Revision NotesExam QuestionsFlashcardsPast Papers
GCSEMathsAQARevision Notes5. ProbabilityCombined & Conditional ProbabilityCombined Conditional Probabilities

Combined Conditional Probabilities (AQA GCSE Maths)

Revision Note

Dan

Author

Dan

Last updated

Did this video help you?

Combined Conditional Probabilities

What are combined conditional probabilities?

  • Many trickier probability questions involve both conditional probabilities and combined probabilities
  • For example drawing multiple counters from a bag of different coloured counters without replacing the counters already drawn
    • Probability questions like this are sometimes called 'without replacement' questions
  • Because multiple counters are being drawn, you need to use combined probabilities to find the probability of, say, drawing two counters of the same colour
  • But because what is drawn one time affects the probabilities for what might be drawn on subsequent times, it is also necessary to consider conditional probability in finding answers

How do I answer combined conditional probability questions?

  • You need to bring combined probability and conditional probability ideas together in an appropriate manner
  • Most of these questions will involve using the AND means multiply (bold cross times) and OR means add (bold plus) rules from combined probability
    • Don't forget that events must be independent to use the AND rule, and mutually exclusive to use the OR rule
  • But the probabilities for second (and subsequent) events will change according to the ideas of conditional probability
    • So when e.g. drawing counters from a bag without replacement, the probabilities for the second counter drawn will depend on which counter was drawn first
  • Consider a bag containing 7 green counters and 3 purple counters
    • One counter is drawn and its colour recorded, then a second counter is drawn without replacing the first counter in the bag
      • What is the probability that both counters are green?
    • The probability that the first counter will be green is 7/10 (number of green divided by total number in bag)
      • But if the first one is green then there are only 6 green tokens left in the bag, and 9 tokens left in total
      • So the probability of the second counter being green (if the first one was green) is 6/9
      • This is the conditional probability part of the question
    • Now to find the probability of '1st counter green AND 2nd counter green' we multiply those two probabilities together
      • straight P open parentheses both space green close parentheses equals 7 over 10 cross times 6 over 9 space equals 42 over 90
      • This is the combined probability part of the question

What about when two things are happening at the same time?

  • It is not always stated in a question that one thing happens first and then another thing happens later
    • For example a question might not say that a counter is drawn from a bag of counters and not replaced, and then a second counter is drawn
    • Instead it might simply say that two counters are drawn from a bag of counters
  • In answering a question like this you can always assume that the relevant things happen one after the other
    • This doesn't change the maths of the question at all, but it makes it a lot easier to answer!
    • So for the example of drawing two counters from a bag
      • you can still just break it down into the probabilities for the 'first counter' and the 'second counter'
      • drawing two counters at the same time is exactly the same as drawing one counter and then a second counter without replacement

Are there any useful shortcuts for combined conditional probability questions?

  • Consider a bag containing 7 green counters and 3 purple counters, from which 2 counters are drawn without replacement
    • For the probability of the two counters being different colours we can use the AND/OR rule with '[1st green AND 2nd purple] OR [1st purple AND 2nd green]' to get

open parentheses 7 over 10 cross times 3 over 9 close parentheses space plus space open parentheses 3 over 10 cross times 7 over 9 close parentheses space equals space 21 over 90 space plus space 21 over 90 space equals space 42 over 90

    • Notice that both 'AND' probabilities are equal (21/90)
      • This is because the same numerators and denominators occur in the original fractions
      • The numerators are swapped around, but this doesn't change the value of the product
    • So another way to do this is to realise that
      • there are two ways for the counters to be different colours (green then purple, or purple then green)
      • and for each of those ways the probability is the same (7/10 times 3/9, or 3/10 times 7/9)
      • therefore the probability is

2 space cross times space open parentheses 7 over 10 cross times 3 over 9 close parentheses space equals space 2 space cross times space 21 over 90 space equals space 42 over 90

  • This method can be especially useful when you have to consider more than just two things
    • For the same bag of counters above, how about the probability of drawing 2 green and 2 purple counters when 4 counters are drawn without replacement?
    • You can do that with the AND/OR method, but you would have to multiply and add together for '[1st green AND 2nd green AND 3rd purple AND 4th purple] OR [1st green AND 2nd purple AND 3rd green AND 4th purple] OR...' (etc., etc.)
      • There would be six products of four fractions each to multiply and then add together
    • Or you can realise that
      • there are 6 ways for '2 green and 2 purple' to happen (GGPP, GPGP, GPPG, PPGG, PGPG, PGGP)
      • and for each of those ways the probability is the same (7/10 times 6/9 times 3/8 times 2/7 -- it doesn't matter which order those numerators are put in, as long as there's a 7, a 6, a 3, and a 2)
      • therefore the probability is

6 space cross times space open parentheses 7 over 10 cross times 6 over 9 cross times 3 over 8 cross times 2 over 7 close parentheses space equals space 6 space cross times space 252 over 5040 space equals space 1512 over 5040 space space open parentheses equals 3 over 10 close parentheses

Examiner Tip

  • In general use whatever the question does for probabilities – decimals, fractions or percentages
    • The only exception is that it can be easier to change percentages to decimals, especially if multiplication is involved
  • When using fractions it is often a good idea NOT to simplify any fractions (except possibly the final answer)
    • 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 bag contains 7 green counters and 3 purple counters.
A counter is taken at random and its colour noted.  The counter is not returned to the box.
Then a second counter is taken at random and its colour noted.  It also is not returned to the box.
Finally a third counter is taken at random, and its colour noted.

