Multistage Events: Arrays & Tree Diagrams
Learn to represent multistage events with arrays and tree diagrams for NSW Year 12 Mathematics Standard 2. When a chance experiment happens in two or more steps — two dice, a coin and a die, drawing two marbles — an array or a tree lays out every outcome so you can count the sample space and read probabilities off it.
You will build a two-way array for a two-stage event, draw a tree diagram for two or more stages, list the outcomes and read branch and combined-outcome probabilities — including how the branch probabilities change when items are drawn without replacement. A core Standard 2 probability skill that leads into the product rule and expected frequency.
Theory
Multistage events happen in two or more steps. This Year 12 Standard 2 (NSW) guide shows how to represent them with an array (two-way table) or a tree diagram — list every outcome, count the sample space, and read the probabilities straight off the diagram.
A multistage event is a chance experiment carried out in two or more steps — tossing a coin then rolling a die, drawing two marbles, spinning a spinner twice. This Year 12 Standard 2 (NSW) topic is about representing those outcomes so you can count and read probabilities from them.
An array (two-way table) is ideal for a two-stage event: one stage runs down the side, the other across the top, and every cell is one outcome (or the sum of the two numbers). A tree diagram handles two or more stages: each stage adds a fan of branches, and every path from the root to a leaf is one outcome.
Once the diagram lists every outcome you read the probability off it. When the outcomes are equally likely, \(P(\text{event})=\dfrac{\text{favourable}}{\text{total}}\). When each branch carries a probability, a combined outcome is found by multiplying along its path from the root to the leaf.
| \(+\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
| \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) |
| \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) |
| \(3\) | \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) |
| \(4\) | \(5\) | \(6\) | \(7\) | \(8\) | \(9\) | \(10\) |
| \(5\) | \(6\) | \(7\) | \(8\) | \(9\) | \(10\) | \(11\) |
| \(6\) | \(7\) | \(8\) | \(9\) | \(10\) | \(11\) | \(12\) |
For equally likely outcomes, count the cells or leaves and use:
The total number of outcomes is the two stage sizes multiplied:
Represent a two-stage event with an array
- Label the rows with the outcomes of stage 1 and the columns with the outcomes of stage 2.
- Fill each cell with the paired outcome (for example \(H5\)) or the sum of the row and column.
- Count the total cells (rows \(\times\) columns) and the cells that match the event.
- Write the probability as \(\dfrac{\text{favourable}}{\text{total}}\) and simplify.
Represent a multistage event with a tree
- Draw a branch for every outcome of stage 1, then grow a fresh fan of branches from each for stage 2 (and so on).
- Write the probability on each branch; on a without replacement tree the second-stage probabilities change.
- List the outcome at the end of each path, and read the probability you need — count leaves, or multiply along a path.
Build the array of all \(2\times 6=12\) outcomes, then count the favourable ones.
| \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) | |
|---|---|---|---|---|---|---|
| \(H\) | \(\text{H}1\) | \(\text{H}2\) | \(\text{H}3\) | \(\text{H}4\) | \(\text{H}5\) | \(\text{H}6\) |
| \(T\) | \(\text{T}1\) | \(\text{T}2\) | \(\text{T}3\) | \(\text{T}4\) | \(\text{T}5\) | \(\text{T}6\) |
| \(\text{total}\) | \(=\) | \(2\times 6 = 12\) |
| \(\text{head and }>4\) | \(:\) | \(\text{H5},\ \text{H6}\) |
| \(P\) | \(=\) | \(\dfrac{2}{12} = \dfrac{1}{6}\) |
Each cell is the sum of its row and column; count the cells equal to \(5\).
| \(+\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|
| \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
| \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
| \(3\) | \(4\) | \(5\) | \(6\) | \(7\) |
| \(4\) | \(5\) | \(6\) | \(7\) | \(8\) |
| \(\text{total}\) | \(=\) | \(4\times 4 = 16\) |
| \(\text{sum}=5\) | \(:\) | \((1,4),(2,3),(3,2),(4,1)\) |
| \(P(\text{sum}=5)\) | \(=\) | \(\dfrac{4}{16} = \dfrac{1}{4}\) |
Grow the tree stage by stage and list the paths; the outcomes are equally likely.
| \(\text{outcomes}\) | \(:\) | \(BC,BS,BF,WC,WS,WF\) |
| \(\text{total}\) | \(=\) | \(2\times 3 = 6\) |
| \(P(WF)\) | \(=\) | \(\dfrac{1}{6}\) |
The first sock is not replaced, so the second-stage branches change; multiply along the red–red path.
| \(P(\text{1st red})\) | \(=\) | \(\dfrac{4}{6}\) |
| \(P(\text{2nd red}\mid\text{red})\) | \(=\) | \(\dfrac{3}{5}\) |
| \(P(RR)\) | \(=\) | \(\dfrac{4}{6}\times\dfrac{3}{5} = \dfrac{2}{5}\) |
Common pitfalls
Frequently asked questions
When should I use an array and when should I use a tree diagram?
Use an array (two-way table) when the event has exactly two stages, such as rolling two dice or a coin and a die — it lays every outcome out in a neat grid. Use a tree diagram when there are two or more stages, or when the branch probabilities change from stage to stage, such as drawing marbles without replacement.
How do I count the total number of outcomes?
Multiply the number of outcomes at each stage. A coin (2) and a die (6) give 2 times 6 equals 12 outcomes; three coins give 2 times 2 times 2 equals 8. On an array it is rows times columns; on a tree it is the number of leaves. Do not add the stage sizes.
Are HT and TH the same outcome?
No. In a multistage event the order matters, so HT (head then tail) and TH (tail then head) are two separate outcomes. That is why the sample space for two coins is HH, HT, TH, TT — four outcomes, not three.
How do I read a probability off a tree diagram?
For a single combined outcome, multiply the probabilities along its path from the root to the leaf. For example, P(HH) for two coins is one half times one half, which is one quarter. If several paths satisfy an event, work out each and add them.
Why do the probabilities change on a without-replacement tree?
Because the item drawn first is not put back, so the second draw is made from fewer items. With 4 red and 2 blue socks, the first red has probability 4/6, but after removing a red sock the second red has probability 3/5 — the numbers on the second-stage branches must be updated.
Is this the same as the product rule P(A and B) = P(A) times P(B)?
It is closely related. Here the focus is on building the array or tree and reading outcomes and probabilities off it. Using the product rule to calculate P(A and B) for independent stages is developed as its own skill in the next subtopic.