Tree Diagrams

What is a tree diagram?

A tree diagram is used to
- show the outcomes of multiple events that happen one after the other
- help calculate probabilities when AND and/or OR’s are involved
Tree diagrams are mostly used when an event only has two outcomes of interest
- e.g. “ Rolling a 6 on a dice” and “ Not rolling a 6 on a dice"
- These outcomes are mutually exclusive (cannot happen at the same time)

How do I draw and label a tree diagram?

The first set of branches will represent the outcomes of the 1 ^st experiment
- in general we can call these outcomes " A" and "not A"
There will be two sets of branches representing the outcomes of the 2 ^nd experiment
- the first set will follow on from " A" in the 1 ^st experiment
- the second set will follow on from "not A" in the 1 ^st experiment
- for the 2 ^nd experiment we can generally call the outcomes " B" and "not B"
Probabilities for each outcome are written along the branches of the tree
At the end of the diagram we can collect together the combinations of the 2 experiments
- " A" and " B"
- " A and not B"
- "not A and B"
- "not A and not B"

3-1-3-fig1-tree-setup

How do I solve probability problems involving tree diagrams?

Interpret questions in terms of AND and/or OR
Draw, or complete a given, tree diagram
- Determine any missing probabilities
  - often using $1 - P (A)$
Write down the outcomes of both events and work out their probabilities
- These are AND statements
- $P (A AND B) = P (A) \times P (B)$
- You may see this as “Multiply along branches”
If more than one outcome is required then add their probabilities
- These are OR statements
- $P (A B OR " not A " " not B ") = P (A B) + P (" not A " " not B ")$
- You may see this as “Add different outcomes”
When you are confident with tree diagrams you can just pull out the outcome(s) you need
- you do not routinely have to work all of them out

How do I use tree diagrams with conditional probability?

Probabilities that depend on a particular thing having happened first in a tree diagram are called conditional probabilities
- For example a team's win and loss probabilities in one game may change depending on whether they won or lost the previous game
  - You might be interested in the probability of them winning a game after having lost the previous one
  - This probability will appear in the tree diagram in the set of branches that follow on from 'lose' in the first set of branches
- Or you might be asked to draw or complete a tree diagram for, say, the situation when two counters are drawn from a bag of different coloured counters without replacement
  - The probabilities on the second set of branches will change depending on which branch has been followed on the first set of branches
  - The denominators in the probabilities for the second set of branches will be one less than the denominators on the first set of branches
  - The numerators on the second set of branches will also change depending on what has happened on the first set of branches
  - See the Worked Example below for an example of this
Conditional probability questions are sometimes (but not always!) introduced by the expression 'given that...'
- For example 'Find the probability that the team win their next game given that they lost their previous game'
Conditional probabilities are sometimes written using the 'straight bar' notation $P (A | B)$
- That is read as 'the probability of A given B'
- For example $P (win | lose)$ would be the probability that the team wins, given that they lost their previous game
- The event after the straight bar occurs first, and the event before the straight bar occurs afterwards

CP Notes fig4 (1), downloadable IGCSE & GCSE Maths revision notes CP Notes fig4 (2), downloadable IGCSE & GCSE Maths revision notes

Exam Tip

It can be tricky to get a tree diagram looking neat and clear on the first first attempt
- it can be worth sketching a rough one first
- just keep an eye on that exam clock!
Tree diagrams make particularly frequent use of the result $P (not A) = 1 - P (A)$
Tree diagrams have built-in checks
- the probabilities for each pair of branches should add up to 1
- the probabilities for all final outcomes should add up to 1
When multiplying along branches with fractions it is often a good idea NOT to simplify any fractions (except possibly the final answer to the question)
- This is because fractions will often need to be added together, which is easier to do if they all have the same denominator

Worked example

A worker will drive through two sets of traffic lights on their way to work.
The probability of the first set of traffic lights being on green is $\frac{5}{7}$ .
The probability of the second set of traffic lights being on green is $\frac{8}{9}$ .

Draw and label a tree diagram including the probabilities of all possible outcomes.

Both sets of lights will either be on green (G) or red (R) (we can ignore yellow/amber for this situation).
We know the probabilities of the traffic lights being on green, so need to work out the probabilities of them being on red.

$P (1^{st} R) = 1 - P (1^{st} G) = 1 - \frac{5}{7} = \frac{2}{7}$
$P (2^{nd} R) = 1 - P (2^{nd} G) = 1 - \frac{8}{9} = \frac{1}{9}$

We also need to work out the combined probabilities of both traffic lights.

Find the probability that both sets of traffic lights are on red.

As we have written the probabilities of the combined events we can write the answer straight down.

$P (both red) = P (R, R) = \frac{2}{63}$

Find the probability that at least one set of traffic lights are on red.

This would be " R AND G" OR " G AND R" OR " R AND R" so we need to add three of the final probabilities.

$P (at least one R) = P (G, R) + P (R, G) + P (R, R) = \frac{5}{63} + \frac{16}{63} + \frac{2}{63} = \frac{23}{63}$

$P (at least one R) = \frac{23}{63}$

Because 'at least one R' is the same as 'not both G', we can also calculate this by subtracting P( G, G) from 1.

$P (at least one R) = 1 - P (G, G) = 1 - \frac{40}{63} = \frac{23}{63}$

Worked example

Liana has 10 pets $-$ 7 guinea pigs (G) and 3 rabbits (R).
Liana is choosing two pets to feature in her latest online video. First she is going to choose at random one of the pets. Once she has carried that pet to her video studio she is going to go back and choose at random a second pet to also feature in the video.

Draw and label a tree diagram including the probabilities of all possible outcomes.

For the 1st pet chosen, there will be a 7/10 probability of choosing a guinea pig, and a 3/10 probability of choosing a rabbit.

If the first pet is a guinea pig, there will only be 6 guinea pigs and 3 rabbits left (9 animals total). So for the second pet the probability of choosing a guinea pig would be 6/9, and probability of choosing a rabbit would be 3/9.

If the first pet is a rabbit, there will only be 7 guinea pigs and 2 rabbits left (9 animals total). So for the second pet the probability of choosing a guinea pig would be 7/9, and probability of choosing a rabbit would be 2/9.

Put these probabilities into the correct places on the tree diagram, and then multiply along the branches to find the probabilities for each outcome.

Find the probability that Liana chooses two rabbits.

As we have already calculated this probability in the table, we can just write the answer down.

$P (two rabbits) = P (R, R) = \frac{6}{90} (= \frac{1}{15})$

Find the probability that Liana chooses two different kinds of animal.

This would be "G AND R" OR "R AND G" so we need to add two of the final probabilities.

$P (two different kinds) = P (G, R) + P (R, G) = \frac{21}{90} + \frac{21}{90} = \frac{42}{90}$

$P (two different kinds) = \frac{42}{90} (= \frac{7}{15})$

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Tree Diagrams

Tree Diagrams

What is a tree diagram?

How do I draw and label a tree diagram?

How do I solve probability problems involving tree diagrams?

How do I use tree diagrams with conditional probability?

Exam Tip

Worked example

Worked example

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