Tree diagrams for two or more events, multiplying along branches and adding between paths, and conditional probability where the second event depends on the first (without replacement) at Higher tier.

What this dot point is asking

Edexcel expects you to use tree diagrams to find the probability of combined events over two or more stages, multiplying along branches and adding between separate paths, and at Higher tier to handle conditional probability where the second event depends on the first (drawing without replacement). Tree diagrams are the clearest way to organise multi-stage probability, and they make the AND and OR rules visual.

Building a tree diagram

A tree diagram shows the stages of an experiment as sets of branches. Each stage's branches cover all its outcomes, so their probabilities sum to $1$.

For two flips of a coin, the first stage branches into heads ($\tfrac{1}{2}$) and tails ($\tfrac{1}{2}$), and each of those branches again into heads and tails. The path heads-then-heads has probability $\tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4}$.

Multiplying along, adding between

The two rules work together on a tree: multiply along a path, then add across paths.

To find the probability of getting exactly one head in two flips, identify the two favourable paths (heads-then-tails and tails-then-heads), find each by multiplying ($\tfrac{1}{4}$ each), then add: $\tfrac{1}{4} + \tfrac{1}{4} = \tfrac{1}{2}$. The phrase "at least one" is often handled fastest by the complement: $P(\text{at least one}) = 1 - P(\text{none})$.

Conditional probability without replacement (Higher)

When an item is not replaced, the second stage depends on the first, so the branch probabilities change.

The "at least one" shortcut

Questions asking for "at least one" of something are usually quickest via the complement. To find the probability of at least one green in the box above, it is easier to find the probability of no green (both yellow) and subtract from $1$, rather than adding several paths.

Independent versus dependent events

The single most important decision on a tree question is whether the events are independent or dependent. Independent events, such as flipping a coin twice or rolling two dice, do not affect each other, so the branch probabilities stay the same at every stage. Dependent events, such as taking counters without replacement, do affect each other, so the probabilities change because the contents change. A quick test is to ask: "after the first event, is the situation the same as before?" If yes, the events are independent; if the total or the makeup has changed, they are dependent and you must adjust the second set of branches.

Try this

Q1. A coin is flipped twice. Work out the probability of two heads. [2 marks]

• Cue. Independent events: $\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$.

Q2. A bag has $3$ red and $2$ blue balls. One is taken and not replaced, then another is taken. Work out the probability both are blue. [3 marks]

• Cue. $\dfrac{2}{5} \times \dfrac{1}{4} = \dfrac{2}{20} = \dfrac{1}{10}$.