Tree Diagrams for Combined Events Revision Notes | GCSE AQA Higher Maths

1. Why Combined Events Need a System

Flipping a coin and rolling a die at the same time gives 12 possible outcomes. Picking two counters from a bag one after another gives even more — and the probabilities shift depending on what you picked first. Without a clear method, it is very easy to miss outcomes or multiply when you should add.

This lesson gives you that system: a set of rules for when to multiply, when to add, and how a tree diagram keeps everything organised. These rules underpin a significant portion of Higher-tier probability questions.

2. The Two Core Rules: Multiply or Add?

Every combined-events question comes down to two decisions. The AND rule says: to find the probability that event A and event B both happen, you multiply their probabilities. The OR rule says: to find the probability that event A or event B happens (where those outcomes cannot both happen at the same time), you add their probabilities.

The notation $P(B|A)$ means "the probability of B given that A has already happened". For independent events, knowing A happened tells you nothing about B, so $P(B|A) = P(B)$ and the formula collapses to a straightforward multiplication.

A reliable memory cue: AND → multiply along a path; OR → add between paths. You will use both rules in almost every tree diagram question.

3. Independent Events and Tree Diagrams

When events are independent, the probabilities on the second set of branches are identical regardless of what happened on the first branch. This is the key visual signal on a tree diagram: all second-stage branches carry the same values.

Notice that all four outcome probabilities add to $\frac{4}{25} + \frac{6}{25} + \frac{6}{25} + \frac{9}{25} = \frac{25}{25} = 1$. This is your built-in check: if the outcomes do not sum to 1, there is an arithmetic error somewhere.

Worked example: Independent Events: Probability from a Tree Diagram

A fair spinner has sections numbered 1 to 5. It is spun twice. Calculate the probability of getting a number greater than 3 on both spins.

1. Numbers greater than 3 from {1, 2, 3, 4, 5} are {4, 5}, so $P(>3) = \frac{2}{5}$ — Count favourable outcomes over total outcomes [M1]
2. The spinner is spun twice independently — the first result does not change the spinner, so the events are independent. — Stating the assumption earns credit in 'explain' questions.
3. $P(>3 \text{ and } >3) = \frac{2}{5} \times \frac{2}{5}$ — AND → multiply along the branch [M1]
4. $\frac{2}{5} \times \frac{2}{5} = \frac{2 \times 2}{5 \times 5} = \frac{4}{25}$ — Multiply numerators and denominators separately [A1]

Answer: $\frac{4}{25}$

4. Dependent Events: Without Replacement

Now for the scenario that catches most students out. When you pick an item from a set and do not replace it, the second pick comes from a smaller set — and the probability changes. These are dependent events, and the second-branch probabilities on your tree diagram must reflect the new totals.

The underlying assumption to state in an exam is: the first item is not replaced before the second is chosen. This is what makes the events dependent. If the question says "without replacement", the denominator on every second branch drops by 1.

Look carefully at the second-stage branches. After picking Red first, only 3 reds remain from 6 counters, so the second branch reads $\frac{3}{6}$. After picking Blue first, all 4 reds remain from 6 counters, so the second branch reads $\frac{4}{6}$. The denominators are all 6 (one fewer than the starting 7), but the numerators differ depending on the first outcome.

Worked example: Dependent Events: Two Picks Without Replacement

A bag contains 5 green and 3 yellow balls. Two balls are picked at random without replacement. Calculate the probability that both balls are the same colour.

1. Total balls = 8. $P(\text{Green first}) = \frac{5}{8}$, $P(\text{Yellow first}) = \frac{3}{8}$ — First-branch denominators are 8 [M1]
2. One green removed: 4 green, 3 yellow, 7 total. $P(\text{Green} | \text{Green first}) = \frac{4}{7}$ — Denominator drops to 7; green numerator drops to 4 [M1]
3. One yellow removed: 5 green, 2 yellow, 7 total. $P(\text{Yellow} | \text{Yellow first}) = \frac{2}{7}$ — Denominator drops to 7; yellow numerator drops to 2
4. $P(GG) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}$ — Multiply along the GG branch
5. $P(YY) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56}$ — Multiply along the YY branch
6. $P(\text{same colour}) = \frac{20}{56} + \frac{6}{56} = \frac{26}{56} = \frac{13}{28}$ — Two separate paths both give 'same colour', so add [A1]

Answer: $\frac{13}{28}$

5. Using the OR Rule Across Multiple Paths

Many exam questions ask for a probability that can be achieved in more than one way — for example, "exactly one red" or "at least one success". Each way is a separate branch path on your tree diagram. You calculate the probability of each path by multiplying along it, then add the results together.

The phrase "at least one" is a classic trigger for either the OR rule across multiple paths, or the complementary approach: $P(\text{at least one}) = 1 - P(\text{none})$. Both methods give the same answer; choose whichever is faster for the specific question.

Worked example: OR Rule: Exactly One Outcome of a Specific Type

Mohammed takes two independent shots at a target. The probability he hits with each shot is $\frac{3}{4}$. Calculate the probability that he hits the target exactly once.

1. Path 1: Hit then Miss (HM). Path 2: Miss then Hit (MH). — List all paths satisfying the condition before calculating [M1]
2. $P(H) = \frac{3}{4}$, $P(M) = 1 - \frac{3}{4} = \frac{1}{4}$ — Probabilities on complementary branches must sum to 1
3. $P(HM) = \frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$ — Multiply along the HM branch (AND rule)
4. $P(MH) = \frac{1}{4} \times \frac{3}{4} = \frac{3}{16}$ — Multiply along the MH branch (AND rule)
5. $P(\text{exactly one hit}) = \frac{3}{16} + \frac{3}{16} = \frac{6}{16} = \frac{3}{8}$ — Two separate paths both satisfy 'exactly one hit', so add [A1]

Answer: $\frac{3}{8}$

6. Practice: Putting It All Together

You have now seen independent events, dependent events, the AND rule, and the OR rule. The next step is combining all of these in a single question — which is exactly what GCSE Higher exam questions do. Work through the practice set below before looking at the solutions.

7. Exam Technique and Common Mistakes

examiners consistently award method marks for a correctly structured tree diagram even when the final answer is wrong. Drawing the tree, labelling all branches with fractions, and showing the multiplication and addition steps separately are all mark-winning moves.

Following these steps in order means you will never confuse when to multiply and when to add — the structure of the tree diagram makes the decision for you. Combined events questions reward method as much as the final answer, so a well-drawn, fully labelled tree is always worth the time.