Probability of Combined Events (AQA Level 3 Mathematical Studies (Core Maths)): Revision Note

Exam code: 1350

Naomi C

Written by: Naomi C

Reviewed by: Dan Finlay

Updated on

Probability of Combined Events

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

  • Independent events are those whose outcome is not dependent on any other event

    • E.g. "Throwing a heads" on a coin and "rolling a 6" on a dice are independent of each other, one event happening does not affect the other

  • Two events are mutually exclusive if they cannot both happen at once

    • For example, when rolling a dice the events “getting a prime number” and “getting a 6” are mutually exclusive

    • Complementary events are mutually exclusive

    • Mutually exclusive events are dependent events

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

      • "The probability of rolling a 6 AND the probability of flipping heads"

    • In general,

      • AND means multiply (cross times), this is used for independent events

      • OR mean add (plus), this 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

    • 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

How do I calculate combined probabilities for independent events?

  • For two events that are independent, you can calculate probabilities for combined events, using the following rules:

    • straight P open parentheses straight A space and space straight B close parentheses equals straight P open parentheses straight A intersection straight B close parentheses equals straight P open parentheses straight A close parentheses cross times straight P open parentheses straight B close parentheses

    • straight P open parentheses straight A apostrophe space and space straight B apostrophe close parentheses equals straight P open parentheses straight A apostrophe intersection straight B apostrophe close parentheses equals straight P open parentheses straight A apostrophe close parentheses cross times straight P open parentheses straight B apostrophe close parentheses

    • On a tree diagram, you would multiply the probabilities along the branches

      • These rules can also be used to test if two events are independent

      • If straight P open parentheses straight A intersection straight B close parentheses equals straight P open parentheses straight A close parentheses cross times straight P open parentheses straight B close parentheses then straight A and straight B are independent

  • Consider the situation of throwing two dice, "getting a 6" on one dice, straight P open parentheses 6 close parentheses equals 1 over 6, "getting an even number" on a second dice, straight P open parentheses even close parentheses equals 1 half

    • straight P open parentheses 6 intersection even close parentheses equals 1 over 6 cross times 1 half equals 1 over 12

    • straight P open parentheses open parentheses not space 6 close parentheses intersection open parentheses not space even close parentheses close parentheses equals open parentheses 1 minus 1 over 6 close parentheses cross times open parentheses 1 minus 1 half close parentheses equals 5 over 6 cross times 1 half equals 5 over 12

  • For two independent events, you can find the probability of different possible overall outcomes by adding their individual probabilities:

    • straight P open parentheses straight A space or space straight B close parentheses equals straight P open parentheses straight A union straight B close parentheses equals straight P open parentheses straight A close parentheses plus straight P open parentheses straight B close parentheses

    • On a tree diagram, you would add the relevant probabilities down the list of final outcomes

  • Consider the previous situation with two dice, and "getting both a 6 and an even" or "not getting a 6 and not getting an even"

    • straight P open parentheses open parentheses 6 intersection even close parentheses union open parentheses open parentheses not space 6 close parentheses intersection open parentheses not space even close parentheses close parentheses close parentheses equals 1 over 12 plus 5 over 12 equals 6 over 12 equals 1 half

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,

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) the two counters are different colours.

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

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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Naomi C

Author: Naomi C

Expertise: Maths Content Creator

Naomi graduated from Durham University in 2007 with a Masters degree in Civil Engineering. She has taught Mathematics in the UK, Malaysia and Switzerland covering GCSE, IGCSE, A-Level and IB. She particularly enjoys applying Mathematics to real life and endeavours to bring creativity to the content she creates.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Maths Subject 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.