Multi-Stage Experiments and Tree Diagrams

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.

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 11At least one

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

$P (at least one) = 1 - P (none)$

Example. A free-throw shooter scores 70% of the time. In 3 shots, $P (at least one goal) = 1 - 0. 3^{3} = 0.973$ .

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 $n > \frac{ln 0.01}{ln 0.9} \approx 43.7$ , 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 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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How many outcomes are there when two dice are rolled?
1. A 12
2. B 36
3. C 6
4. D 18
Show answer
B: 36
$6 \times 6$ .
A coin is tossed twice. What is $P (two heads)$ ?
1. A $\frac{1}{2}$
2. B $\frac{1}{4}$
3. C $\frac{1}{3}$
4. D $\frac{3}{4}$
Show answer
B: $\frac{1}{4}$
$\frac{1}{2} \times \frac{1}{2}$ .
Two dice are rolled. What is $P (double six)$ ?
1. A $\frac{1}{36}$
2. B $\frac{1}{6}$
3. C $\frac{2}{36}$
4. D $\frac{1}{12}$
Show answer
A: $\frac{1}{36}$
One outcome out of 36.

Multi-Stage Experiments and Tree Diagrams

Starting point: Years 9 and 10

Listing outcomes

Tree diagrams

With and without replacement

Quick summary

Quiz

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