Tree Diagrams
- Construct tree diagrams for two-event experiments (independent and dependent)
- Calculate probabilities using the multiply-along, add-branches method
- Handle without-replacement problems where probabilities change on the second draw
- Use the complement rule to find P(at least one) = 1 − P(none)
- Extract conditional probabilities from completed tree diagrams
🔑 Core Concepts
A. Structure of a Tree Diagram
A tree diagram maps every possible sequence of outcomes from a multi-stage experiment. Each branch represents one possible outcome; the probability of that outcome labels the branch. Reading left-to-right traces a single scenario through the experiment.
B. Independent Events (With Replacement)
Two events are independent if the outcome of the first does not affect the probability of the second. This happens whenever you replace a selected item before making a second selection. On the tree diagram, every second-stage branch carries the same probability regardless of which first-stage branch you are on.
C. Dependent Events (Without Replacement)
When items are not replaced, the pool from which the second item is drawn is smaller. Both the numerator and denominator of the probability fraction may change depending on which branch you took on the first draw.
D. Complement Rule — P(at least one)
Finding "at least one" directly often means adding many separate routes. The complement method is almost always faster: P(at least one success) = 1 − P(zero successes). P(none) is a single path along the tree — multiply its branches.
E. Conditional Probability from Tree Diagrams
Conditional probability $P(B \mid A)$ asks: "Given that A has already happened, what is the probability of B?" On a tree diagram, you restrict your view to the branches where A occurred, then look at B's sub-branches.
F. Three-Event Tree Diagrams
For three-stage experiments, the tree gains a third set of branches. The rules are identical: multiply along branches for AND, add across branches for OR. The total number of end-paths = (branches at stage 1) × (branches at stage 2) × (branches at stage 3). All end-path probabilities must sum to 1.
G. Frequency Trees vs Probability Trees
A frequency tree (also called a tree diagram with frequencies) uses actual counts rather than probabilities on branches. They are read differently: divide a node's branch-count by the node's total to find a probability.
| Feature | Probability Tree | Frequency Tree |
|---|---|---|
| Branch labels | Fractions/decimals (0–1) | Integer counts |
| Node values sum | Sum of branches = 1 | Sum of branches = parent count |
| Finding probability | Multiply along branches | Divide branch count by root total |
| Typical use | Given probabilities | Given survey/experiment data |
| AQA exam use | Main tree questions | Sometimes in data/statistics context |
H. Algebraic Probabilities in Trees (Grade 9)
At Grade 9, the probability on a branch may be given as an algebraic expression such as $x$ or $\frac{x}{10}$. Since all branches from a node sum to 1, you can form an equation to find $x$, then substitute back to read off all probabilities.
🗺️ Visual Notes
- Branch from each node for every outcome
- All branches at a node sum to 1
- Label every branch with its probability
- Total end-paths must also sum to 1
- Multiply along a single path
- Gives P(sequence of outcomes)
- Works for both independent & dependent
- Denominator changes without replacement
- Add end-path probabilities
- Identify all paths meeting the condition
- End-paths are mutually exclusive
- So simple addition is valid
- Denominator becomes $n-1$ on 2nd draw
- Numerator decreases if same type drawn
- Different 2nd-stage fractions on each branch
- Check: each node's branches still sum to 1
- P(at least 1) = 1 − P(none)
- P(none) = single path (all failures)
- Saves adding multiple paths
- Most efficient for Grade 9 questions
- Three-event trees
- Conditional probability $P(B|A)$
- Algebraic probabilities on branches
- P(exactly $k$ successes)
Decision Process: Solving Any Tree Diagram Problem
Annotated Two-Event Tree (With Replacement)
With vs Without Replacement — Probability Comparison
Bag: 3 red, 5 blue (8 total). Two draws.
| Outcome | With Replacement | Without Replacement |
|---|---|---|
| P(Red, Red) | $\frac{3}{8} \times \frac{3}{8} = \frac{9}{64}$ | $\frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28}$ |
| P(Red, Blue) | $\frac{3}{8} \times \frac{5}{8} = \frac{15}{64}$ | $\frac{3}{8} \times \frac{5}{7} = \frac{15}{56}$ |
| P(Blue, Red) | $\frac{5}{8} \times \frac{3}{8} = \frac{15}{64}$ | $\frac{5}{8} \times \frac{3}{7} = \frac{15}{56}$ |
| P(Blue, Blue) | $\frac{5}{8} \times \frac{5}{8} = \frac{25}{64}$ | $\frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$ |
| Sum | $\frac{64}{64} = 1$ ✓ | $\frac{56}{56} = 1$ ✓ |
✏️ Worked Examples
- If Green drawn 1st: 3G, 6Y remain. $P(G|G) = \frac{3}{9}$, $P(Y|G) = \frac{6}{9}$.
- If Yellow drawn 1st: 4G, 5Y remain. $P(G|Y) = \frac{4}{9}$, $P(Y|Y) = \frac{5}{9}$.
The complement method is far quicker than listing all 7 paths — this is the Grade 9 approach examiners expect.
