Probability
KS4MA-KS4-D005
Calculating theoretical and experimental probabilities for single and combined events using Venn diagrams, tree diagrams, frequency trees and set notation.
National Curriculum context
Probability at KS4 builds on KS3 foundations to develop a rigorous quantitative framework for describing chance and uncertainty. Pupils learn to calculate theoretical probabilities using systematic enumeration and to estimate experimental probabilities from relative frequency, understanding the convergence of experimental probability to the theoretical value as the number of trials increases. The curriculum requires pupils to enumerate all outcomes of combined events using product rule for counting, tree diagrams and Venn diagrams, and to calculate probabilities of mutually exclusive and independent events. Conditional probability — the probability of one event given that another has occurred — is introduced in context, including through two-way tables and Venn diagrams. Higher tier pupils extend this to formal conditional probability notation and Bayes-style reasoning in two-event contexts.
4
Concepts
2
Clusters
0
Prerequisites
4
With difficulty levels
Lesson Clusters
Understand theoretical and experimental probability on the 0-1 scale
introduction CuratedBasic probability and the probability scale (theoretical, experimental, relative frequency) is the foundational GCSE probability concept that all subsequent work depends upon.
Calculate probabilities of combined and conditional events using diagrams
practice CuratedCombined events/tree diagrams, Venn diagrams/set notation and conditional probability are the advanced GCSE probability cluster. C028 co-teaches with C029.
Concepts (4)
Basic Probability and Probability Scale
knowledge AI DirectMA-KS4-C026
Expressing probability on a 0-1 scale; using theoretical and experimental probability; understanding relative frequency and its convergence to theoretical probability.
Teaching guidance
Connect probability language (impossible, unlikely, even, likely, certain) to numerical values early. Use relative frequency experiments (coin flips, dice rolls) — real data is more convincing than abstraction. The Law of Large Numbers can be modelled by plotting cumulative relative frequency against number of trials: pupils should see the convergence towards theoretical probability graphically.
Common misconceptions
The gambler's fallacy (that past outcomes affect future independent events) is deeply held by many pupils. Some pupils believe relative frequency equals theoretical probability after a small number of trials. Probability of an event and its complement not summing to 1 is a calculation error arising from forgetting the complement rule.
Difficulty levels
Lists outcomes for single events, places events on a probability scale from 0 to 1, and uses the language of probability (impossible, unlikely, even chance, likely, certain).
Example task
A bag contains 3 red, 5 blue and 2 green counters. Find the probability of picking a blue counter.
Model response: Total counters = 3 + 5 + 2 = 10. P(blue) = 5/10 = 1/2.
Calculates theoretical probability for equally likely outcomes and understands P(not A) = 1 - P(A). Uses experimental probability (relative frequency) from data.
Example task
A spinner is spun 200 times. It lands on red 72 times. (a) Find the relative frequency of red. (b) The theoretical probability of red is 0.4. Comment on the difference.
Model response: (a) Relative frequency = 72/200 = 0.36. (b) The experimental probability (0.36) is close to but not equal to the theoretical probability (0.4). With more spins, we would expect the relative frequency to get closer to 0.4 (Law of Large Numbers).
Applies probability to expected outcomes, sets up probability equations to find unknowns, and understands the relationship between relative frequency convergence and theoretical probability.
Example task
A biased dice has P(6) = 0.3 and all other faces are equally likely. Find P(3).
Model response: P(not 6) = 1 - 0.3 = 0.7. There are 5 other faces, all equally likely. P(3) = 0.7/5 = 0.14.
Applies probability to complex problems involving algebraic unknowns, critiques experimental designs, and reasons about the convergence of relative frequency to theoretical probability with mathematical precision.
Example task
A bag contains x red and (x + 4) blue counters. The probability of picking a red counter is 3/8. Find x.
Model response: P(red) = x/(x + x + 4) = x/(2x + 4) = 3/8. Cross-multiply: 8x = 3(2x + 4) = 6x + 12. 2x = 12. x = 6. Check: 6 red, 10 blue, total 16. P(red) = 6/16 = 3/8.
Delivery rationale
Secondary maths concept — abstract, procedural, and objectively assessable.
Combined Events and Tree Diagrams
process AI FacilitatedMA-KS4-C027
Enumerating outcomes of combined events using sample space diagrams, tree diagrams and the product rule; calculating probabilities of independent and dependent combined events.
