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 ( $\times$ ) and is used for independent events
  - OR mean add ( $+$ ) 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. $P (rolling a 6) = \frac{1}{6}$ so $P (NOT rolling a 6) = 1 - \frac{1}{6} = \frac{5}{6}$
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

both counters are red,

ii)

the two counters are different colours.

This is an "AND" question: 1st counter red AND 2nd counter red.

$\begin{array}{rcl} P (both red) & = & P (R) \times P (R) \\ = & \frac{8}{11} \times \frac{8}{11} \\ = & \frac{64}{121} \end{array}$

$\begin{array}{rcl} P (both red) & = & \frac{64}{121} \end{array}$

ii)

This is an "AND" and "OR" question: [ 1st red AND 2nd green ] OR [ 1st green AND 2nd red ].

$\begin{array}{rcl} P (one of each) & = & [P (R) \times P (G)] + [P (G) + P (R)] \\ = & \frac{8}{11} \times \frac{3}{11} + \frac{3}{11} \times \frac{8}{11} \\ = & \frac{24}{121} + \frac{24}{121} \\ = & \frac{48}{121} \end{array}$

$\begin{array}{rcl} P (one of each) & = & \frac{48}{121} \end{array}$

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 $2 \times (\frac{8}{11} \times \frac{3}{11}) = 2 \times \frac{24}{121} = \frac{48}{121} .$

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.

$P (at least 1 win) = 1 - P (0 wins)$

Use the same fact to work out the probability of a loss.

$\begin{array}{rcl} P (lose) & = & 1 - P (win) \\ = & 1 - 0.26 \\ = & 0.74 \end{array}$

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 $0.74 \times 0.74 \times 0.74 \times 0.74 = {(0.74)}^{4}$ .

$\begin{array}{rcl} P (at least 1 win) & = & 1 - P (0 wins) \\ = & 1 - P (4 loses) \\ = & 1 - {(0.74)}^{4} \\ = & 0.700 134 . . . \end{array}$

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.

Combined Probability (AQA GCSE Maths)

Revision Note