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AP®StatisticsCollege BoardStudy GuidesUnit 4: Probability, Random Variables & Probability DistributionsProbabilityProbabilities of Combined Events using Tree Diagrams

Probabilities of Combined Events using Tree Diagrams (College Board AP® Statistics): Study Guide

Dan Finlay

Written by: Dan Finlay

Reviewed by: Lucy Kirkham

Updated on

Tree diagrams

What is a tree diagram for probability?

  • A tree diagram is a way to visualize probabilities and conditional probabilities when dealing with combined events

  • Tree diagrams show how the occurrence of an event affects the probability of another event occurring

  • The first set of branches is labeled with the probabilities of mutually exclusive events

    • This is typically just a given event and its complement

      • e.g. rolling a 6 and not rolling a 6

    • However, there can be more than two events

      • e.g. rolling a 1, rolling a 2, rolling a 3, etc.

    • The probabilities on these branches add up to 1

  • The second sets of branches are labeled with the conditional probabilities given that one of the previous events has occurred

    • The conditional probabilities on the branches for each set of events add up to 1

  • Tree diagrams can have more than two sets of events

    • However, in these cases, there are more efficient ways to find the probabilities

      • See the student guide on finding probabilities of combined events using the rules

A tree diagram showing probabilities of winning or losing a game based on whether or not a six is rolled.The probabilities are on the branches.
Example of a tree diagram to show the probability of winning a game based on whether or not a six is rolled

Examiner Tips and Tricks

Be careful with the probabilities on the second set of branches, these are conditional probabilities. In the above example, 3 over 5 is the probability that a person wins a game given that they rolled a six. It is not the probability that the person wins the game.

How can I find probabilities from a tree diagram?

  • You can find the probability of the intersection of two events by multiplying the probabilities on their branches

    • This is just the multiplication rule P open parentheses A intersection B close parentheses equals P open parentheses A close parentheses times P open parentheses B vertical line A close parentheses

Probability tree diagram showing outcomes for rolling a six. Branches: rolling a six or not, then winning or losing. Shows calculations for each scenario's probability.
Example of a tree diagram with the probabilities of each intersection

  • To find the probability of an event in the second set of branches:

    • Identify any intersections which involve that event

    • Add together the probabilities of those intersections

      • e.g. P open parentheses wins close parentheses equals P open parentheses six space and space wins close parentheses plus P open parentheses not space straight a space six space and space wins close parentheses

  • It can be quicker to find the probability of the complement of an event

    • If you are asked to find the probability that A or B occurs

      • find the probability that neither A nor B occurs

      • then subtract this from 1

How can I find conditional probabilities from a tree diagram?

  • Suppose the tree diagram has the events A and A apostrophe on its first set of branches and the events B and B apostrophe on its second set

    • e.g. let A be the event that a 6 is rolled and B be the event that the game is won

  • The conditional probabilities P open parentheses B vertical line A close parentheses, P open parentheses B vertical line A apostrophe close parentheses, P open parentheses B apostrophe vertical line A close parentheses and P open parentheses B apostrophe vertical line A apostrophe close parentheses are labeled on the second set of branches

    • e.g. the probability that the game is won given that a six is rolled is labeled on the relevant branch

  • The conditional probabilities P open parentheses A vertical line B close parentheses, P open parentheses A vertical line B apostrophe close parentheses, P open parentheses A apostrophe vertical line B close parentheses and P open parentheses A apostrophe vertical line B apostrophe close parentheses can be found using the conditional probability formula

    • P open parentheses A vertical line B close parentheses equals fraction numerator P open parentheses A intersection B close parentheses over denominator P open parentheses B close parentheses end fraction

      • e.g. you have to use this formula to find the probability that a six is rolled given that the game is won

    • Remember you can partition the event B using A and A apostrophe

      • P open parentheses A vertical line B close parentheses equals fraction numerator P open parentheses A intersection B close parentheses over denominator P open parentheses A intersection B close parentheses plus P open parentheses A apostrophe intersection B close parentheses end fraction

Worked Example

To get to work, Jen either rides the subway or takes a cab. The probability that Jen is late for work depends on how she travels there. An incomplete tree diagram of the possible outcomes for Jen being late to work or not is shown below.

A probability tree diagram showing transportation to work and outcomes. Paths: rides subway or takes cab; late or on time. Contains conditional probabilities. Some probabilities are missing.

(a) Complete the tree diagram by writing the correct probability in each of the six empty boxes.

Answer:

The probabilities on each pair of branches add up to 1
Calculate the three missing probabilities on the branches

table row cell P open parentheses takes space straight a space cab close parentheses end cell equals cell 1 minus 0.34 equals 0.66 end cell row cell P open parentheses arrives space on space time vertical line rides space the space subway close parentheses end cell equals cell 1 minus 0.25 equals 0.75 end cell row cell P open parentheses arrives space on space time vertical line takes space straight a space cab close parentheses end cell equals cell 1 minus 0.45 equals 0.55 end cell end table

The probability of each outcome is found by multiplying the probabilities on the branches that lead to that outcome
Calculate the three missing probabilities for the outcomes

table row cell P open parentheses rides space the space subway space and space arrives space on space time close parentheses end cell equals cell 0.34 times 0.75 equals 0.255 end cell row cell P open parentheses takes space straight a space cab space and space late space for space work close parentheses end cell equals cell 0.66 times 0.45 equals 0.297 end cell row cell P open parentheses takes space straight a space cab space and space arrives space on space time close parentheses end cell equals cell 0.66 times 0.55 equals 0.363 end cell end table

The completed probability tree.

(b) Calculate the probability that Jen is late for work.

Answer:

Add together the probabilities of the outcomes that involve Jen being late for work

table row cell P open parentheses late space for space work close parentheses end cell equals cell P open parentheses rides space the space subway space and space is space late space for space work close parentheses plus P open parentheses takes space straight a space cab space and space is space late space for space work close parentheses end cell row blank equals cell 0.085 plus 0.297 end cell row blank equals cell 0.382 end cell end table

(c) Given that Jen is late for work, find the probability that she rode the subway.

Answer:

The conditional probability required is P open parentheses rides space the space subway vertical line late space for space work close parentheses
Use the conditional probability formula and the answer from the previous part

table row cell P open parentheses rides space the space subway vertical line late space for space work close parentheses end cell equals cell fraction numerator P open parentheses rides space the space subway space and space late space for space work close parentheses over denominator P open parentheses late space for space work close parentheses end fraction end cell row blank equals cell fraction numerator 0.085 over denominator 0.382 end fraction end cell row blank equals cell 0.2225... end cell end table

Given that Jen is late for work, the probability that she rode the subway is 0.223

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    Author
    Reviewer
    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.

    Lucy Kirkham

    Author: Lucy Kirkham

    Expertise: Head of STEM

    Lucy has been a passionate Maths teacher for over 12 years, teaching maths across the UK and abroad helping to engage, interest and develop confidence in the subject at all levels.Working as a Head of Department and then Director of Maths, Lucy has advised schools and academy trusts in both Scotland and the East Midlands, where her role was to support and coach teachers to improve Maths teaching for all.