Experimental Probability
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Theory
Relative frequency is how often an outcome actually happened, divided by the number of trials. It is the experimental probability. This Year 9 maths page shows how to find it from counts and tables, estimate probabilities of "and", inclusive "or" and exclusive "or" from data, predict an expected frequency, and design and run a simulation to estimate a probability.
The frequency of an outcome is the number of times it happened. The relative frequency is that frequency divided by the total number of trials. It is also called the experimental or observed probability.
The theoretical probability is the probability expected under fair conditions, worked out from equally likely outcomes. For a fair die, \(P(6) = \dfrac{1}{6}\).
Relative frequency and theoretical probability are rarely exactly equal. With more trials, the relative frequency usually gets closer to the theoretical probability, so more trials give a better estimate.
If, after many trials, the relative frequency is far from the theoretical probability, the object may be biased (not fair).
A simulation imitates a chance experiment with something easier to repeat: coins, dice, spinners, random digits or a spreadsheet formula such as =RANDBETWEEN(1,10). Each outcome of the simulation stands for an outcome of the real situation, with the same probability.
Simulations are useful when a probability cannot be found exactly from equally likely outcomes, such as the chance that a thrown drawing pin lands point up, or when the exact working is very long, such as the chance of collecting all six cards in ten cereal boxes, or when a real experiment would take too long. Many repeats give a relative frequency that estimates the probability.
From data about two events \(A\) and \(B\) (for example, a table or a Venn diagram of counts):
How to work with relative frequency
- Find the number of trials. In a table, add the whole frequency column.
- Find the frequency of the event. Add every row that fits it.
- Divide frequency by trials. Give a decimal, or simplify the fraction.
- Predict by multiplying a relative frequency by a new number of trials.
- Compare with the theoretical probability. A small gap is normal; a big gap after many trials suggests bias.
How to design and run a simulation
- State the probability of each real outcome.
- Choose a device with the same probabilities, and say which results stand for which outcomes.
- Decide what one trial is (one day, one shot, one family) and run many trials.
- Count the trials where the event happened and divide by the number of trials.
- Compare with the theoretical probability if there is one. More trials give a more reliable estimate.
| People in car | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Frequency | 18 | 12 | 6 | 4 |
Add the frequency row to get the number of trials. Then add the rows that fit the event: \(3\) people or \(4\) people.
| \(\text{cars}\) | \(=\) | \(18 + 12 + 6 + 4\) |
| \(\) | \(=\) | \(40\) |
| \(\text{more than } 2\) | \(=\) | \(6 + 4 = 10\) |
| \(\text{relative frequency}\) | \(=\) | \(\dfrac{10}{40} = \mathbf{0.25}\) or \(\mathbf{\dfrac{1}{4}}\) |
"More than \(2\)" does not include \(2\) itself.
(i) How many faulty parts would you expect in \(2500\) parts?
(ii) A batch had \(45\) faulty parts. About how many parts were in the batch?
(i) Multiply the relative frequency by the number of trials.
| \(\text{expected}\) | \(=\) | \(0.12 \times 2500 = \mathbf{300}\) |
(ii) Work backwards: divide the frequency by the relative frequency.
| \(0.12 \times n\) | \(=\) | \(45\) |
| \(n\) | \(=\) | \(45 \div 0.12 = \mathbf{375}\) |
Both answers are estimates. A real batch will usually be close to these, not exactly on them.
(i) a bike and a skateboard
(ii) a bike or a skateboard (or both)
(iii) a bike or a skateboard, but not both.
Divide each count by the \(160\) students surveyed.
(i) "And" is the overlap.
| \(P(\text{both})\) | \(\approx\) | \(\dfrac{24}{160}\) |
| \(\) | \(=\) | \(\mathbf{0.15}\) |
(ii) Inclusive or: add the two groups, then take the overlap off once.
| \(n(\text{or})\) | \(=\) | \(72 + 50 - 24 = 98\) |
| \(P\) | \(\approx\) | \(\dfrac{98}{160} = \mathbf{0.6125}\) |
(iii) Exclusive or: take the overlap off twice.
| \(n\) | \(=\) | \(72 + 50 - 2 \times 24 = 74\) |
| \(P\) | \(\approx\) | \(\dfrac{74}{160} = \mathbf{0.4625}\) |
Check: bike only \(48\) plus skateboard only \(26\) is \(74\) ✓
(i) Which digits could stand for a save?
(ii) Here are the \(25\) digits: 34167 21106 84038 85421 43044. How many saves do they show?
(iii) Find the relative frequency of a save and compare it with \(0.4\).
(i) \(40\%\) of the \(10\) digits is \(4\) digits. Let \(0, 1, 2, 3\) be a save and \(4\) to \(9\) a goal.
(ii) Count the digits \(0\), \(1\), \(2\) and \(3\) in each group of five: \(2, 4, 2, 2, 2\).
| \(\text{saves}\) | \(=\) | \(2 + 4 + 2 + 2 + 2\) |
| \(\) | \(=\) | \(\mathbf{12}\) |
(iii) Divide by the number of trials.
| \(\text{rel. freq.}\) | \(=\) | \(\dfrac{12}{25}\) |
| \(\) | \(=\) | \(\mathbf{0.48}\) |
\(0.48\) is a little above \(0.4\). With only \(25\) trials a gap like this is normal; a few hundred trials would usually give a value closer to \(0.4\).
Common pitfalls
Frequently asked questions
What is relative frequency?
Relative frequency is the number of times an outcome happens divided by the total number of trials. It is the experimental probability of that outcome.
What is the difference between theoretical and experimental probability?
Theoretical probability is what should happen under fair conditions, worked out from equally likely outcomes. Experimental probability, or relative frequency, is what actually happened in trials.
Why do more trials give a better estimate?
With only a few trials, chance can push the results a long way from the true probability. Over many trials the relative frequency settles close to the true probability.
What is a simulation in probability?
A simulation copies a chance situation with a device that is easy to repeat, such as random digits, dice or a spreadsheet. Each result stands for a real outcome with the same probability, and the relative frequency estimates the probability.
How do you estimate an or probability from data?
For inclusive or, add the two counts, take off the number in both, and divide by the total. For or but not both, take the number in both off twice.