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Probability Tree Diagrams (Cambridge O Level Maths)

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Roger

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Roger

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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 1st experiment
    • in general we can call these outcomes "A" and "not A"
  • There will be two sets of branches representing the outcomes of the 2nd experiment
    • the first set will follow on from "A" in the 1st experiment
    • the second set will follow on from "not A" in the 1st experiment
    • for the 2nd 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 minus straight P left parenthesis A right parenthesis
  • Write down the outcomes of both events and work out their probabilities
    • These are AND statements
    • straight P left parenthesis A space bold AND bold space B right parenthesis equals straight P left parenthesis A right parenthesis cross times straight P left parenthesis B right parenthesis
    • You may see this as “Multiply along branches”
  • If more than one outcome is required then add their probabilities
    • These are OR statements
    • straight P left parenthesis A B space bold OR space " not space A " " not space B " right parenthesis equals straight P left parenthesis A B right parenthesis plus straight P left parenthesis " not space A " " not space B " right parenthesis
    • 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 straight P open parentheses A vertical line B close parentheses  
    • That is read as 'the probability of A given B'
    • For example straight P open parentheses win vertical line lose close parentheses 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 straight P left parenthesis not space A italic right parenthesis italic equals 1 italic minus P italic left parenthesis A italic right parenthesis
  • 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 5 over 7.
The probability of the second set of traffic lights being on green is 8 over 9.

a)

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.

straight P open parentheses 1 to the power of st space R close parentheses equals 1 minus straight P open parentheses 1 to the power of st space G close parentheses equals 1 minus 5 over 7 equals 2 over 7
straight P open parentheses 2 to the power of nd space R close parentheses equals 1 minus straight P open parentheses 2 to the power of nd space G close parentheses equals 1 minus 8 over 9 equals 1 over 9

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

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

b)

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.

c)

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.

straight P open parentheses at space least space one space R close parentheses space equals space straight P open parentheses G comma space R close parentheses space plus space straight P open parentheses R comma space G close parentheses space plus space straight P open parentheses R comma space R close parentheses space equals space 5 over 63 space plus space 16 over 63 space plus space 2 over 63 space equals space 23 over 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.

straight P stretchy left parenthesis at space least space one space R stretchy right parenthesis space equals space 1 space minus space straight P open parentheses G comma space G close parentheses space equals space 1 space minus space 40 over 63 space equals space 23 over 63 

Worked example

Liana has 10 pets minus 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.

a)

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

b)

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.

c)

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.

straight P open parentheses two space different space kinds close parentheses space equals space straight P open parentheses straight G comma space straight R close parentheses space plus space straight P open parentheses straight R comma space straight G close parentheses space equals space 21 over 90 plus 21 over 90 equals 42 over 90

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 (cross times) and is used for independent events
      • OR mean add (plus) 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.  straight P open parentheses rolling space straight a space 6 close parentheses equals 1 over 6  so  straight P open parentheses NOT space rolling space straight a space 6 close parentheses equals 1 minus 1 over 6 equals 5 over 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

i)

both counters are red,

ii)

the two counters are different colours.

i)

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

table row cell straight P open parentheses both space red close parentheses end cell equals cell straight P open parentheses R close parentheses cross times straight P open parentheses R close parentheses end cell row blank equals cell 8 over 11 cross times 8 over 11 end cell row blank equals cell 64 over 121 end cell end table

table row cell bold P stretchy left parenthesis both space red stretchy right parenthesis end cell bold equals cell bold 64 over bold 121 end cell end table

ii)

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

table row cell straight P open parentheses one space of space each close parentheses end cell equals cell open square brackets straight P open parentheses R close parentheses cross times straight P open parentheses G close parentheses close square brackets plus open square brackets straight P open parentheses G close parentheses plus straight P open parentheses R close parentheses close square brackets end cell row blank equals cell 8 over 11 cross times 3 over 11 plus 3 over 11 cross times 8 over 11 end cell row blank equals cell 24 over 121 plus 24 over 121 end cell row blank equals cell 48 over 121 end cell end table

table row cell bold P stretchy left parenthesis one space of space each stretchy right parenthesis end cell bold equals cell bold 48 over bold 121 end cell end table

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 space cross times space open parentheses 8 over 11 cross times 3 over 11 close parentheses space equals space 2 space cross times space 24 over 121 space equals space 48 over 121 space.

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.

straight P open parentheses at space least space 1 space win close parentheses equals 1 minus straight P open parentheses 0 space wins close parentheses

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

table attributes columnalign right center left columnspacing 0px end attributes row cell straight P open parentheses lose close parentheses end cell equals cell 1 minus straight P open parentheses win close parentheses end cell row blank equals cell 1 minus 0.26 end cell row blank equals cell 0.74 end cell end table

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 cross times 0.74 cross times 0.74 cross times 0.74 equals open parentheses 0.74 close parentheses to the power of 4.

table row cell straight P open parentheses at space least space 1 space win close parentheses end cell equals cell 1 minus straight P open parentheses 0 space wins close parentheses end cell row blank equals cell 1 minus straight P open parentheses 4 space loses close parentheses end cell row blank equals cell 1 minus open parentheses 0.74 close parentheses to the power of 4 end cell row blank equals cell 0.700 space 134 space... end cell end table

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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    Roger

    Author: Roger

    Expertise: Maths

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.