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Combined Probability

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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