Calculating Probabilities from Tree Diagrams Revision Notes | GCSE AQA Higher Maths

1. Why Tree Diagrams Are a Probability Superpower

Imagine you flip a coin and then spin a spinner. How many things could happen? Two events, each with their own outcomes — and the combinations multiply fast. Tree diagrams give you a structured way to track every possibility without missing a single one.

At GCSE, tree diagrams appear in questions about bags of counters, biased spinners, and selections from groups. The method is always the same: multiply along branches to find the probability of a combined outcome, and add across branches when you want the probability of more than one route leading to the same result.

2. Reading a Tree Diagram: The Core Rules

Every set of branches that sprouts from the same point must have probabilities that add up to exactly 1. This is your built-in check — if your branches from one point sum to anything other than 1, something has gone wrong before you even start calculating.

3. Independent Events: Worked Example

When events are independent, the probabilities on the second set of branches stay the same regardless of what happened on the first branch. A classic example: spinning a biased spinner twice, where the spinner has no memory of its previous result.

Notice that the second-level branches are identical regardless of which first branch you took. That is the hallmark of independent events — the second set of probabilities does not change. Now watch how the multiply-then-add method works on this tree.

Worked example: Finding P(at least one Red) from the spinner tree

A biased spinner lands on Red with probability 0.3 and Blue with probability 0.7. It is spun twice. Calculate the probability of getting at least one Red.

1. Red, Red: $0.3 \times 0.3 = 0.09$ Red, Blue: $0.3 \times 0.7 = 0.21$ Blue, Red: $0.7 \times 0.3 = 0.21$ Blue, Blue: $0.7 \times 0.7 = 0.49$ — Multiply along each path. Check: $0.09 + 0.21 + 0.21 + 0.49 = 1.00$ [M1 — correct multiplication on at least one branch]
2. Red, Red $\to 0.09$ Red, Blue $\to 0.21$ Blue, Red $\to 0.21$ — Three of the four paths contain at least one Red. The only path excluded is Blue, Blue.
3. $P(\text{at least one Red}) = 0.09 + 0.21 + 0.21 = 0.51$ — Add across the three qualifying branches. [A1 — correct final answer]
4. $P(\text{at least one Red}) = 1 - P(\text{Blue, Blue}) = 1 - 0.49 = 0.51$ ✓ — The complement method gives the same answer. On a calculator paper, this shortcut saves time and reduces arithmetic errors.

Answer: $P(\text{at least one Red}) = 0.51$

4. Dependent Events: Without Replacement

Here is where tree diagrams earn their keep. When you pick items from a bag without replacing them, the probabilities on the second set of branches change depending on what was picked first. This is the dependent events scenario — and it is where many students lose marks by copying the same fractions onto every branch.

The key adjustment: after the first pick, the total number of items in the bag decreases by one, and the count of whichever colour was picked also decreases by one. Every second-level branch must reflect this updated count.

Worked example: Dependent Events: Two Counters from a Bag Without Replacement

A bag contains 5 red counters and 3 blue counters. Two counters are taken out one at a time without replacement. Calculate the probability that both counters are the same colour.

1. $P(\text{Red first}) = \frac{5}{8}$ $P(\text{Blue first}) = \frac{3}{8}$ — 8 counters in total at the start. [M1 — correct first-level probabilities]
2. After Red: 4 red, 3 blue remain (7 total) $P(\text{Red} \mid \text{Red first}) = \frac{4}{7}$ $P(\text{Blue} \mid \text{Red first}) = \frac{3}{7}$ — One red counter has been removed, so the red count drops from 5 to 4 and the total drops from 8 to 7. [M1 — adjusted denominator]
3. After Blue: 5 red, 2 blue remain (7 total) $P(\text{Red} \mid \text{Blue first}) = \frac{5}{7}$ $P(\text{Blue} \mid \text{Blue first}) = \frac{2}{7}$ — One blue counter has been removed, so the blue count drops from 3 to 2 and the total drops from 8 to 7.
4. $P(\text{Red, Red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$ — Multiply along the Red-Red path. $5 \times 4 = 20$ and $8 \times 7 = 56$, then simplify by dividing both by 4.
5. $P(\text{Blue, Blue}) = \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}$ — Multiply along the Blue-Blue path. $3 \times 2 = 6$ and $8 \times 7 = 56$, then simplify by dividing both by 2.
6. $P(\text{same colour}) = \frac{5}{14} + \frac{3}{28}$ $= \frac{10}{28} + \frac{3}{28} = \frac{13}{28}$ — Convert $\frac{5}{14}$ to $\frac{10}{28}$ to get a common denominator, then add. [A1 — correct final answer]

Answer: $P(\text{same colour}) = \frac{13}{28}$

6. Deliberate Practice

These questions cover both independent and dependent scenarios. Work through each one using the multiply-then-add method, and check your branch probabilities sum to 1 at each node before calculating.

7. Exam Technique and Common Mistakes

Tree diagram questions are reliable mark-earners — but only if you avoid the traps that appear in almost every exam sitting. The mistakes below are drawn from the most common errors seen in GCSE probability questions.

Tree diagrams reward methodical working. Every mark on a tree diagram question — from writing correct branch probabilities to showing multiplication and addition steps — is available independently. Even if your final answer is wrong, clear working earns method marks.