Combined Probability

What is meant by combined probabilities?

This means there is more than one event to consider when finding probabilities
- Events may be independent or mutually exclusive
- An event may follow on from a previous event
  - e.g. Rolling a dice, followed by flipping a coin

How do I work with and calculate combined probabilities?

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
Recall that the sum of the probabilities of all the possible outcomes is 1
- This is useful 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
- They are most useful when there is more than one event, but only two outcomes for each
  - e.g. The probability of being stopped at one set of traffic lights and also being stopped at a second set of lights
- For many questions it is quicker 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

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

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

(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 20%.

If the game is played 3 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

$\begin{array}{rcl} P (lose) & = & 1 - P (win) \\ = & 1 - 0.2 \\ = & 0.8 \end{array}$

The probability of three losses is an "AND" statement; lose AND lose AND lose
Assuming the probability of losing doesn't change, this is $0.8 \times 0.8 \times 0.8 = {(0.8)}^{3}$

$\begin{array}{rcl} P (at least 1 win) & = & 1 - P (0 wins) \\ = & 1 - P (3 loses) \\ = & 1 - {(0.8)}^{3} \\ = & 0.488 \end{array}$

P(at least 1 win) = 0.488

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 (Edexcel GCSE Maths: Foundation)

Revision Note