# Multi-Stage Experiments and Tree Diagrams

> Two-step experiments, grids and tree diagrams, with and without replacement, and the "at least one" trick.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/tree-diagrams/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The lolly jar
At a birthday party, a jar holds 4 red and 6 green lollies. Leo reaches in without looking and takes one, eats it, then takes another. What is the chance both are red?

The first pick changes what is left in the jar. A **tree diagram** keeps track: one set of branches for the first pick, then a new set for the second pick with the jar's new numbers. Multiply along the branches and the answer appears.

**In short:** A tree diagram shows every outcome of an experiment with more than one stage. Multiply the probabilities along each branch, then add the paths that fit the event, and remember the second-stage numbers change when items are not replaced.

## Starting point: Years 9 and 10

### Listing outcomes

When two things happen, list every combination. For a coin and a die there are $2 \times 6 = 12$ outcomes. For two dice, use a grid of $6 \times 6 = 36$ outcomes.

### Worked example: Two dice
What is the chance the total of two dice is 7? The pairs (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) give 7. That is $\dfrac{6}{36} = \dfrac{1}{6}$.

### Tree diagrams

Draw one set of branches for each stage. Write the probability on each branch.

*Diagram: Tree diagram for picking two lollies without replacement from 4 red and 6 green*
*First pick, then second pick. The second set of branches uses the 9 lollies left.*

### Key idea: Multiply along, add down
- To find the probability of one path, **multiply** along its branches.
- If several paths fit the event, **add** their probabilities.

**Both red:** $\dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}$.

**One of each colour:** RG or GR, so $\dfrac{4}{10} \times \dfrac{6}{9} + \dfrac{6}{10} \times \dfrac{4}{9} = \dfrac{48}{90} = \dfrac{8}{15}$.

### Calculator tip: Multiply branches with the fraction key
Multiply along the branches with the fraction key so the answer stays exact: `2/5 × 3/4` gives $\frac{3}{10}$. Add the results of different paths the same way. Press `S⇔D` if a decimal is wanted.

### With and without replacement

- **With replacement:** the lolly goes back. The second set of branches is the same as the first.
- **Without replacement:** the lolly is eaten. The second stage has one fewer item, so the fractions change.

### Watch out: Check each set of branches adds to 1
Branches from one point must add to 1. If they do not, a probability is wrong.

**Mind map**

```text
Multi-stage experiments
  Count outcomes
    Lists and grids
    Multiply the options
  Tree diagrams
    Probability on each branch
    Multiply along
    Add matching paths
  Replacement
    With: same each stage
    Without: numbers change
  Shortcuts
    At least one: 1 - P(none)
```

### Year 11: The multiplication rule and conditional branches
The second-stage branches of a tree are **conditional probabilities**. In the lolly tree, $\dfrac{3}{9}$ is $P(\text{second red} \mid \text{first red})$. So

$$P(A \cap B) = P(A) \times P(B \mid A)$$

If $A$ and $B$ are independent, $P(B \mid A) = P(B)$, and this becomes $P(A) \times P(B)$.

**Working backwards.** Given that the second lolly is red, what is the chance the first was red? $P(\text{second red}) = \dfrac{12}{90} + \dfrac{24}{90} = \dfrac{36}{90}$. So $P(\text{first red} \mid \text{second red}) = \dfrac{12/90}{36/90} = \dfrac{1}{3}$.

### Year 11: At least one
"At least one" questions usually have many paths. Use the complement: the only way to miss is "none".

$$P(\text{at least one}) = 1 - P(\text{none})$$

**Example.** A free-throw shooter scores 70% of the time. In 3 shots, $P(\text{at least one goal}) = 1 - 0.3^3 = 0.973$.

**First success on a given try.** If each try succeeds with probability 0.7, the first success is on the 3rd try only if the first two fail: $0.3 \times 0.3 \times 0.7 = 0.063$.

**Least number needed.** For more than a 99% chance that at least one of $n$ people is left-handed (probability 0.1 each), solve $1 - 0.9^n > 0.99$: $0.9^n < 0.01$, so $n > \dfrac{\ln 0.01}{\ln 0.9} \approx 43.7$, and the group needs 44 people.

**Different boxes.** When one item comes from each of two boxes, the stages are independent: multiply the probabilities from each box, and add the paths that fit (such as "red then blue" plus "blue then red").

### Year 12: From trees to random variables
Let $X$ be the number of red lollies Leo gets. From the tree:

| $x$ | 0 | 1 | 2 |
|---|---|---|---|
| $P(X = x)$ | $\dfrac{30}{90}$ | $\dfrac{48}{90}$ | $\dfrac{12}{90}$ |

The probabilities add to 1. This is a **probability distribution**, the starting point for the next chapters. Its expected value is $0 \times \dfrac{30}{90} + 1 \times \dfrac{48}{90} + 2 \times \dfrac{12}{90} = 0.8$ red lollies.

## Quick summary

- Count outcomes for two stages by multiplying: a die and a coin give 12.
- In a tree, multiply along the branches, then add the paths that fit.
- Without replacement, the second-stage numbers change.
- Second-stage branches are conditional probabilities.
- $P(\text{at least one}) = 1 - P(\text{none})$.

## Quiz

Try these questions from the chapter quiz.

1. How many outcomes are there when two dice are rolled?
   - A. 12
   - B. 36
   - C. 6
   - D. 18
   - Answer: B. 36. $6 \times 6$.

2. A coin is tossed twice. What is $P(\text{two heads})$?
   - A. $\dfrac{1}{2}$
   - B. $\dfrac{1}{4}$
   - C. $\dfrac{1}{3}$
   - D. $\dfrac{3}{4}$
   - Answer: B. $\dfrac{1}{4}$. $\dfrac{1}{2} \times \dfrac{1}{2}$.

3. Two dice are rolled. What is $P(\text{double six})$?
   - A. $\dfrac{1}{36}$
   - B. $\dfrac{1}{6}$
   - C. $\dfrac{2}{36}$
   - D. $\dfrac{1}{12}$
   - Answer: A. $\dfrac{1}{36}$. One outcome out of 36.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/tree-diagrams/#quiz