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Probability Tree Diagrams (CIE IGCSE Maths: Extended)

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Roger

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20 February 2024

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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

Examiner 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.

AD8TrDKf_cie-igcse-we-4-2-3-tree-diagram-image

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.

Eymcf2I-_question

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.

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

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

Combined Probability

What do we mean by combined probabilities?

In general this means there is more than one event to bear in mind when considering probabilities
- these events may be independent or mutually exclusive
- they may involve an event that follows on from a previous event
  - e.g. Rolling a dice, followed by flipping a coin

How do I work with and calculate combined probabilities?

In your head, try to rephrase each question as an AND and/or OR probability statement
- e.g. The probability of rolling a 6 followed by flipping heads would be "the probability of rolling a 6 AND the probability of flipping heads"
- In general,
  - AND means multiply ( $\times$ ) and is used for independent events
  - OR mean add ( $+$ ) and is used for mutually exclusive events
The fact that all probabilities sum to 1 is often used in combined probability questions
- In particular when we are interested in an event "happening" or "not happening"
  - e.g. $P (rolling a 6) = \frac{1}{6}$ so $P (NOT rolling a 6) = 1 - \frac{1}{6} = \frac{5}{6}$
Tree diagrams can be useful for calculating combined probabilities
- especially when there is more than one event but you are only concerned with two outcomes from each
  - e.g. The probability of being stopped at one set of traffic lights and also being stopped at a second set of lights
- however unless a question specifically tells you to, you don't have to draw a diagram
- for many questions it is quicker simply to consider the possible options and apply the AND and OR rules without drawing a diagram

Worked example

A box contains 3 blue counters and 8 red counters.
A counter is taken at random and its colour noted.
The counter is put back into the box.
A second counter is then taken at random, and its colour noted.

Work out the probability that

both counters are red,

ii)

the two counters are different colours.

This is an "AND" question: 1st counter red AND 2nd counter red.

$\begin{array}{rcl} P (both red) & = & P (R) \times P (R) \\ = & \frac{8}{11} \times \frac{8}{11} \\ = & \frac{64}{121} \end{array}$

$\begin{array}{rcl} P (both red) & = & \frac{64}{121} \end{array}$

ii)

This is an "AND" and "OR" question: [ 1st red AND 2nd green ] OR [ 1st green AND 2nd red ].

$\begin{array}{rcl} P (one of each) & = & [P (R) \times P (G)] + [P (G) + P (R)] \\ = & \frac{8}{11} \times \frac{3}{11} + \frac{3}{11} \times \frac{8}{11} \\ = & \frac{24}{121} + \frac{24}{121} \\ = & \frac{48}{121} \end{array}$

$\begin{array}{rcl} P (one of each) & = & \frac{48}{121} \end{array}$

In the second line of working in part (ii) we are multiplying the same two fractions together twice, just 'the other way round'.

It would be possible to write that instead as $2 \times (\frac{8}{11} \times \frac{3}{11}) = 2 \times \frac{24}{121} = \frac{48}{121} .$

That sort of 'shortcut' is often possible in questions like this.

Worked example

The probability of winning a fairground game is known to be 26%.

If the game is played 4 times find the probability that there is at least one win.
Write down an assumption you have made.

At least one win is the opposite to no losses so use the fact that the sum of all probabilities is 1.

$P (at least 1 win) = 1 - P (0 wins)$

Use the same fact to work out the probability of a loss.

$\begin{array}{rcl} P (lose) & = & 1 - P (win) \\ = & 1 - 0.26 \\ = & 0.74 \end{array}$

The probability of four losses is an "AND" statement; lose AND lose AND lose AND lose.
Assuming the probability of losing doesn't change, this is $0.74 \times 0.74 \times 0.74 \times 0.74 = {(0.74)}^{4}$ .

$\begin{array}{rcl} P (at least 1 win) & = & 1 - P (0 wins) \\ = & 1 - P (4 loses) \\ = & 1 - {(0.74)}^{4} \\ = & 0.700 134 . . . \end{array}$

P(at least 1 win) = 0.7001 (4 d.p.)

The assumption that we made was that the probability of winning/losing doesn't change between games.
Mathematically this is described as each game being independent.
I.e., the outcome of one game does not affect the outcome of the next (or any other) game.

It has been assumed that the outcome of each game is independent.

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Probability Tree Diagrams (CIE IGCSE Maths: Extended)

Revision Note

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?

What do we mean by combined probabilities?

How do I work with and calculate combined probabilities?

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Author: Roger