27. Multi-Stage Experiments and Tree Diagrams

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.

Tree diagrams

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

First pick, then second pick. The second set of branches uses the 9 lollies left.

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

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

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.

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 12From trees to random variables

Let $X$ be the number of red lollies Leo gets. From the tree:

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

The probabilities add to 1. This is a probability distribution, the starting point for the next chapters. Its expected value is $0 \times \frac{30}{90} + 1 \times \frac{48}{90} + 2 \times \frac{12}{90} = 0.8$ 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.
$P (at least one) = 1 - P (none)$ .

Quiz

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