Comparing Simple and Compound Events
Theory
A simple event is one result of one step, such as green on one spin. A related compound event joins several steps, such as green on both of two spins, or green on at least one. This Year 9 maths page shows how to compare them, first with theoretical probabilities and then with simulations using digital tools such as a spreadsheet or random digits. It is Year 9 content in the Australian Curriculum. The Victorian Curriculum also uses simulations with digital tools, with a different focus: estimating probabilities that cannot be found exactly. Curriculum: VC2M9P03 (different focus) / AC9M9P03.
A simple event involves one step of an experiment: a head on one toss, a six on one roll, green on one spin.
A compound event combines two or more steps. With and every part must happen ("green on both spins"). With or at least one part must happen ("green on at least one spin").
Compare a simple event with a related compound event. An and event is never more likely than the simple event: green on both spins needs green on the first spin and then green again. An or event is never less likely: green on at least one spin includes every time the first spin is green, and more.
For independent steps, multiply: \(P(\text{A and A}) = P(A) \times P(A)\). Multiplying by a number less than \(1\) makes the probability smaller, and each extra "and" makes it smaller again.
A simulation repeats the experiment many times with a digital tool. In a spreadsheet, =RANDBETWEEN(1,3) gives \(1\), \(2\) or \(3\) with equal chance, so it can stand for a spinner with three equal sectors. One row is one trial. Comparing the relative frequencies of the simple and compound events shows the same pattern as the theory, and gets closer to the theory as the number of trials grows.
For a simple event \(A\) with probability \(p\), repeated in independent steps:
=RANDBETWEEN(1,6) rolls a die; =IF(AND(…),1,0) marks an "and" event; =IF(OR(…),1,0) marks an "or" event; =COUNTIF(C2:C301,1) counts them and =AVERAGE(C2:C301) gives the relative frequency.How to compare a simple and a compound event
- Name the simple event and the compound event (and / or).
- Find the theoretical probabilities: list the outcomes in an array or tree, or multiply for independent steps.
- Simulate many trials: one row (or one group of digits) per trial, with columns for each step.
- Count the trials where each event happened and divide by the number of trials.
- Compare: the "and" event should be less likely than the simple event, and the "or" event more likely. Check how close each relative frequency is to its theoretical probability.
Describing the results
- Say which is more likely, and give the numbers.
- Explain why using and / or: the compound "and" event needs the simple event to happen first.
- Comment on the trials: small simulations can be a long way from the theory; more trials give a better estimate.
(i) green on the first spin
(ii) green on both spins
(iii) green on at least one spin.
(iv) Put the three events in order, from least to most likely.
(i) One of three equal sectors.
| \(P(\text{green first})\) | \(=\) | \(\boldsymbol{\dfrac{1}{3}}\) |
(ii) The spins are independent, so multiply.
| \(P(\text{both})\) | \(=\) | \(\dfrac{1}{3} \times \dfrac{1}{3}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{1}{9}}\) |
(iii) The complement is "no green on either spin".
| \(P(\text{at least one})\) | \(=\) | \(1 - \dfrac{2}{3} \times \dfrac{2}{3}\) |
| \(\) | \(=\) | \(1 - \dfrac{4}{9} = \boldsymbol{\dfrac{5}{9}}\) |
(iv) \(\dfrac{1}{9} \lt \dfrac{3}{9} \lt \dfrac{5}{9}\): both spins, then the first spin, then at least one. The array in the Diagrams tab shows the same \(1\), \(3\) and \(5\) cells out of \(9\).
=RANDBETWEEN(1,3) in two columns, with \(1\) for green, in \(300\) rows. The first spin was green in \(104\) rows, both spins in \(31\) rows and at least one spin in \(168\) rows.(i) Find the three relative frequencies, correct to \(3\) decimal places.
(ii) Compare them with the theoretical probabilities.
