Probability

This lesson covers the fundamental concepts of probability at A-Level, including the rules for combining events, Venn diagrams, and tree diagrams. Probability is the mathematical framework for quantifying uncertainty and forms the basis for all statistical inference.

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Basic Probability Concepts

The probability of an event $A$A is a number between 0 and 1 (inclusive):

$0 \leq P(A) \leq 1$0≤P(A)≤1

Probability	Meaning
$P(A) = 0$P(A)=0	Event $A$A is impossible
$P(A) = 1$P(A)=1	Event $A$A is certain
$P(A) = 0.5$P(A)=0.5	Event $A$A is equally likely to occur or not

The sample space $S$S is the set of all possible outcomes. The complement of $A$A, written $A'$A′, is the event that $A$A does not occur:

$P(A') = 1 - P(A)$P(A′)=1−P(A)

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Venn Diagrams

Venn diagrams are used to visualise events and their relationships. For two events $A$A and $B$B:

• $A \cap B$A∩B — the intersection: both $A$A and $B$B occur.
• $A \cup B$A∪B — the union: $A$A or $B$B (or both) occur.
• $A'$A′ — the complement: $A$A does not occur.

The Addition Rule

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$P(A∪B)=P(A)+P(B)−P(A∩B)

For mutually exclusive events ($A \cap B = \emptyset$A∩B=∅):

$P(A \cup B) = P(A) + P(B)$P(A∪B)=P(A)+P(B)

Exam Tip: Always draw a Venn diagram when solving probability problems involving two or more events. Fill in the intersection first, then work outwards. The total of all regions must equal 1 if probabilities are used (or the total number if frequencies are used).

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Tree Diagrams

Tree diagrams display sequences of events. Each branch represents a possible outcome, and the probability is written along the branch.

Rules:

• The probabilities on branches from the same node must sum to 1.
• Multiply along branches to find the probability of a sequence of events.
• Add the results of separate branches to find the probability of combined outcomes.

Example: A bag contains 3 red and 2 blue balls. Two balls are drawn without replacement.

First draw: $P(R_1) = \frac{3}{5}$P(R1​)=53​, $P(B_1) = \frac{2}{5}$P(B1​)=52​.

Second draw (given Red first): $P(R_2) = \frac{2}{4}$P(R2​)=42​, $P(B_2) = \frac{2}{4}$P(B2​)=42​.

$P(\text{both red}) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$P(both red)=53​×42​=206​=103​

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Mutually Exclusive and Independent Events

Mutually Exclusive Events

Two events are mutually exclusive if they cannot occur at the same time:

$P(A \cap B) = 0$P(A∩B)=0

Independent Events

Two events are independent if the occurrence of one does not affect the probability of the other:

$P(A \cap B) = P(A) \times P(B)$P(A∩B)=P(A)×P(B)

Equivalently, $P(A | B) = P(A)$P(A∣B)=P(A).

Exam Tip: To determine whether events are independent, check if $P(A \cap B) = P(A) \times P(B)$P(A∩B)=P(A)×P(B). If this equation holds, the events are independent. If not, they are dependent. Never assume independence — always verify.

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Summary

• Probability is a number between 0 and 1 representing the likelihood of an event.
• Use the addition rule for $P(A \cup B)$P(A∪B) and the multiplication rule for independent events.
• Mutually exclusive events cannot happen together: $P(A \cap B) = 0$P(A∩B)=0.
• Independent events do not affect each other: $P(A \cap B) = P(A) \times P(B)$P(A∩B)=P(A)×P(B).
• Use Venn diagrams for two or three overlapping events and tree diagrams for sequential events.
• Always check whether your probabilities sum correctly and are consistent with the given information.

Exam Tip: Probability questions often require careful reading. Identify whether events are independent or mutually exclusive before applying any formula. Always state the rule you are using and show your working clearly.

A-Level Deep Dive: Probability

Spec mapping

AQA 7357 specification, Paper 3 — Statistics, Section Q (Probability) covers mutually exclusive and independent events when calculating probabilities; link to discrete and continuous distributions; understand and use conditional probability, including the use of tree diagrams, Venn diagrams, two-way tables (refer to the official specification document for exact wording). Section Q is the conceptual gateway to the entire statistics paper. Every later section depends on it: Section R (Statistical distributions: binomial, normal) uses independence to justify product structure; Section S (Statistical hypothesis testing) rests on $P(\text{type I error})$P(type I error) and conditional reasoning under $H_0$H0​; Section P (Statistical sampling) uses simple-random-sample probabilities. The AQA formula booklet supplies binomial and normal formulae but does not list $P(A\cup B) = P(A) + P(B) - P(A\cap B)$P(A∪B)=P(A)+P(B)−P(A∩B), the conditional probability identity, or Bayes' rule — these must be memorised.

