What a sample space is
When you do something with an uncertain result, such as flipping a coin or rolling a die, each thing that could happen is an outcome. Collect every possible outcome together and you have the sample space. Nothing fancy: it is just a complete list.
- Sample space
- The set of all possible outcomes of an experiment, often written inside curly brackets. Flipping one coin has sample space \(\{H, T\}\); rolling one die has \(\{1, 2, 3, 4, 5, 6\}\).
Once the outcomes are equally likely, probability follows straight away:
So the whole job is really counting. Get an honest, complete list of the sample space and you can count both the top and the bottom of that fraction.
Listing outcomes systematically
The danger with listing is missing one or writing one twice. The cure is to be systematic: fix the first part, run through every option for the second part, then move the first part on by one. That way you never lose your place.
Flipping two coins, hold the first coin at heads and list both options for the second coin, then switch the first coin to tails and repeat:
Four outcomes, each equally likely. Notice \(HT\) and \(TH\) are different: the order matters because the coins are two separate events.
Count before you list
If each part has its own number of options, multiply them to predict the total. Two coins give \(2 \times 2 = 4\); a coin and a die give \(2 \times 6 = 12\). If your list does not reach that total, you have missed one.
Sample space diagrams for two events
For two events, a table beats a plain list. A sample space diagram puts the outcomes of one event along the top and the outcomes of the other down the side, then fills each cell with the combined result.
Rolling two fair dice and adding the scores gives \(6 \times 6 = 36\) equally likely outcomes. The grid below shows every total:
| + | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Every one of the \(36\) cells is a single equally likely outcome, even though the totals themselves (like \(7\)) turn up more than once.
Reading probabilities off the grid
With the grid built, any probability is just a matter of counting the cells that match your event and dividing by \(36\).
Two fair dice are rolled and the scores added. Using the grid, find the probability that the total is 8.
Where this is assessed
Producing a complete, correctly counted sample space is Criterion A (Knowing and understanding). A clearly labelled grid with rows and columns headed is Criterion C (Communicating). Using it to answer a real question, such as fairness of a game, reaches into Criterion D (Applying in real-life contexts).
Check yourself
Try each one before opening the answer.
1. A coin is flipped and a fair die is rolled. How many outcomes are in the sample space? +
The coin has \(2\) outcomes and the die has \(6\), so the sample space has \(2 \times 6 = \) \(12\) outcomes: \(H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6\).
2. Using the two-dice grid, find the probability the total is 7. +
The cells showing \(7\) are \((1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\), which is \(6\) cells out of \(36\). So \(P(\text{total } 7) = \dfrac{6}{36} = \) \(\dfrac{1}{6}\).
3. Two fair dice are rolled. Find the probability the total is greater than 9. +
Greater than \(9\) means a total of \(10\), \(11\) or \(12\). Count the cells: \(10\) appears \(3\) times, \(11\) appears \(2\) times, \(12\) appears \(1\) time, giving \(3 + 2 + 1 = 6\) cells. So \(P(\text{total} > 9) = \dfrac{6}{36} = \) \(\dfrac{1}{6}\).
Part of the Standard Maths library. Spotted an error or want a topic added? That feedback makes the notes better.