Conditional Probability

What is conditional probability?

Conditional probability refers to situations where the probability of an event changes or is dependent on other events having already happened
For example, drawing names from a hat, without replacement
- If there are 10 (different) names in a hat to start with
- the first name drawn has the probability of $\frac{1}{10}$ of being a particular name
- the second name drawn has probability $\frac{1}{9}$ of being a particular name
  - or, if this particular name was the first one to be drawn, it would have probability 0 of being drawn second
- The probability has changed depending on what has happened already
Conditional probabilities often occur in the context of Venn diagrams, tree diagrams or two-way tables
- however questions may also be given in words only
- in such cases it may sometimes be easier to understand what is happening by drawing one of these diagrams
- unless a question tells you to though, drawing a diagram is not essential
- for many questions it is quicker simply to consider the possible options without drawing a diagram
Conditional probability questions are often in the form of "given that" questions
- e.g. Find the probability it will rain today given that it rained yesterday
  - It makes sense that whether or not it rained yesterday would affect the probability of whether or not it rains today
  - The phrase "given that" is not always used in conditional probability questions
  - Like AND/OR, you will need to interpret the phrases used in questions
Conditional probabilities are sometimes written using the 'straight bar' notation $P (A | B)$
- That is read as 'the probability of A given B'
- For example $P (passes | no revision)$ would indicate the probability that a student passes his exams, given that he has done no revision
- That probability is likely to be quite different from $P (passes | lots of revision)$ !

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 then set aside and not put back into the box.
A second counter is then taken at random, and its colour noted.

Write down the probability that

the second counter is red, given that the first counter was red

ii)

the second counter is blue, given that the first counter was red

iii)

the second counter is red, given that the first counter was blue

iv)

the second counter is blue, given that the first counter was blue.

If the first counter was red, then only 7 red counters remain in the box.
There are still 3 blue counters, and 10 counters in total.

$\begin{array}{rcl} \frac{7}{10} \end{array}$

ii)

If the first counter was red, then only 7 red counters remain in the box.
There are still 3 blue counters, and 10 counters in total.

$\begin{array}{rcl} \frac{3}{10} \end{array}$

iii)

If the first counter was blue, then only 2 blue counters remain in the box.
There are still 8 red counters, and 10 counters in total.

$\frac{8}{10} (= \frac{4}{5})$

iv)

If the first counter was blue, then only 2 blue counters remain in the box.
There are still 8 red counters, and 10 counters in total.

$\frac{2}{10} (= \frac{1}{5})$

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 ( $\times$ ) and OR means add ( $+$ ) 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
  - $P (both green) = \frac{7}{10} \times \frac{6}{9} = \frac{42}{90}$
  - 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

$(\frac{7}{10} \times \frac{3}{9}) + (\frac{3}{10} \times \frac{7}{9}) = \frac{21}{90} + \frac{21}{90} = \frac{42}{90}$

- 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

$2 \times (\frac{7}{10} \times \frac{3}{9}) = 2 \times \frac{21}{90} = \frac{42}{90}$

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

$6 \times (\frac{7}{10} \times \frac{6}{9} \times \frac{3}{8} \times \frac{2}{7}) = 6 \times \frac{252}{5040} = \frac{1512}{5040} (= \frac{3}{10})$

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

all three counters are purple

ii)

exactly one of the three counters is purple

iii)

at least one of the three counters is purple

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.

$\begin{array}{rcl} P (all 3 purple) & = & \frac{3}{10} \times \frac{2}{9} \times \frac{1}{8} \\ = & \frac{6}{720} \end{array}$

$\begin{array}{rcl} P (all 3 purple) & = & \frac{6}{720} (= \frac{1}{120}) \end{array}$

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.

$\begin{array}{rcl} P (one purple) & = & (\frac{3}{10} \times \frac{7}{9} \times \frac{6}{8}) + (\frac{7}{10} \times \frac{3}{9} \times \frac{6}{8}) + (\frac{7}{10} \times \frac{6}{9} \times \frac{3}{8}) \\ = & \frac{126}{720} + \frac{126}{720} + \frac{126}{720} \\ = & \frac{378}{720} \end{array}$

$\begin{array}{rcl} P (one purple) & = & \frac{378}{720} (= \frac{21}{40}) \end{array}$

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 $3 \times (\frac{3}{10} \times \frac{7}{9} \times \frac{6}{8}) = 3 \times \frac{126}{720} = \frac{378}{720}$ .

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).

$\begin{array}{rcl} P (all 3 green) & = & \frac{7}{10} \times \frac{6}{9} \times \frac{5}{8} \\ = & \frac{210}{720} \\ P (at least 1 purple) & = & 1 - \frac{210}{720} \\ = & \frac{720}{720} - \frac{210}{720} \\ = & \frac{510}{720} \end{array}$

$\begin{array}{rcl} P (at least one purple) & = & \frac{510}{720} (= \frac{17}{24}) \end{array}$

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.

$(\frac{9}{30} \times \frac{17}{29}) + (\frac{17}{30} \times \frac{9}{29}) + (\frac{9}{30} \times \frac{4}{29}) + (\frac{4}{30} \times \frac{9}{29}) + (\frac{17}{30} \times \frac{4}{29}) + (\frac{4}{30} \times \frac{17}{29}) = \frac{514}{870}$

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

$2 \times (\frac{9}{30} \times \frac{17}{29}) + 2 \times (\frac{9}{30} \times \frac{4}{29}) + 2 \times (\frac{17}{30} \times \frac{4}{29}) = \frac{514}{870}$

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).

$\begin{array}{rcl} P (both bags are different) & = & (\frac{9}{30} \times \frac{21}{29}) + (\frac{17}{30} \times \frac{13}{29}) + (\frac{4}{30} \times \frac{26}{29}) \\ = & \frac{189}{870} + \frac{221}{870} + \frac{104}{870} \\ = & \frac{514}{870} \end{array}$

$P (both bags are different) = \frac{514}{870} (= \frac{257}{435})$

Conditional Probability (CIE IGCSE Maths: Extended)

Revision Note