Basic Concepts
The probability of an event is the number of favourable outcomes divided by the total number of equally likely outcomes, a value from 0 to 1, and the probabilities of all outcomes add to 1. This subtopic revises the Stage 4 basics at the start of the NSW Year 10 Advanced maths probability topic (NESA outcome MA4‑PRO‑C‑01).
You will find probabilities from bags of marbles, dice, cards and letters, including words with repeated letters. You will use complementary events, find a missing probability so the total is 1, work out how many items are in a jar, and compare a relative frequency with the expected number. Harder questions use algebra, such as probabilities written as \(x\) and \(2x\), and explain why a set of probabilities is impossible. It is part of Year 10 Advanced Probability.
Every question with a fully worked solution.
- Basic Concepts - Video - Probability, Sample Spaces, and the Complement Rule Watch
Theory
For equally likely outcomes, probability \(=\dfrac{\text{favourable outcomes}}{\text{total outcomes}}\). This Year 10 Advanced maths page revises probability from earlier years one step further: complements, conditions that overlap, probabilities that add to 1, working backwards to a count, expected numbers and relative frequency.
A chance experiment has outcomes that cannot be predicted, such as rolling a die. The set of all possible outcomes is the sample space. An event is a set of outcomes, such as "an even number".
When all outcomes are equally likely, \(P(E)=\dfrac{n(E)}{n(S)}\), the number of outcomes in the event divided by the number in the sample space. Every probability is between \(0\) (impossible) and \(1\) (certain).
The complement of \(E\), written \(\overline{E}\) or \(E'\), is "\(E\) does not happen". \(P(\overline{E})=1-P(E)\). The probabilities of all the outcomes of an experiment add to \(1\).
The expected number of times an event occurs in \(n\) trials is \(n\times P(E)\). Relative frequency is the observed fraction, \(\dfrac{\text{times it happened}}{\text{number of trials}}\). It estimates the probability when the outcomes are not known to be equally likely, and the estimate improves with more trials.
| Situation | Tool | Example |
|---|---|---|
| "not" | complement | \(P(\text{not a } 6)=1-\dfrac{1}{6}=\dfrac{5}{6}\) |
| "a multiple of \(a\) and of \(b\)" | multiples of the LCM | multiples of \(4\) and \(10\) are multiples of \(20\) |
| "\(A\) but not \(B\)" | count \(A\), subtract the overlap | factors of \(30\) on a die that are not factors of \(4\): \(3, 5, 6\) |
| missing probability | all outcomes add to \(1\) | \(0.25+0.4+p=1\Rightarrow p=0.35\) |
| missing count | \(\dfrac{\text{part}}{\text{total}}=P\) | \(P=0.15\) and \(9\) items gives \(9\div 0.15=60\) in total |
| spinner sector | angle \(=P\times 360^\circ\) | \(P=\dfrac{1}{8}\) gives \(45^\circ\) |
How to find a probability
- List or count the sample space and check the outcomes are equally likely.
- Count the favourable outcomes. For overlapping conditions, subtract or add the overlap so nothing is counted twice.
- Use the complement when "not" is quicker to count.
- Write the fraction and simplify it. Give a decimal or percentage only if asked.
Other question types
- Missing probability: let it be a pronumeral, add every probability, set the sum to \(1\) and solve (Example 2).
- Missing count: write \(\dfrac{\text{count}}{\text{total}}=P\) with a pronumeral and solve (Example 3). If items are added or removed, update both the count and the total.
- Is it possible? If the probabilities given add to more or less than \(1\), the experiment as described cannot exist.
- Expected number and relative frequency: multiply the probability by the number of trials; compare with what happened to judge whether something is fair (Example 4).
- Counting odd or even numbers: list a few small cases, such as \(1\) to \(5\) and \(1\) to \(7\), find the pattern in terms of the last number, then write the count as an expression.
