(M_2 | D) = \dfrac{0.020}{0.038} = \dfrac{20}{38} = \dfrac{10}{19}$

A1 — exact fraction $\tfrac{10}{19}$ or decimal $0.526$ (3 s.f.).

A1 — final form simplified, with reasoning shown. Note the answer is larger than $P(M_2) = 0.4$ — observing the defect raises our credence that $M_2$ produced the component, even though $M_2$ is the minority producer, because $M_2$ is more error-prone.

Total: 8 marks (B2 M3 A3, split as shown).

Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Question (6 marks): Events $A$ and $B$ are such that $P(A) = 0.4$ , $P(B) = 0.5$ , and $P(A \cup B) = 0.7$ .

(a) Find $P(A \cap B)$ . (2)

(b) Determine, with reasoning, whether $A$ and $B$ are independent. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — applying the addition rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ rearranged.
A1 (AO1.1b) — $P(A \cap B) = 0.4 + 0.5 - 0.7 = 0.2$ .

(b)

M1 (AO2.1) — testing the independence criterion $P(A) \cdot P(B) = 0.4 \times 0.5 = 0.20$ and comparing with $P(A \cap B) = 0.20$ .
A1 (AO2.4) — concluding "since $P(A \cap B) = P(A) P(B)$ , the events are independent" with the equality stated explicitly.

(c)

M1 (AO1.1b) — recognising $P(A \cap B') = P(A) - P(A \cap B) = 0.4 - 0.2 = 0.2$ and $P(B') = 1 - 0.5 = 0.5$ .
A1 (AO1.1b) — $P(A | B') = 0.2 / 0.5 = 0.4$ . Observing that this equals $P(A)$ confirms independence from a different angle.

Total: 6 marks split AO1 = 4, AO2 = 2. Independence questions reliably carry an AO2 reasoning mark — Edexcel rewards the justification step (stating the criterion and the equality) as much as the arithmetic.

Synoptic links

Connects to:

Section 4 — Discrete distributions (binomial): $X \sim B(n, p)$ models $n$ independent Bernoulli trials with success probability $p$ . The independence assumption is exactly the multiplication rule $P(A \cap B) = P(A)P(B)$ applied across all trials. Without confident probability foundations the binomial formula $P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}$ is mechanical rather than meaningful.
Section 4 — Normal distribution: $P(a < X < b)$ for $X \sim N(\mu, \sigma^2)$ is a continuous probability — but the rules of the algebra of events (complements, unions, conditional probabilities) carry over unchanged, e.g. $P(X > a | X > b) = P(X > a)/P(X > b)$ for $a > b$ .
Section 6 — Hypothesis testing: the p-value is $P(\text{observed or more extreme} | H_0)$ — a conditional probability where the conditioning event is "the null hypothesis is true". Misinterpreting p-values as $P(H_0 | \text{data})$ is the same conditional-flip error as confusing $P(A|B)$ with $P(B|A)$ .
Section 1 — Statistical sampling and combinatorics (GCSE link): counting outcomes via $\binom{n}{r}$ underwrites tree-diagram leaf probabilities. Equally-likely outcomes reduce probability calculation to combinatorial counting — the original Laplace definition.
Section 2 — Statistical inference and large-data set: the language of "given that" used in conditional probability is the same language used when reading sub-tables of the LDS — $P(\text{rainfall} > 10 | \text{Camborne, July})$ is read off the LDS by filtering rows.

Mark-scheme literacy

Probability questions on 9MA0-03 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Computing tree-branch products, applying the addition rule, substituting into $P(A
AO2 (reasoning / interpretation)	25–35%	Justifying independence, interpreting a conditional probability in context, explaining why a Bayes-style answer differs from naïve intuition
AO3 (problem-solving / modelling)	10–20%	Modelling a real-world scenario as a tree or Venn diagram, identifying the events and assumptions, evaluating whether independence is reasonable

Probability

Probability

Basic Probability Rules

The probability scale

The addition rule

Venn Diagrams

Setting up a Venn diagram

Example

Tree Diagrams

Rules for tree diagrams

Example

Conditional Probability

The formula

Rearranging the formula

Example

Mutually Exclusive Events

Properties

Testing for mutual exclusivity

Independent Events

Properties

Testing for independence

Example

Example (not independent)

Combining Conditional Probability with Tree Diagrams

Summary

A-Level Deep Dive: Probability

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Synoptic links

Mark-scheme literacy

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