Drawing and using tree diagrams for combined events, conditional probability, and using Venn diagrams with set notation.

What this dot point is asking

AQA wants you to draw and use tree diagrams for two-stage events (with and without replacement), handle conditional probability, and use Venn diagrams with set notation to organise overlapping groups. These are the main tools for combined-event problems at Higher tier, and they reward careful, systematic working over guesswork.

Tree diagrams for combined events

A tree diagram lays out a sequence of events as branches. Each branch carries a probability, and the branches from any point sum to $1$. To find the probability of a particular path, multiply along its branches; to find the probability of an event that can happen several ways, add the relevant path probabilities.

For two flips of a fair coin, the path "head then tail" has probability $\tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4}$. The probability of exactly one head adds the "head then tail" and "tail then head" paths: $\tfrac{1}{4} + \tfrac{1}{4} = \tfrac{1}{2}$.

With and without replacement

When an item is replaced before the next draw, the probabilities stay the same on every stage (independent events). When it is not replaced, both the favourable count and the total drop, so the second-stage probabilities change. This is conditional probability: the chance of the second event depends on the first.

Venn diagrams and set notation

A Venn diagram uses overlapping circles to sort items into sets. Always fill the overlap (the intersection) first, then subtract it from each total to get the "only" regions, and place anything in neither group outside the circles.

To read a probability from a filled Venn diagram, count the items in the region you want and divide by the total. Conditional probabilities ("given that the student studies French") restrict the total to one circle: divide the relevant region by the size of that circle.

The "at least one" shortcut

A frequent tree-diagram question asks for the probability of "at least one" of something across several trials. Listing every favourable path is slow and error-prone; the efficient method uses the complement. The probability of at least one success is $1$ minus the probability of no successes. For three independent attempts each succeeding with probability $0.2$, the chance of at least one success is $1 - 0.8^3 = 1 - 0.512 = 0.488$. Spotting the "at least one means one minus none" shortcut turns a multi-branch problem into a single line of working.

Conditional probability in context

Conditional probability questions are increasingly common at Higher tier and often arrive through a two-way table or a Venn diagram rather than a tree. The phrase "given that" is the signal: it tells you to shrink the sample space to a particular group. If $30$ of $50$ people own a car and $12$ of those car owners also own a bike, then the probability that a randomly chosen car owner owns a bike is $\dfrac{12}{30} = 0.4$, using the car owners as the new total rather than all $50$ people. Recognising which figure becomes the denominator is the whole skill.