Probability — A-Level Mathematics Stats Study Guide

For: A-Level Mathematics candidates sitting Paper 5 (Probability & Statistics 1).

Covers: Sample spaces with equally likely outcomes, conditional probability, independent vs mutually exclusive events, tree and Venn diagrams, and Bayes-style reverse conditioning for A-Level Mathematics Paper 5.

You should already know: Basic probability, summation, integration (Pure 1 calculus).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Probability?

Probability is the numerical measure of the likelihood that a specific event occurs, ranging from 0 (impossible) to 1 (certain). It is used across the A-Level Mathematics Statistics 1 syllabus to model real-world random processes, from dice rolls to survey sampling, and forms the foundation for all later statistics topics including permutations, combinations, and probability distributions. The standard notation $P(A)$ denotes the probability of event $A$ occurring, with the complement rule $P(A^{\prime })=1-P(A)$ giving the probability that $A$ does not occur.

2. Sample space and equally likely outcomes

The sample space $S$ of a random experiment is the set of all possible distinct outcomes of the experiment, while an event is any subset of $S$. For experiments where all outcomes are equally likely (e.g., fair dice, random selection from a group), the probability of an event $A$ is calculated as the ratio of favorable outcomes to total possible outcomes: $$P(A)=\frac{n(A)}{n(S)}$$ where $n(A)$ is the number of outcomes in event $A$, and $n(S)$ is the total number of outcomes in the sample space. This formula only applies when outcomes are equally likely; it cannot be used for weighted dice, biased coins, or non-random selection.

Worked example

You roll two fair 6-sided dice. What is the probability the sum of the two numbers is 7?

• Total outcomes: $n(S)=6\times 6=36$ (each die is independent, so multiply outcomes)
• Favorable outcomes (sum = 7): $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$, so $n(A)=6$
• $P(\text{sum}=7)=\frac{6}{36}=\frac{1}{6}$

Exam tip: Examiners often pair this rule with counting problems; always confirm outcomes are equally likely before using the ratio formula, and avoid double-counting ordered outcomes (e.g., $(1,2)$ and $(2,1)$ are distinct for two separate dice).

3. Conditional probability — $P(A|B)=\frac{P(A\cap B)}{P(B)}$

Conditional probability is the probability of event $A$ occurring given that we already know event $B$ has occurred. Knowing $B$ is true reduces the effective sample space to only outcomes where $B$ occurs, hence we divide by $P(B)$ (which must be greater than 0 for the formula to be valid). Rearranging the formula gives the multiplication rule for joint probability: $$P(A\cap B)=P(A|B)P(B)=P(B|A)P(A)$$ where $A\cap B$ denotes the intersection of events $A$ and $B$ (both $A$ and $B$ occur).

Worked example

A bag has 5 red balls and 3 blue balls. You draw two balls without replacement. What is the probability the second ball is red given the first is red?

• Let $A$ = second ball red, $B$ = first ball red
• $P(B)=\frac{5}{8}$, $P(A\cap B)=\frac{5}{8}\times \frac{4}{7}=\frac{20}{56}$
• $P(A|B)=\frac{\frac{20}{56}}{\frac{5}{8}}=\frac{4}{7}$ This matches the intuitive result: after drawing one red ball, 4 red balls remain out of 7 total balls.

Exam tip: Always explicitly label which event is the condition first; mixing up $A$ and $B$ is the most common error on conditional probability questions.

4. Independent vs mutually exclusive events

These two properties describe entirely different relationships between events, and are almost never true at the same time (unless one event has probability 0):

• Mutually exclusive (disjoint) events: $A$ and $B$ cannot occur simultaneously, so $A\cap B=\varnothing$, and $P(A\cap B)=0$. For mutually exclusive events, the addition rule simplifies to $P(A\cup B)=P(A)+P(B)$, where $A\cup B$ is the union of $A$ and $B$ (either $A$ or $B$ or both occur).
• Independent events: The occurrence of one event does not affect the probability of the other, so $P(A|B)=P(A)$. Substituting into the conditional probability formula gives the independence test: $P(A\cap B)=P(A)P(B)$.

Worked example

You roll one fair 6-sided die. Let $A$ = roll an even number, $B$ = roll a 3, $C$ = roll a number greater than 2.

