Sample Space & Theoretical Probability
Master sample space and theoretical probability for NSW Year 12 Mathematics Standard 2. The sample space is the list of all possible outcomes, and the theoretical probability of an event is the number of favourable outcomes divided by the total number of outcomes, \(P(E)=\dfrac{n(E)}{n(S)}\).
You will list sample spaces and count outcomes — including two-dice arrays and frequency tables — and write each probability from \(0\) (impossible) to \(1\) (certain) as a fraction, decimal or percentage, using the complement when it is easier. These are core Standard 2 probability skills for dice, coins, spinners and cards.
Theory
Sample space and theoretical probability is part of the probability topic in Year 12 Standard 2 (NSW). The sample space lists every possible outcome; the theoretical probability of an event is \(P(E)=\dfrac{n(E)}{n(S)}\), the favourable outcomes over the total. This guide shows how to count outcomes with lists and arrays, and write a probability from \(0\) to \(1\) as a fraction, decimal or percentage.
The sample space \(S\) of a chance experiment is the set of every possible outcome. The number of outcomes in it is written \(n(S)\) — for one roll of a die, \(S=\{1,2,3,4,5,6\}\) and \(n(S)=6\).
An event is the outcome, or group of outcomes, you are interested in. When the outcomes are equally likely, the theoretical probability of an event is the number of favourable outcomes divided by the total: \(P(E)=\dfrac{n(E)}{n(S)}\).
Every probability lies between \(0\) and \(1\): \(0\) means impossible and \(1\) means certain. You can write it as a fraction, decimal or percentage. In this Year 12 Standard 2 (NSW) topic you list sample spaces (including two-way arrays for two dice) and use the complement \(P(\text{not }A)=1-P(A)\).
For a sample space of equally likely outcomes, with \(n(E)\) favourable and \(n(S)\) in total:
Every probability is bounded, and the outcome probabilities of a sample space add to \(1\):
How to find a theoretical probability
- List the sample space — every possible outcome — and count them for \(n(S)\). For two dice, use a \(6\times6\) array.
- Identify the favourable outcomes for the event and count them for \(n(E)\).
- Divide: \(P(E)=\dfrac{n(E)}{n(S)}\).
- Simplify the fraction, or convert to a decimal or percentage, and check the answer lies between \(0\) and \(1\).
List the sample space, count the even outcomes, then divide by \(6\).
| \(n(S)\) | \(=\) | \(6\) |
| \(\text{even outcomes}\) | \(=\) | \(\{2,4,6\}\) |
| \(n(E)\) | \(=\) | \(3\) |
| \(P(\text{even})\) | \(=\) | \(\dfrac{3}{6}=\dfrac{1}{2}\) |
So \(P(\text{even})=\dfrac{1}{2}\) (which is \(0.5\)).
| + | 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 |
There are \(6\times6\) equally likely outcomes; count the cells equal to \(7\).
| \(n(S)\) | \(=\) | \(6\times6 = 36\) |
| \(\text{sums of }7\) | \(=\) | \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\) |
| \(n(E)\) | \(=\) | \(6\) |
| \(P(\text{sum}=7)\) | \(=\) | \(\dfrac{6}{36}=\dfrac{1}{6}\) |
The six shaded cells give \(P(\text{sum}=7)=\dfrac{1}{6}\).
| Colour | Red | Blue | Green |
|---|---|---|---|
| Number | 10 | 14 | 16 |
Divide the blue count by the total of \(40\), then write it as a percentage.
| \(n(S)\) | \(=\) | \(10+14+16 = 40\) |
| \(n(\text{blue})\) | \(=\) | \(14\) |
| \(P(\text{blue})\) | \(=\) | \(\dfrac{14}{40}=0.35\) |
| \(\) | \(=\) | \(35\%\) |
So \(P(\text{blue})=35\%\).
Find \(P(\text{heart})\) first, then subtract it from \(1\).
| \(P(\text{heart})\) | \(=\) | \(\dfrac{13}{52}=\dfrac{1}{4}\) |
| \(P(\text{not heart})\) | \(=\) | \(1-\dfrac{1}{4}\) |
| \(\) | \(=\) | \(\dfrac{3}{4}\) |
There are \(39\) non-hearts, so \(P(\text{not heart})=\dfrac{3}{4}\).
Common pitfalls
Frequently asked questions
What is a sample space?
The sample space is the set of all possible outcomes of a chance experiment. For one roll of a die it is 1, 2, 3, 4, 5, 6, so the number of outcomes n(S) is 6. For two dice you can list all 36 outcomes in a 6 by 6 array.
How do you calculate theoretical probability?
Count the favourable outcomes for the event, count the total outcomes in the sample space, then divide. P(E) equals n(E) over n(S). This works when every outcome is equally likely, such as a fair die or a well-shuffled deck of cards.
Can a probability be greater than 1?
No. Every probability is between 0 and 1, where 0 means the event is impossible and 1 means it is certain. If your answer is more than 1 or negative, you have counted the outcomes incorrectly.
How do you write a probability as a percentage?
Work out the probability as a fraction, convert it to a decimal by dividing, then multiply by 100 to get a percentage. For example, 14 out of 40 is 0.35, which is 35 percent.
What is the complement of an event?
The complement is 'the event not happening'. Because all outcome probabilities add to 1, P(not A) equals 1 minus P(A). For a card that is not a heart, subtract the probability of a heart, one quarter, from 1 to get three quarters.