Work out the probability that

i)

all three counters are purple

ii)

exactly one of the three counters is purple

iii)

at least one of the three counters is purple

i)

This is an "AND" question: 1st purple AND 2nd purple AND 3rd purple.
Each time a purple is chosen there is one less purple left, so the numerator goes down by one each time.
And each time a counter is chosen the total number remaining in the bag is one less, so the denominator also goes down by one each time.

table row cell straight P open parentheses all space 3 space purple close parentheses space end cell equals cell space 3 over 10 space cross times space 2 over 9 space cross times space 1 over 8 end cell row blank equals cell space 6 over 720 end cell end table

ii)

This is an "AND" and "OR" question: [ 1st purple AND 2nd green AND 3rd green ] OR [ 1st green AND 2nd purple AND 3rd green ] OR [ 1st green AND 2nd green AND 3rd purple ].
The denominator needs to go down by 1 each time.
And the numerators need to change each time based on how many of each colour are left after previous counters have been taken.

table row cell straight P open parentheses one space purple close parentheses space end cell equals cell space open parentheses 3 over 10 cross times 7 over 9 cross times 6 over 8 close parentheses space plus space open parentheses 7 over 10 cross times 3 over 9 cross times 6 over 8 close parentheses space plus space open parentheses 7 over 10 cross times 6 over 9 cross times 3 over 8 close parentheses end cell row blank equals cell space 126 over 720 space plus space 126 over 720 space plus space 126 over 720 end cell row blank equals cell space 378 over 720 end cell end table

It would be quicker to do this using the 'shortcut method'.
There are 3 ways to get exactly 1 purple counter (PGG, GPG, and GGP).

So the probability is  3 space cross times space open parentheses 3 over 10 cross times 7 over 9 cross times 6 over 8 close parentheses space equals space 3 space cross times space 126 over 720 space equals space 378 over 720. 

iii)

The easiest way to do this is to realise that 'at least 1 purple' is the same as 'NOT all green'.
So find the 'all 3 green' probability and subtract it from 1.
'All three green' is an "AND" question: 1st green AND 2nd green AND 3rd green.
The numerators and denominators will change just as in part (i).

table row cell straight P open parentheses all space 3 space green close parentheses space end cell equals cell space 7 over 10 space cross times space 6 over 9 space cross times space 5 over 8 end cell row blank equals cell space 210 over 720 end cell row blank blank blank row cell straight P open parentheses at space least space 1 space purple close parentheses space end cell equals cell space 1 space minus space 210 over 720 end cell row blank equals cell space 720 over 720 space minus space 210 over 720 end cell row blank equals cell space 510 over 720 end cell end table

Worked example

A large box contains 30 bags of crisps.  There are 9 bags of ready salted crisps (R), 17 bags of salt and vinegar crisps (V), and 4 bags of cheese and onion crisps (C).
Aram takes at random two bags of crisps from the box.
Work out the probability that the two bags he takes are of different types.

There are several ways to do this.

The long way is as an AND/OR question considering all the different options: '[R AND V] OR [V AND R] OR [R AND C] OR [C AND R] OR [V AND C] OR [C AND V]'.
Be sure to change the numerators and denominators in the fractions to match what is left after each choice.

 open parentheses 9 over 30 cross times 17 over 29 close parentheses plus open parentheses 17 over 30 cross times 9 over 29 close parentheses plus open parentheses 9 over 30 cross times 4 over 29 close parentheses plus open parentheses 4 over 30 cross times 9 over 29 close parentheses plus open parentheses 17 over 30 cross times 4 over 29 close parentheses plus open parentheses 4 over 30 cross times 17 over 29 close parentheses equals 514 over 870 

You could also use the 'shortcut method' to simplify this slightly:
'2cross times[R AND V] + 2cross times[R AND C] + 2cross times[V AND C]'.

 2 cross times open parentheses 9 over 30 cross times 17 over 29 close parentheses space plus space 2 cross times open parentheses 9 over 30 cross times 4 over 29 close parentheses space plus space 2 cross times open parentheses 17 over 30 cross times 4 over 29 close parentheses space equals space 514 over 870 

The simplest way, however, is to do this as an AND/OR question with the following breakdown:
'[R AND not R] OR [V AND not V] OR [C AND not C]'
For [R AND not R] there is 9/30 probability of the first bag being R.  Then there are 17+4=21 V and C bags left in the box, and 29 bags in total left in the box.  So there is a 21/29 probability of the second bag NOT being R.
Do the same thing to find the numerators and denominators for [V AND not V] and [C AND not C].
Note that for each of the products, the numerators add up to 30 (the total number of crisps in the box to start).

  table row cell straight P open parentheses both space bags space are space different close parentheses space end cell equals cell space open parentheses 9 over 30 cross times 21 over 29 close parentheses space plus space open parentheses 17 over 30 cross times 13 over 29 close parentheses space plus space open parentheses 4 over 30 cross times 26 over 29 close parentheses end cell row blank equals cell space 189 over 870 space plus space 221 over 870 space plus space 104 over 870 end cell row blank equals cell space 514 over 870 end cell end table 

 bold P stretchy left parenthesis both space bags space are space different stretchy right parenthesis bold equals bold 514 over bold 870 bold space bold space stretchy left parenthesis equals 257 over 435 stretchy right parenthesis

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourselfFlashcards

    Did this page help you?

    Previous:Conditional ProbabilityNext:Mean, Median & Mode
    Dan

    Author: Dan

    Expertise: Maths

    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.