(i) Divide each count by the \(300\) trials.
| \(\text{first green}\) | \(:\) | \(\dfrac{104}{300} \approx \mathbf{0.347}\) |
| \(\text{both}\) | \(:\) | \(\dfrac{31}{300} \approx \mathbf{0.103}\) |
| \(\text{at least one}\) | \(:\) | \(\dfrac{168}{300} = \mathbf{0.560}\) |
(ii) The theory gives \(\dfrac{1}{3} \approx 0.333\), \(\dfrac{1}{9} \approx 0.111\) and \(\dfrac{5}{9} \approx 0.556\).
| Event | Simulation | Theory |
|---|---|---|
| first green | 0.347 | 0.333 |
| both | 0.103 | 0.111 |
| at least one | 0.560 | 0.556 |
Each result is within about \(0.015\) of the theory, and the order is the same: "both" is the least likely and "at least one" the most likely.
93 85 70 83 93 47 35 79 43 21
78 53 75 36 63 18 74 31 65 46
(i) Find the relative frequency of the morning bus being on time.
(ii) Find the relative frequency of both buses being on time.
(iii) Compare each with its theoretical probability.
(i) The morning bus is late when the first digit is \(8\) or \(9\): 93, 85, 83, 93. That is \(4\) of the \(20\) days.
| \(\text{morning on time}\) | \(=\) | \(\dfrac{16}{20} = \mathbf{0.8}\) |
(ii) Both on time needs both digits \(0\) to \(7\). Of the \(16\) days with the morning bus on time, the second digit is \(8\) or \(9\) on 79, 78, 18.
| \(\text{both on time}\) | \(=\) | \(\dfrac{13}{20} = \mathbf{0.65}\) |
(iii) Theory: \(P(\text{morning}) = 0.8\) and \(P(\text{both}) = 0.8 \times 0.8 = 0.64\). The simple event matches exactly here, and the compound event, \(0.65\), is very close to \(0.64\), even after only \(20\) trials. In both the simulation and the theory, both buses on time is less likely than the morning bus on time.
(i) Find the theoretical probability of winning, and compare it with \(P(\text{first die greater than } 4)\).
(ii) A simulation of \(90\) games gave \(13\) wins. Find the relative frequency.
(iii) Predict the number of wins in \(450\) games, using the theory and using the simulation. Which is better?
(i) A number greater than \(4\) is \(5\) or \(6\): \(\dfrac{2}{6} = \dfrac{1}{3}\) for one die.
| \(P(\text{win})\) | \(=\) | \(\dfrac{1}{3} \times \dfrac{1}{3}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{1}{9}}\) |
The win needs the simple event \(\left(\dfrac{1}{3}\right)\) and then the same again, so it is less likely.
(ii) Divide the wins by the games.
| \(\text{rel. freq.}\) | \(=\) | \(\dfrac{13}{90}\) |
| \(\) | \(\approx\) | \(\mathbf{0.144}\) |
(iii) Multiply each probability by \(450\).
| \(\text{theory}\) | \(:\) | \(\dfrac{1}{9} \times 450 = \mathbf{50}\) |
| \(\text{simulation}\) | \(:\) | \(\dfrac{13}{90} \times 450 = \mathbf{65}\) |
The theory is better: the dice are fair, so \(\dfrac{1}{9}\) is exact, while \(90\) games is a small simulation.
Common pitfalls
Frequently asked questions
What is the difference between a simple event and a compound event?
A simple event is one result of one step, such as a head on one toss. A compound event combines steps with and or or, such as a head on both of two tosses, or a head on at least one.
Why is a compound and event less likely than the simple event?
To happen, it needs the simple event and then something more. For independent steps you multiply by a probability less than \(1\), which makes the answer smaller.
Why is an at least one event more likely than the simple event?
It includes every trial where the simple event happens on the first step, plus the trials where it happens only on a later step.
How do you simulate a compound event in a spreadsheet?
Put a random formula such as =RANDBETWEEN(1,6) in one column for each step and fill down many rows. Use IF with AND or OR to mark the event in each row, then average that column for the relative frequency.
How many trials should a simulation have?
As many as you can. Hundreds of trials usually give relative frequencies close to the theoretical probabilities; twenty or thirty trials can be well away from them.