This module builds on: M5 Probability.
Sample space diagrams
Sample space diagrams are a visual way of recording the possible outcomes of two events, which can then be used to calculate probabilities.
The tables include the possible outcomes of one event listed across and one event listed down.
Example
Two dice are rolled at the same time and their scores are added together. Find the probability of the sum of the two dice equalling 7.
Below is a table with the outcome of rolling die 1 across the top and die 2 down the left hand side.
| + | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 1 + 1 = 2 | 1 + 2 = 3 | 1 + 3 = 4 | 1 + 4 = 5 | 1 + 5 = 6 | 1 +6 = 7 |
| 2 | 2 + 1 = 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 |
The sample space diagram shows there are 6 ways of making a 7, out of a total of 36 possible outcomes.
Therefore, the probability of rolling two dice and the sum being 7 is \(\frac{6}{36} = \frac{1}{6}\).
Two tetrahedral (four-sided) dice are thrown.
Copy and complete the following table, which shows the sum of their scores:
a) What is the most likely outcome?
b) What is the probability that the sum of the scores will be \(3\)?
c) What is the probability that the sum of the scores will be greater than \(5\)?
Answer
a) \(5\) is the most likely outcome
b) The probability of getting the sum \(3 = \frac{2}{16} = \frac{1}{8}\)
c) The probability of a sum greater than \(5 = \frac{6}{16} = \frac{3}{8}\)
Relative frequency
Relative frequency is probability calculated from the outcomes of an experiment.
The theoretical probability of getting a head when you flip a coin is \(\frac{1}{2}\), but if a coin was actually flipped 100 times you may not get exactly 50 heads, although it should be close to this amount.
If a coin was flipped a hundred times, the amount of times a head actually did appear would be the relative frequency, so if there were 59 heads and 41 tails the relative frequency of flipping a head would be \(\frac{59}{100}\) (or 0.59 or 59%).
Relative frequency is used when probability is being estimated using the outcomes of an experiment or trial, when theoretical probability cannot be used.
For example, when using a biased die, the probability of getting each number is no longer \(\frac{1}{6}\). To be able to assign a probability to each number, an experiment would need to be conducted. From the experimental results, the relative frequency could be calculated.
The more times that an experiment has been carried out, the more reliable the relative frequency is as an estimate of the probability.
##Example
Ella rolls a dice and records how many times she scores a six. Find the probability that Ella rolls a six on her dice.
Ella’s results will give different estimates of the probability, depending on which total is selected.
For example, in the first 10 rolls, the probability of scoring 6 is \(\frac{2}{10} = 0.2\), but in the first 20 rolls, the probability of scoring 6 is \(\frac{3}{20} = 0.15\).
The most accurate estimate of the probability is found by using the highest number of rolls, which gives \(\frac{9}{50} = 0.18\).
Question
Marcus is testing whether a six-sided dice is biased or not. He rolls the dice 20 times and works out the relative frequency of each outcome.
| Score on dice | 1 | 2 | 3 | 4 | 5 | 6 |
| Relative frequency | 0.1 | 0.15 | 0.3 | 0.2 | 0.15 |
Marcus has forgotten to record the relative frequency of obtaining a 4 on the dice. What is this relative frequency?
Answer
Relative frequencies for all outcomes add up to 1.
The relative frequency of obtaining a 4 is \(1 – (0.1 + 0.15 + 0.3 + 0.2 + 0.15) = 0.1\).
How many times did the dice show a 3?
Answer
Marcus rolled the dice 20 times.
relative frequency of 3 \(= 0.3\)
number of 3s \(= 20 \times 0.3 = 6\)
Marcus suspects that the dice is biased. How could he ensure a more accurate result if he were to experiment again?
Answer
He could increase the number of trials. The more trials, the more accurate the result.