• $A$ and $B$ are mutually exclusive: you cannot roll an even number and a 3 at the same time, so $P(A\cap B)=0$.
• $A$ and $C$ are independent: $P(A)=\frac{1}{2}$, $P(C)=\frac{4}{6}=\frac{2}{3}$, $P(A\cap C)=P(\text{roll 4,6})=\frac{2}{6}=\frac{1}{3}=P(A)P(C)$. They are not mutually exclusive, as rolling a 4 satisfies both events.

Exam tip: Examiners almost always ask you to distinguish these two properties; you will lose marks if you state independent events cannot occur together.

5. Tree diagrams and Venn diagrams

These visual tools simplify probability calculations by organizing outcomes and event relationships explicitly, and examiners award marks for correctly labeled diagrams even if your final answer is incorrect.

• Tree diagrams: Best for sequential, multi-stage events (e.g., drawing without replacement, repeated trials). Each branch represents an outcome, with the probability of the outcome written on the branch. Multiply probabilities along connected branches to get joint probabilities, and add probabilities across branches to get total probabilities for composite events.
• Venn diagrams: Best for visualizing relationships between 2-3 events. Each circle represents an event, with overlaps representing intersections, and the area outside all circles representing the complement of the union of all events. They are particularly useful for solving problems involving survey data or grouped event probabilities.

Worked example (tree diagram)

A biased coin has $P(\text{Heads})=\frac{2}{3}$. Flip the coin twice. What is the probability of getting at least one head?

• Tree structure: First flip heads ($\frac{2}{3}$) → second flip heads ($\frac{2}{3}$) or tails ($\frac{1}{3}$); first flip tails ($\frac{1}{3}$) → second flip heads ($\frac{2}{3}$) or tails ($\frac{1}{3}$)
• Joint probabilities: $P(HH)=\frac{4}{9}$, $P(HT)=\frac{2}{9}$, $P(TH)=\frac{2}{9}$, $P(TT)=\frac{1}{9}$
• $P(\text{at least one head})=1-P(TT)=1-\frac{1}{9}=\frac{8}{9}$

Worked example (Venn diagram)

100 students are surveyed: 40 study Maths, 30 study Physics, 15 study both. What is the probability a randomly selected student studies Physics given they study Maths?

• Venn circle for Maths: 15 in the overlap with Physics, 25 study only Maths
• Venn circle for Physics: 15 in the overlap, 15 study only Physics
• 45 students study neither subject
• $P(\text{Physics}|\text{Maths})=\frac{\text{overlap count}}{\text{total Maths count}}=\frac{15}{40}=\frac{3}{8}$

6. Bayes-style reverse conditioning

Bayes' rule lets you calculate $P(A|B)$ when you only have conditional probabilities in the reverse direction ($P(B|A)$ and $P(B|A^{\prime })$). It is derived directly from rearranging the conditional probability formula, combined with the law of total probability to calculate $P(B)$: $$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$$ $$P(B)=P(B|A)P(A)+P(B|A^{\prime })P(A^{\prime })$$ where $A^{\prime }$ is the complement of event $A$. These questions are labeled "reverse conditioning" because they ask you to work backwards from an observed outcome to find the probability of an earlier cause.

Worked example

A disease test has a 95% true positive rate ($P(\text{positive}|\text{sick})=0.95$) and a 2% false positive rate ($P(\text{positive}|\text{not sick})=0.02$). 1% of the population has the disease. What is the probability you are sick if you test positive?

• Let $A$ = sick, $B$ = test positive
• $P(A)=0.01$, $P(A^{\prime })=0.99$, $P(B|A)=0.95$, $P(B|A^{\prime })=0.02$
• $P(B)=(0.95\times 0.01)+(0.02\times 0.99)=0.0095+0.0198=0.0293$
• $P(A|B)=\frac{0.95\times 0.01}{0.0293}\approx 0.324$ (32.4% chance of being sick despite a positive test)

