Grade 11 · Statistics 1 · Cambridge A-Level 9709 · Age 16–17
Probability is the mathematical study of uncertainty. In Cambridge A-Level Statistics 1 (9709), you build a rigorous framework for calculating the likelihood of events — from simple sample spaces to multi-stage tree diagrams, conditional probability, and counting arrangements. These ideas underpin statistical inference, which you will use throughout A-Level and beyond.
List all equally likely outcomes; P = favourable/total
P(A') = 1 − P(A) — often the fastest route
P(A∪B) = P(A)+P(B)−P(A∩B)
P(A∩B) = P(A)×P(B) iff A, B independent
P(A|B) = P(A∩B) / P(B)
Multiply along branches; add between branches
ⁿPᵣ = n!/(n−r)! ordered selections
n!/p!q!... for identical objects
Probability is a number between 0 and 1 inclusive. P = 0 means impossible; P = 1 means certain. For any event A:
When every outcome in the sample space S is equally likely:
The sample space S is the set of all possible outcomes. For two dice, list pairs (a,b) where a is the first die and b the second — giving 36 equally likely outcomes.
The complement A' (read "A prime" or "not A") is the event that A does not occur. Since every outcome is either in A or not in A:
A Venn diagram represents events as overlapping circles inside a rectangle (the sample space). The overlapping region is A∩B (both A and B). The total shaded area for A∪B covers everything in A or B.
We subtract P(A∩B) to avoid counting the overlap twice.
Events A and B are mutually exclusive if they cannot both occur, so their intersection is empty:
This holds for any two events. If they are mutually exclusive, P(A∩B) = 0 and the rule simplifies to P(A)+P(B).
Events A and B are independent if knowledge that B occurred gives no information about whether A occurred:
To verify whether two events are independent from a table or Venn diagram, check whether P(A∩B) = P(A) × P(B).
A two-way (contingency) table shows frequencies for two categorical variables. Read off probabilities by dividing cell counts by the total.
| Male | Female | Total | |
|---|---|---|---|
| Left-handed | 10 | 15 | 25 |
| Right-handed | 45 | 30 | 75 |
| Total | 55 | 45 | 100 |
Calculating P(at least one) directly involves many cases. The complement method is almost always faster:
The conditional probability of A given B is the probability that A occurs, knowing that B has already occurred:
Read P(A|B) as "probability of A given B". The vertical bar | means "given".
If A and B are independent, then knowing B occurred tells us nothing about A:
This is just the conditional probability definition rearranged. It holds for any events (not just independent ones).
A tree diagram represents a multi-stage experiment. Each branch shows an outcome and its probability. The rules are:
Sometimes we know the result and want to work backwards to find the cause. This uses the law of total probability plus the definition of conditional probability.
n factorial (written n!) is the product of all positive integers from 1 to n:
The number of ways to arrange n distinct objects in a line is n!.
The number of ordered selections of r objects from n distinct objects:
For arrangements in a circle, one object is fixed to remove rotational equivalences:
When some objects are identical, divide by the factorial of the count of each repeated object to avoid overcounting:
Eight fully worked examples covering all key probability topics for Cambridge A-Level 9709 Statistics 1.
These are the errors Cambridge examiners see most frequently. Knowing why they are wrong helps you avoid them under time pressure.
| Concept | Formula | Notes / When to Use |
|---|---|---|
| Classical probability | P(A) = favourable / total | Equally likely outcomes only |
| Complement | P(A') = 1 − P(A) | Use for "not A" or "at least one" |
| Addition rule | P(A∪B) = P(A)+P(B)−P(A∩B) | Always valid for two events |
| Mutually exclusive | P(A∩B) = 0 → P(A∪B) = P(A)+P(B) | Cannot both occur simultaneously |
| Independence (condition) | P(A∩B) = P(A)×P(B) | Verify by checking this equation |
| Independence (conditional) | P(A|B) = P(A) and P(B|A) = P(B) | Alternative independence check |
| Conditional probability | P(A|B) = P(A∩B) / P(B) | Requires P(B) > 0 |
| Multiplication rule (general) | P(A∩B) = P(A|B)×P(B) | Works for any events (dependent or independent) |
| Law of total probability | P(A) = P(A|B)P(B) + P(A|B')P(B') | When B partitions the sample space |
| Bayes-style reversal | P(B|A) = P(A|B)P(B) / P(A) | Reverse conditioning |
| At least one | P(≥1) = 1 − P(none) | Complement is almost always faster |
| Tree: multiply along | P(path) = product of branch probs | For joint probability of a sequence |
| Tree: add between | P(event) = sum of path probs satisfying it | For mutually exclusive paths |
| Factorial | n! = n(n−1)(n−2)···1, 0!=1 | Fundamental counting |
| Arrangements (line) | n! | All n objects distinct |
| Permutations | ⁿPᵣ = n!/(n−r)! | Ordered selection of r from n |
| Circular arrangements | (n−1)! | Fix one object; arrange rest |
| Identical objects | n! / (p!q!r!...) | Divide by repeated object factorials |
| Combinations | ⁿCᵣ = n! / r!(n−r)! | Unordered selection; often needed with probability |
Three proofs that build rigorous understanding and may appear in deeper exam questions or A-Level interviews.
Enter values for P(A), P(B) and P(A∩B) — then click Compute to see P(A∪B), P(A|B), P(B|A) and a live Venn diagram.
8 Cambridge S1-style questions with mark schemes. Attempt each before revealing the solution.
Events A and B are such that P(A) = 0.45, P(B) = 0.30, P(A∪B) = 0.60. Find P(A∩B) and state, with a reason, whether A and B are mutually exclusive.
A bag contains 6 red and 4 green balls. Two balls are drawn at random without replacement. Find the probability that both are the same colour.
In a class of 30 students, 18 play football (F) and 12 play tennis (T), and 5 play both. A student is selected at random. Find P(F|T') — the probability that the student plays football given they do not play tennis.
Box A contains 3 red and 2 white counters. Box B contains 1 red and 4 white counters. A box is chosen at random, and then a counter is drawn. Given the counter is white, find the probability it came from Box B.
Five letters from the word CHANCE are arranged in a row (all six letters: C,H,A,N,C,E). How many distinct arrangements are there of all six letters?
Three cards are drawn without replacement from a standard pack of 52. Find the probability that exactly two are hearts.
Events A and B are independent with P(A) = 0.3 and P(B) = 0.4. Find P(A∪B) and P(A'∩B').
In a group of 80 people, each likes tea (T) or coffee (C) or both. 50 like tea, 40 like coffee, and x like both. (i) Find x. (ii) A person is chosen at random. Find P(T|C). (iii) Determine whether T and C are independent.
5 questions drawn from Cambridge A-Level 9709 Statistics 1 past papers on Probability. Attempt under timed conditions before revealing the mark scheme.
Events A and B are such that P(A) = 0.6, P(B|A) = 0.3, and P(B|A') = 0.5. Find P(A|B).
A bag contains 5 blue and 3 yellow counters. Three counters are drawn without replacement. Find the probability that (i) all three are blue, (ii) at least one is yellow.
The letters of the word PROBABLE are arranged in a random order. Find the probability that the two letters B are not next to each other.
A fair six-sided die is thrown twice. Find the probability that the sum of the two scores is greater than 9.
A committee of 5 is chosen from 9 people: 4 men and 5 women. Find the number of ways the committee can be formed if it must include at least 3 women. Hence find the probability that a randomly formed committee of 5 includes at least 3 women.