❓ Exam Questions
A spinner has three equal sections labelled 1, 2, 3. It is spun once. Write down the probability that the spinner lands on 2.
Mark scheme: B1 for $\frac{1}{3}$ (or equivalent decimal 0.33… or percentage 33.3%).
A fair coin is flipped and a fair six-sided die is rolled. Using a tree diagram or otherwise, find the probability of getting a Head and a 6.
$$P(\text{Head and 6}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$$ Mark scheme: M1 for multiplying the two probabilities. A1 for $\frac{1}{12}$.
A bag contains 5 red balls and 3 white balls. One ball is picked at random, its colour noted, and it is not replaced. A second ball is then picked. Complete the tree diagram and find the probability that the two balls are different colours.
Second draw (dependent):
After Red: P(R|R) = $\frac{4}{7}$, P(W|R) = $\frac{3}{7}$.
After White: P(R|W) = $\frac{5}{7}$, P(W|W) = $\frac{2}{7}$.
Different colours = RW or WR:
$$P(\text{RW}) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}$$ $$P(\text{WR}) = \frac{3}{8} \times \frac{5}{7} = \frac{15}{56}$$ $$P(\text{different}) = \frac{15}{56} + \frac{15}{56} = \frac{30}{56} = \frac{15}{28}$$ Mark scheme: B1 for correct second-stage probabilities (denominators of 7). M1 for identifying and multiplying both paths. A1 for $\frac{15}{28}$.
The probability that Amara is late to school on any given day is 0.15, independently. Find the probability that Amara is late on at least one day in a two-day period (Monday and Tuesday).
P(not late on one day) = 1 − 0.15 = 0.85
P(not late on either day) = 0.85 × 0.85 = 0.7225
P(at least one day late) = 1 − 0.7225 = 0.2775
Method 2 — Direct (all 3 qualifying paths):
P(Late, Not late) = 0.15 × 0.85 = 0.1275
P(Not late, Late) = 0.85 × 0.15 = 0.1275
P(Late, Late) = 0.15 × 0.15 = 0.0225
Total = 0.1275 + 0.1275 + 0.0225 = 0.2775 ✓
Mark scheme: M1 for P(not late) = 0.85. M1 for 0.85² or equivalent. M1 for 1 − their P(none). A1 for 0.2775.
A box contains $n$ chocolates: 4 are dark chocolate and the rest are milk chocolate. Two chocolates are taken at random without replacement. The probability that both chocolates are dark is $\dfrac{2}{15}$. Find the value of $n$.
$$P(\text{dark 1st}) = \frac{4}{n}, \quad P(\text{dark 2nd} \mid \text{dark 1st}) = \frac{3}{n-1}$$ $$P(\text{both dark}) = \frac{4}{n} \times \frac{3}{n-1} = \frac{12}{n(n-1)}$$
Step 2 — Set equal to $\frac{2}{15}$:
$$\frac{12}{n(n-1)} = \frac{2}{15}$$
Step 3 — Cross-multiply and solve:
$$12 \times 15 = 2 \times n(n-1)$$ $$180 = 2n(n-1)$$ $$n(n-1) = 90$$ $$n^2 - n - 90 = 0$$ $$(n-10)(n+9) = 0$$ $$n = 10 \quad (\text{since } n > 0)$$
Mark scheme: M1 setting up $\frac{4}{n} \times \frac{3}{n-1}$. M1 equating to $\frac{2}{15}$. M1 forming quadratic. M1 factorising. A1 for $n = 10$. B1 for rejecting negative root with reason.
In a school survey, 200 students were asked whether they study Maths (M) and whether they study a Science (S). 130 study Maths; of those, 90 also study a Science. 30 students study neither. Use a frequency tree to find the probability that a randomly chosen student studies a Science but not Maths.
Total = 200. Study Maths: 130. Don't study Maths: 70.
Study Maths AND Science: 90. Study Maths only: 40.
Study neither: 30. So study Science but not Maths: 200 − 130 − 30 = 40.
(Check: Science only = 40. Total studying Science = 90 + 40 = 130.)
$$P(\text{Science but not Maths}) = \frac{40}{200} = \frac{1}{5}$$
Mark scheme: M1 setting up frequency tree correctly. B1 Science not Maths = 40. M1 forming probability. A1 $\frac{1}{5}$.
⭐ Grade 9 Model Answers
This question requires forming a probability expression with algebra, solving a quadratic, and justifying the rejection of a negative root. These are the hallmarks of a Grade 9 question.
Why each part earns marks:
• Setting up $\frac{4}{n} \times \frac{3}{n-1}$ correctly → demonstrates understanding of without-replacement AND the multiply rule (M1).
• Equating to $\frac{2}{15}$ → shows equation-forming skill (M1).
• Correct expansion and rearrangement to $n^2 - n - 90 = 0$ → algebraic manipulation (M1).
• Correct factorisation → quadratic solving (M1).
• $n = 10$ → final answer (A1).
• Explicit rejection of $n = -9$ with reason → shows mathematical rigour, the mark that separates Grade 8 from Grade 9 (B1).