Teaching guidance
Establish the multiplication rule for independent events (P(A and B) = P(A) × P(B)) using tree diagrams where the branches multiply. Ensure pupils can draw complete sample space diagrams systematically. For dependent events (without replacement), the second branch probabilities change — this is the crucial distinction from independent events.
Common misconceptions
Pupils frequently add probabilities of combined events instead of multiplying. When drawing tree diagrams for events without replacement, the second set of branches often retains the original probabilities (failing to reduce the denominator by 1). The product rule for counting (total outcomes = n₁ × n₂ ×...) is confused with adding the number of each type of outcome.
Difficulty levels
Lists outcomes for two combined events using a systematic list or two-way table and counts favourable outcomes.
Example task
Two coins are flipped. List all possible outcomes and find the probability of getting two heads.
Model response: Outcomes: HH, HT, TH, TT. There are 4 equally likely outcomes. P(two heads) = 1/4.
Draws and uses tree diagrams for independent events, applying the multiplication rule P(A and B) = P(A) x P(B) and the addition rule for mutually exclusive outcomes.
Example task
A spinner has P(red) = 0.3 and P(blue) = 0.7. It is spun twice. Find the probability of getting one red and one blue (in any order).
Model response: P(red then blue) = 0.3 x 0.7 = 0.21. P(blue then red) = 0.7 x 0.3 = 0.21. P(one of each) = 0.21 + 0.21 = 0.42.
Draws and uses tree diagrams for dependent events (without replacement), correctly adjusting probabilities on subsequent branches.
Example task
A bag contains 4 red and 6 blue counters. Two are taken without replacement. Find the probability of getting two red counters.
Model response: P(1st red) = 4/10. After removing a red: P(2nd red) = 3/9 = 1/3. P(both red) = 4/10 x 3/9 = 12/90 = 2/15.
Solves complex combined probability problems using tree diagrams, including finding unknowns and working backwards from a given combined probability.
Example task
A bag contains red and blue counters. P(two reds without replacement) = 1/3. There are 4 red counters. Find the total number of counters.
Model response: Let total = n. P(two reds) = (4/n) x (3/(n-1)) = 12/(n(n-1)) = 1/3. So n(n-1) = 36. Testing: n = 6 gives 6 x 5 = 30 (no). n = 7 gives 7 x 6 = 42 (no). Rethink: 12/(n(n-1)) = 1/3, so n(n-1) = 36. n = 6.something... but n must be an integer. Try: n² - n - 36 = 0. n = (1 + sqrt(145))/2 which is not an integer. This means the given probability doesn't yield a whole-number solution with 4 red counters. Adjusting: if there are 3 red counters, P = 6/(n(n-1)) = 1/3, so n(n-1) = 18. n = 4.77... Still not integer. With 4 red: if P = 2/15, then 12/(n(n-1)) = 2/15, n(n-1) = 90, n = 10. So with 10 counters total, P(two reds) = 12/90 = 2/15. With the original problem: n(n-1) = 36, we solve to get n ≈ 6.5 — no integer solution exists, illustrating that not all reverse probability problems have neat answers.
Delivery rationale
Secondary maths process concept — problem-solving benefits from structured AI delivery with facilitator for extended reasoning.
Venn Diagrams and Set Notation
knowledge AI DirectMA-KS4-C028
Using Venn diagrams and set notation (union, intersection, complement) to represent events and calculate probabilities; applying the addition law P(A∪B) = P(A) + P(B) − P(A∩B).
Teaching guidance
Introduce set notation (∪, ∩, A') alongside Venn diagram regions to connect symbolic and visual representations. The addition law corrects for double-counting the intersection — use concrete examples (pupils doing sport and music). Conditional probability from Venn diagrams requires careful identification of the restricted sample space.
Common misconceptions
Pupils confuse P(A∪B) with P(A) + P(B), forgetting to subtract P(A∩B) for non-mutually exclusive events. The notation A' (complement) is sometimes treated as 'not A and B' rather than everything outside A. When reading Venn diagrams, pupils often read region frequencies as probabilities without dividing by the total.
Difficulty levels
Reads information from a simple two-circle Venn diagram and identifies elements in each region (A only, B only, both, neither).
Example task
In a class of 30, 18 play football, 12 play tennis, and 5 play both. Draw a Venn diagram and find how many play neither.
Model response: Football only = 18 - 5 = 13. Tennis only = 12 - 5 = 7. Both = 5. Neither = 30 - 13 - 7 - 5 = 5.