Exam tip: Reverse conditioning questions always give you conditional probabilities in the opposite direction of what is asked; list out all given probabilities first to confirm it is a Bayes' rule question, and never skip calculating the full $P(B)$ term.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using the equally likely outcome ratio formula for biased events (e.g., weighted dice, biased coins). Why: Students assume all outcomes are equally likely by default. Correct move: Only use $\frac{n(A)}{n(S)}$ if the question explicitly states outcomes are equally likely (fair dice, random selection, etc.).
• Wrong move: Confusing independent and mutually exclusive events, using $P(A\cap B)=P(A)P(B)$ for mutually exclusive events. Why: Students mix up the two properties because both describe relationships between events. Correct move: First check if events can occur at the same time: if no, they are mutually exclusive, $P(A\cap B)=0$. If they can occur together, test for independence with the product rule.
• Wrong move: Forgetting to reduce the sample space for conditional probability in without replacement problems. Why: Students use the original total number of items for both draws. Correct move: Always adjust the denominator and numerator for sequential draws without replacement, or use the conditional probability formula explicitly.
• Wrong move: Omitting the complement term when calculating $P(B)$ for Bayes' rule questions. Why: Students only use the $P(B|A)P(A)$ term and forget to add $P(B|A^{\prime })P(A^{\prime })$. Correct move: Write out the full law of total probability for $P(B)$ before substituting into Bayes' formula.
• Wrong move: Double-counting unordered outcomes in sample space calculations. Why: Students count $(1,2)$ and $(2,1)$ as the same outcome for two distinct dice. Correct move: If items are distinguishable (separate dice, different balls), ordered outcomes are distinct, so count both.

8. Practice Questions (A-Level Mathematics Paper 5 Style)

Question 1

A fair 8-sided die is rolled twice. (a) Find the probability that the sum of the two rolls is greater than 12. (b) Find the probability that the first roll is a 5 given that the sum is greater than 12.

Solution

(a) Total outcomes $n(S)=8\times 8=64$. Favorable outcomes (sum > 12): Sum = 13: $(5,8),(6,7),(7,6),(8,5)$; Sum =14: $(6,8),(7,7),(8,6)$; Sum=15: $(7,8),(8,7)$; Sum=16: $(8,8)$. Total $n(A)=4+3+2+1=10$. $$P(\text{sum}>12)=\frac{10}{64}=\frac{5}{32}$$ (b) Let $A$ = first roll is 5, $B$ = sum >12. $P(A\cap B)=P(\text{first=5, second=8})=\frac{1}{64}$. $$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{\frac{1}{64}}{\frac{5}{32}}=\frac{1}{10}$$

Question 2

Events $A$ and $B$ are such that $P(A)=0.6$, $P(B)=0.3$, $P(A\cup B)=0.72$. (a) Find $P(A\cap B)$. (b) Determine whether $A$ and $B$ are independent. (c) Determine whether $A$ and $B$ are mutually exclusive.

Solution

(a) Use the general addition rule: $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ $$0.72=0.6+0.3-P(A\cap B)⟹P(A\cap B)=0.9-0.72=0.18$$ (b) Test independence: $P(A)P(B)=0.6\times 0.3=0.18=P(A\cap B)$, so $A$ and $B$ are independent. (c) $P(A\cap B)=0.18\ne 0$, so $A$ and $B$ are not mutually exclusive.

Question 3

A factory has two machines producing light bulbs. Machine X produces 60% of the bulbs, and 2% of its output is defective. Machine Y produces 40% of the bulbs, and 5% of its output is defective. A bulb is selected at random and found to be defective. Find the probability it was produced by Machine X.

Solution

Let $A$ = bulb produced by X, $B$ = bulb is defective. $P(A)=0.6$, $P(A^{\prime })=0.4$, $P(B|A)=0.02$, $P(B|A^{\prime })=0.05$. First calculate total $P(B)$ using the law of total probability: $$P(B)=(0.6\times 0.02)+(0.4\times 0.05)=0.012+0.02=0.032$$ Apply Bayes' rule: $$P(A|B)=\frac{P(B|A)P(A)}{P(B)}=\frac{0.012}{0.032}=\frac{3}{8}=0.375$$

9. Quick Reference Cheatsheet

10. What's Next

Probability is the foundational topic for all remaining content in A-Level Mathematics Paper 5: you will use these rules to solve permutation and combination counting problems, model discrete probability distributions including the binomial and geometric distributions, and calculate expected values for random variables. A strong grasp of conditional probability and Bayes' rule is also required for more advanced statistics topics in Paper 6 if you are taking the full A Level, including hypothesis testing and continuous probability distributions.

If you struggle with any of the concepts, practice questions, or worked examples in this guide, you can ask Ollie, our AI tutor, for personalized explanations, additional practice problems, or step-by-step walkthroughs tailored to your learning pace at any time on the homepage. You can also access our full library of A-Level Mathematics Paper 5 study guides and topic quizzes to test your mastery before your exam.

Aligned with the Cambridge International AS & A Level Mathematics 9709 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.