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How do you find probabilities of combined and conditional events?

Tree diagrams, Venn diagrams, sample space diagrams, independent and conditional probability, and set notation.

A focused answer to AQA GCSE Statistics on tree and Venn diagrams, covering sample space diagrams, calculating combined probabilities from tree diagrams, using Venn diagrams and set notation, and conditional probability.

Generated by Claude Opus 4.810 min answer

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  1. What this dot point is asking
  2. Sample space diagrams
  3. Tree diagrams
  4. Venn diagrams and set notation
  5. Conditional probability

What this dot point is asking

AQA wants you to use sample space diagrams, tree diagrams and Venn diagrams to find probabilities of combined events, distinguish independent from conditional probability, and use basic set notation. These diagrams are the main tools for the longer probability questions on both papers.

Sample space diagrams

The strength of a sample space diagram is that, when every cell is equally likely, probability is just counting: the chance of a total of 77 on two dice is the 66 cells that sum to 77 out of 3636, giving 636=16\frac{6}{36} = \frac{1}{6}. They become unwieldy for more than two stages, where a tree diagram is clearer.

Tree diagrams

Venn diagrams and set notation

Always fill the intersection first, then subtract to get the "only" regions, then place anything in neither set. The regions must total the whole group. The union ABA \cup B counts everything in AA or BB (including the overlap once), which is why the general addition rule subtracts the intersection: P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B).

Conditional probability

To read a conditional probability from a Venn diagram, restrict attention to the "given" set: P(FrenchGerman)P(\text{French} \mid \text{German}) is the number who do both divided by the number who do German, not by the whole class. The key shift is the denominator: a conditional probability narrows the sample space to only those cases where the given event happened, so you divide by the size of that group rather than by the whole population.

Choosing the right diagram speeds you up. A tree diagram is best for a sequence of stages, especially "without replacement" problems where probabilities change at each step, because the branch structure makes the changing fractions explicit. A Venn diagram is best for overlapping categories, because it lays out the "both", "only", and "neither" regions you need for unions, intersections and conditional probabilities. A sample space diagram suits two simple events with equally likely outcomes, where counting cells is quickest.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20205 marksA bag holds 55 green and 33 yellow counters. Two are drawn without replacement. (a) Draw a tree diagram for the two draws. (b) Calculate the probability that both counters are the same colour.
Show worked answer →

First draw: P(G)=58P(G) = \frac{5}{8}, P(Y)=38P(Y) = \frac{3}{8}.

Second draw depends on the first. Two greens: 58×47=2056\frac{5}{8} \times \frac{4}{7} = \frac{20}{56}. Two yellows: 38×27=656\frac{3}{8} \times \frac{2}{7} = \frac{6}{56}.

Same colour == both green or both yellow (mutually exclusive), so add: 2056+656=2656=1328\frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28}.

Markers reward correct second-draw fractions (denominator 77, reduced numerators), multiplying along each branch, and adding the two same-colour branches.

AQA 20224 marksIn a class of 3030 students, 1818 study French, 1414 study German, and 66 study both. (a) Complete a Venn diagram for these data. (b) A student is chosen at random. Find the probability the student studies French but not German.
Show worked answer →

Intersection (both) =6= 6. French only =186=12= 18 - 6 = 12. German only =146=8= 14 - 6 = 8. Neither =30(12+6+8)=4= 30 - (12 + 6 + 8) = 4.

French but not German is the "French only" region: P=1230=25P = \frac{12}{30} = \frac{2}{5}.

Markers reward filling the intersection first, subtracting to get the "only" regions, checking the regions total 3030, and reading 1230\frac{12}{30} for French but not German.

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