(i) \(P(\text{a multiple of } 4 \text{ but not of } 6)\)
(ii) \(P(\text{not a perfect square})\)
(i) Multiples of \(4\): \(60\div 4=15\). Those that are also multiples of \(6\) are multiples of the LCM \(12\): \(12, 24, 36, 48, 60\), so \(5\).
| \(P\) | \(=\) | \(\dfrac{15-5}{60}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{1}{6}}\) |
(ii) The perfect squares are \(1, 4, 9, 16, 25, 36, 49\): \(7\) of them. Use the complement.
| \(P(\text{not a square})\) | \(=\) | \(1-\dfrac{7}{60}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{53}{60}}\) |
(i) Find \(P(C)\) and \(P(B)\).
(ii) How many times would you expect \(B\) in \(150\) spins?
(i) Let \(P(C)=c\), so \(P(B)=2c\). The four probabilities add to \(1\).
| \(0.3+2c+c+0.1\) | \(=\) | \(1\) |
| \(3c\) | \(=\) | \(0.6\) |
| \(c\) | \(=\) | \(0.2\) |
\(P(C)=\boldsymbol{0.2}\) and \(P(B)=\boldsymbol{0.4}\). Check: \(0.3+0.4+0.2+0.1=1\).
(ii) Expected number \(=\) probability \(\times\) number of trials.
| \(0.4\times 150\) | \(=\) | \(\boldsymbol{60}\) |
(i) How many green marbles are there?
(ii) Six red marbles are removed. Find the new probability of green.
(i) Let there be \(g\) green marbles, so \(18+g\) altogether.
| \(\dfrac{g}{18+g}\) | \(=\) | \(\dfrac{2}{5}\) |
| \(5g\) | \(=\) | \(2(18+g)\) |
| \(3g\) | \(=\) | \(36\) |
| \(g\) | \(=\) | \(\boldsymbol{12}\) |
(ii) Now there are \(12\) red and \(12\) green, \(24\) in total.
| \(P(\text{green})\) | \(=\) | \(\dfrac{12}{24}=\boldsymbol{\dfrac{1}{2}}\) |
Removing marbles changes the total as well as the count, so recalculate both.
(i) Find the relative frequency of a six.
(ii) How many sixes would a fair die be expected to give in \(300\) rolls?
(iii) Use the relative frequency to estimate the number of sixes in \(1200\) rolls.
(i) Relative frequency \(=\dfrac{\text{number of times}}{\text{number of trials}}\).
| \(\dfrac{72}{300}\) | \(=\) | \(\boldsymbol{0.24}\) |
(ii) For a fair die \(P(6)=\dfrac{1}{6}\), so expect \(\dfrac{1}{6}\times 300=\boldsymbol{50}\).
(iii) \(0.24\times 1200=\boldsymbol{288}\).
\(72\) sixes is well above the \(50\) expected, so this die may be biased. More rolls would give a more reliable estimate.
Common pitfalls
Frequently asked questions
How do you calculate probability?
For equally likely outcomes, divide the number of favourable outcomes by the total number of outcomes. For example, \(P(\text{less than } 3)\) on a die is \(\dfrac{2}{6}=\dfrac{1}{3}\).
What is a complementary event?
The complement of an event is everything in the sample space that is not in the event. An event and its complement have probabilities that add to \(1\), so \(P(\overline{E})=1-P(E)\).
What is the difference between probability and relative frequency?
Probability is what should happen in the long run, worked out from equally likely outcomes. Relative frequency is what did happen in an experiment: the number of times an outcome occurred divided by the number of trials.
How do you find the expected number of outcomes?
Multiply the probability by the number of trials. With a probability of \(0.2\) and \(80\) trials, you would expect the outcome \(16\) times.
How do you find the total number of items from a probability?
Write \(\dfrac{\text{count}}{\text{total}}=P\) and solve for the total. If \(9\) items have a probability of \(0.15\), the total is \(9\div 0.15=60\).