📋 Revision Sheet
| Tree Diagram | Branching diagram showing all outcome sequences |
| Independent Events | P(A) unchanged by whether B occurred |
| Dependent Events | One event's outcome changes P of the next |
| Without Replacement | Item removed; pool shrinks for next draw |
| Conditional Probability | P(B|A) — probability of B given A has occurred |
| Frequency Tree | Tree with counts rather than probabilities |
| Complement | P(event) = 1 − P(event does not happen) |
AND (multiply along branch):
$P(A \cap B) = P(A) \times P(B \mid A)$
OR (add qualifying paths):
$P(A \cup B) = \sum P(\text{qualifying paths})$
Complement:
$P(\text{at least 1}) = 1 - P(\text{none})$
Conditional Probability:
$P(B \mid A) = \dfrac{P(A \cap B)}{P(A)}$
Without Replacement (2nd draw):
denominator $= n - 1$
- AND → × — "and" sounds like multiplication
- OR → + — "or" sounds like addition
- All branches from a node → sum to 1
- All end-paths → sum to 1
- Without replacement: −1 downstairs
- At least one = 1 − none ("Everything but the failure path")
- Conditional: restrict to A, then look at B
- Frequency tree: divide count by parent to get P
- Label every single branch — marks are given for correct labels
- Check all branch-sets sum to 1 before calculating
- For "at least one" problems, always consider the complement first
- State your method clearly — examiners award method marks
- For "without replacement", write $\frac{k}{n-1}$ not $\frac{k}{n}$ on second stage
- For algebraic trees, set up an equation and solve properly
- Reject negative/impossible solutions with a stated reason
- Final answers: simplify fractions unless told otherwise
🔄 Flashcards
Click each card to flip it and reveal the answer.
✗ Common Mistakes
What students do wrong: They write P(A and B) = P(A) + P(B), getting a value greater than 1 for common probabilities.
Why marks are lost: The answer is wrong and the method mark for "multiply along branch" is lost.
How to avoid it: AND = × (along branch). You only add when combining separate path results (OR). Memorise: "walk along the branch, multiply; collect separate paths, add."
What students do wrong: Writing $\frac{3}{8}$ on every second-stage branch even in a without-replacement problem (keeping the denominator as 8 instead of 7).
Why marks are lost: Both the method mark for correct second-stage branches and the accuracy mark are lost.
How to avoid it: Without replacement → write "−1 downstairs" on the 2nd draw. Also check: after Green from {3G,5B}, the remaining total is 7, not 8.
What students do wrong: They reduce the denominator to $n-1$ but leave the numerator the same for the branch where the same type was drawn. For example, after drawing Red from {3R, 5B}, they write P(R|R) = 3/7 instead of 2/7.
Why marks are lost: Incorrect probabilities cascade through the entire calculation, losing all subsequent accuracy marks.
How to avoid it: Ask yourself: "Has the type I drew first decreased in the bag?" If yes, subtract 1 from the numerator too.
What students do wrong: They list all 7 out of 8 paths (in a three-event tree) and try to add them all, often making arithmetic errors or missing a path.
Why marks are lost: Even if the method is correct, an arithmetic slip means the final accuracy mark is dropped.
How to avoid it: Use the complement: P(at least one) = 1 − P(none). "None" is always just one single path. Faster, cleaner, fewer errors.
What students do wrong: They get $\frac{30}{56}$ and leave it unsimplified, or they give a decimal when a fraction is required.
Why marks are lost: The accuracy mark often requires "correct and simplified" form. AQA exam papers typically say "give your answer as a fraction in its simplest form."
How to avoid it: Always check whether the fraction simplifies (find HCF of numerator and denominator). $\frac{30}{56} = \frac{15}{28}$ since HCF(30, 56) = 2.
What students do wrong: They solve the quadratic, find $n = 10$ or $n = -9$, and circle both answers or simply write $n = 10$ without mentioning the negative root.
Why marks are lost: There is often a dedicated mark (B1) for stating that the negative root is rejected because $n$ must be positive (as a count or probability).
How to avoid it: Always write "reject $n = -9$ since $n$ is a number of items and must be positive." One sentence earns the mark.
✅ Final Checklist
Click each item to mark it as complete. Your progress is saved automatically.
0 / 14- I can draw a two-event tree diagram, labelling all branches correctly
- I can verify that all branches from each node sum to 1
- I understand what "independent events" means and when it applies
- I can set up a without-replacement tree, reducing the denominator to $n-1$
- I remember to reduce the numerator when the same type is drawn again
- I can multiply along branches to find P(A AND B)
- I can identify all qualifying paths and add them to find P(A OR B)
- I can use the complement to find P(at least one) = 1 − P(none)
- I can draw and read a three-event tree diagram
- I can find conditional probability P(B|A) from a completed tree
- I can set up an equation when branch probabilities are given algebraically
- I know how to reject impossible solutions (negative counts) with a reason
- I can read and complete a frequency tree, converting counts to probabilities
- I can simplify all probability fractions and check my final answer is between 0 and 1