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Worked example with full mark scheme

Question (8 marks):

A diagnostic test for a rare condition is taken by a screened population. The condition affects $2\%$2% of the population. The test correctly identifies the condition in $95\%$95% of those who have it (sensitivity), and correctly returns negative in $90\%$90% of those who do not (specificity).

(a) Draw a tree diagram representing the joint distribution of condition status and test result. (2)

(b) Find the probability that a randomly chosen person tests positive. (3)

(c) Given that a person tests positive, find the probability that they actually have the condition. Comment briefly on the result. (3)

Solution with mark scheme:

(a) The tree diagram has two stages: condition status $C$C vs $C'$C′, then test result $T^+$T+ vs $T^-$T−.

• Branch $C$C: probability $0.02$0.02. From $C$C: $T^+$T+ with $0.95$0.95, $T^-$T− with $0.05$0.05.
• Branch $C'$C′: probability $0.98$0.98. From $C'$C′: $T^+$T+ with $0.10$0.10, $T^-$T− with $0.90$0.90.

B1 — first-stage branches with correct probabilities $0.02$0.02 and $0.98$0.98 summing to $1$1.

B1 — second-stage conditional branches correctly labelled, with each pair summing to $1$1.

(b) Step 1 — identify the two routes to $T^+$T+.

A positive test arises either as a true positive ($C\cap T^+$C∩T+) or as a false positive ($C'\cap T^+$C′∩T+). These are mutually exclusive, so:

$P(T^+) = P(C\cap T^+) + P(C'\cap T^+)$P(T+)=P(C∩T+)+P(C′∩T+)

M1 — recognising mutually exclusive decomposition (the law of total probability).

Step 2 — multiply along each branch.

$P(C\cap T^+) = P(C)\,P(T^+\mid C) = 0.02 \times 0.95 = 0.019$P(C∩T+)=P(C)P(T+∣C)=0.02×0.95=0.019

$P(C'\cap T^+) = P(C')\,P(T^+\mid C') = 0.98 \times 0.10 = 0.098$P(C′∩T+)=P(C′)P(T+∣C′)=0.98×0.10=0.098

M1 — correct multiplication along both branches using $P(A\cap B) = P(A)P(B\mid A)$P(A∩B)=P(A)P(B∣A).

Step 3 — sum.

$P(T^+) = 0.019 + 0.098 = 0.117$P(T+)=0.019+0.098=0.117

A1 — correct numerical answer $0.117$0.117.

(c) Step 1 — identify the conditional probability required.

We want $P(C\mid T^+) = \dfrac{P(C\cap T^+)}{P(T^+)}$P(C∣T+)=P(T+)P(C∩T+)​.

M1 — quoting and applying the conditional probability definition (this is Bayes' rule in tree form).

Step 2 — substitute.

$P(C\mid T^+) = \dfrac{0.019}{0.117} = 0.1624...\approx 0.162$P(C∣T+)=0.1170.019​=0.1624...≈0.162

A1 — correct numerical answer to $3$3 s.f.

Step 3 — comment.

Despite a positive result, the probability of actually having the condition is only about $16\%$16%. Because the base rate is small ($2\%$2%), false positives drawn from the much larger healthy population dominate the positive-test pool.

A1 — comment that explicitly references the low base rate (or equivalently the dominance of false positives), not merely "it's surprising".

Total: 8 marks (B2 M3 A3).

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Specimen question modelled on the AQA Paper 3 format

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

(a) Find $P(A\cap B)$P(A∩B). (2)

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

(c) Find $P(A\mid B')$P(A∣B′). (2)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — applying $P(A\cup B) = P(A) + P(B) - P(A\cap B)$P(A∪B)=P(A)+P(B)−P(A∩B), so $0.7 = 0.4 + 0.5 - P(A\cap B)$0.7=0.4+0.5−P(A∩B).
• A1 (AO1.1b) — $P(A\cap B) = 0.2$P(A∩B)=0.2.

(b)

• M1 (AO2.1) — comparing $P(A\cap B)$P(A∩B) with $P(A)\,P(B) = 0.4\times 0.5 = 0.20$P(A)P(B)=0.4×0.5=0.20.
• A1 (AO2.4) — concluding independent because $P(A\cap B) = P(A)\,P(B) = 0.20$P(A∩B)=P(A)P(B)=0.20. The conclusion must follow the comparison; merely asserting independence without the product test scores zero.

(c)

• M1 (AO1.1b) — using $P(A\mid B') = \dfrac{P(A\cap B')}{P(B')}$P(A∣B′)=P(B′)P(A∩B′)​ with $P(A\cap B') = P(A) - P(A\cap B) = 0.4 - 0.2 = 0.2$P(A∩B′)=P(A)−P(A∩B)=0.4−0.2=0.2 and $P(B') = 1 - 0.5 = 0.5$P(B′)=1−0.5=0.5.
• A1 (AO1.1b) — $P(A\mid B') = \dfrac{0.2}{0.5} = 0.4$P(A∣B′)=0.50.2​=0.4.