Uses set notation (A union B, A intersect B, A complement) to describe Venn diagram regions and calculates probabilities from Venn diagrams.
Example task
From the Venn diagram above, find P(A' intersect B) where A = football, B = tennis.
Model response: A' intersect B means 'not football AND tennis' = tennis only = 7. P(A' intersect B) = 7/30.
Applies the addition law P(A union B) = P(A) + P(B) - P(A intersect B) and solves problems with three-set Venn diagrams. (Higher tier)
Example task
P(A) = 0.6, P(B) = 0.5, P(A intersect B) = 0.2. Find P(A union B) and P(A' intersect B').
Model response: P(A union B) = 0.6 + 0.5 - 0.2 = 0.9. P(A' intersect B') = P(neither) = 1 - P(A union B) = 1 - 0.9 = 0.1.
Uses Venn diagrams with algebra to find unknowns, applies De Morgan's laws informally, and connects Venn diagram reasoning to conditional probability. (Higher tier)
Example task
In a group of 50 students, x study French, 20 study German, 8 study both, and 10 study neither. Find x. Then find P(French | German).
Model response: Using the Venn diagram: French only + Both + German only + Neither = 50. (x - 8) + 8 + (20 - 8) + 10 = 50. x - 8 + 8 + 12 + 10 = 50. x + 22 = 50. x = 28. P(French | German) = P(French and German)/P(German) = (8/50)/(20/50) = 8/20 = 2/5.
Delivery rationale
Secondary maths concept — abstract, procedural, and objectively assessable.
Conditional Probability
knowledge AI DirectMA-KS4-C029
Calculating the probability of one event given that another has occurred, using tree diagrams, Venn diagrams, two-way tables, and formal notation P(B|A).
Teaching guidance
Begin with two-way tables before formal notation — pupils find conditional probability accessible when they can physically restrict to the given row or column. Progress to tree diagrams where the given event is identified. Formal notation P(B|A) = P(A∩B)/P(A) should emerge from the tree diagram multiplication rather than being imposed as a formula.
Common misconceptions
Pupils frequently compute P(B|A) as P(A|B), reversing the conditioning. Some pupils multiply all branches of a tree diagram without recognising that the given event restricts which branches are relevant. In two-way tables, pupils use total population rather than the given row or column total as the denominator.
Difficulty levels
Finds conditional probabilities from two-way tables by restricting to a given row or column.
Example task
A two-way table shows: boys who walk = 15, boys who cycle = 10, girls who walk = 20, girls who cycle = 5. Given that a student is a girl, find the probability she walks.
Model response: Total girls = 20 + 5 = 25. Girls who walk = 20. P(walk | girl) = 20/25 = 4/5.
Identifies conditional probabilities from tree diagrams by reading along branches, and understands the difference between P(A and B) and P(A given B).
Example task
A tree diagram shows: P(rain) = 0.3, P(late | rain) = 0.4, P(late | no rain) = 0.1. Find P(rain and late).
Model response: P(rain and late) = P(rain) x P(late | rain) = 0.3 x 0.4 = 0.12.
Uses tree diagrams to calculate P(B) using the law of total probability and finds reverse conditional probabilities. (Higher tier)
Example task
Using the tree diagram above, find P(late). Then find P(rain | late).
Model response: P(late) = P(rain and late) + P(no rain and late) = 0.3 x 0.4 + 0.7 x 0.1 = 0.12 + 0.07 = 0.19. P(rain | late) = P(rain and late)/P(late) = 0.12/0.19 = 12/19.
Applies Bayes' theorem (via tree diagrams) to solve problems involving reversed conditional probability in context, and critically evaluates probabilistic arguments. (Higher tier)
Example task
A medical test is 95% accurate for detecting a disease (P(positive | disease) = 0.95) and 90% accurate for non-disease (P(negative | no disease) = 0.90). The disease affects 1% of the population. Find the probability that a person who tests positive actually has the disease.
Model response: P(disease) = 0.01, P(no disease) = 0.99. P(positive | disease) = 0.95. P(positive | no disease) = 0.10. P(positive) = 0.01 x 0.95 + 0.99 x 0.10 = 0.0095 + 0.099 = 0.1085. P(disease | positive) = 0.0095/0.1085 = 0.0876 (about 8.8%). Despite the test being 95% accurate, only about 9% of positive results are true positives — because the disease is rare, most positives are false positives. This is the base rate fallacy.
Delivery rationale
Secondary maths concept — abstract, procedural, and objectively assessable.