Total: 6 marks split AO1 = 4, AO2 = 2. A typical AQA Section Q opener: short, structurally diverse (union, independence test, conditional), and front-loaded with formula recall.

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Synoptic links

Connects to:

1. Section R — Binomial distribution: $X\sim B(n,p)$X∼B(n,p) models $n$n independent trials with constant success probability $p$p. Independence is exactly the assumption $P(\text{success on trial }i\cap \text{success on trial }j) = p^2$P(success on trial i∩success on trial j)=p2. Without Section Q's independence concept, the binomial PMF $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$P(X=k)=(kn​)pk(1−p)n−k has no justification.

2. Section R — Normal distribution: the normal arises as the limiting distribution of sums of independent random variables (Central Limit Theorem, sketched at A-Level). Conditional probabilities under the normal — e.g. $P(X > 5\mid X > 3)$P(X>5∣X>3) — are computed using $P(A\mid B) = P(A\cap B)/P(B)$P(A∣B)=P(A∩B)/P(B) from Section Q.

3. Section S — Hypothesis testing: the significance level is the conditional probability $P(\text{reject }H_0\mid H_0\text{ true})$P(reject H0​∣H0​ true). Type I error, Type II error, and $p$p-values are all conditional probabilities, and reasoning about them requires fluency in the $P(A\mid B)$P(A∣B) vs $P(B\mid A)$P(B∣A) distinction.

4. Combinatorics (cross-paper): counting equally likely outcomes $P(\text{event}) = \dfrac{\#\text{favourable}}{\#\text{total}}$P(event)=#total#favourable​ relies on $\binom{n}{r}$(rn​), $n!$n!, and the multiplication principle. Hypergeometric-style problems (drawing without replacement from a finite urn) sit at this intersection.

5. Bayes' rule (extension): $P(C\mid T^+) = \dfrac{P(T^+\mid C)\,P(C)}{P(T^+)}$P(C∣T+)=P(T+)P(T+∣C)P(C)​ is the foundation of medical screening, spam filtering, and every modern statistical-inference framework. The disease-screening worked example above is its canonical A-Level appearance.

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Mark-scheme literacy

Probability questions on AQA 7357 split AO marks across all three objectives:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Applying $P(A\cup B)$P(A∪B), $P(A\cap B)$P(A∩B), $P(A\mid B)$P(A∣B) formulae; multiplying along tree branches; reading two-way tables
AO2 (reasoning / interpretation)	25–35%	Justifying independence via the product test; commenting on conditional results in context; identifying mutually exclusive structure
AO3 (problem-solving / modelling)	10–20%	Setting up the tree or Venn from a wordy stem; choosing between conditional and joint formulations; modelling a real screening or quality-control context

Examiner-rewarded phrasing: "since $A$A and $B$B are mutually exclusive, $P(A\cap B) = 0$P(A∩B)=0"; "by the law of total probability, $P(T^+) = P(C)P(T^+\mid C) + P(C')P(T^+\mid C')$P(T+)=P(C)P(T+∣C)+P(C′)P(T+∣C′)"; "comparing $P(A\cap B)$P(A∩B) with $P(A)P(B)$P(A)P(B) shows the events are not independent". Phrases that lose marks: "the probability is $0.162$0.162, so the test is unreliable" (no reference to base rate — comment scored zero); writing $P(C\mid T^+) = P(T^+\mid C)$P(C∣T+)=P(T+∣C) (the inversion fallacy); declaring independence on the basis of disjoint Venn regions ("mutually exclusive" is not "independent" — in fact the opposite, when both events have positive probability).

A specific AQA pattern to watch: questions that supply marginal probabilities and one joint probability, then ask three short parts (union, conditional, independence test). Read the constraints carefully — "given that $A$A has occurred" pins you to $P(\cdot\mid A)$P(⋅∣A), not $P(\cdot \cap A)$P(⋅∩A).

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Grade-band model answers

3-mark question

Question: A bag contains $5$5 red and $3$3 blue counters. Two are drawn without replacement. Find the probability that both are red.

Grade C response (~170 words):

The probability the first is red is $\tfrac{5}{8}$85​. After removing one red, $4$4 red and $3$3 blue remain, so the probability the second is red is $\tfrac{4}{7}$74​.

So $P(\text{both red}) = \tfrac{5}{8}\times \tfrac{4}{7} = \tfrac{20}{56} = \tfrac{5}{14}$P(both red)=85​×74​=5620​=145​.