Probability Review
Every question with a fully worked solution.
- Probability Review - Video - Probability, Sample Spaces, and the Complement Rule Watch
Theory
Probability measures how likely an event is, from \(0\) (impossible) to \(1\) (certain). For equally likely outcomes it is the number of favourable outcomes divided by the total in the sample space. This Year 9 maths page revises the probability you met in Years 7 and 8: complements, outcomes that overlap, two-step experiments and expected numbers.
A chance experiment is something with an uncertain result, like rolling a die. The sample space is the list of all possible outcomes. For one roll of a die it is \(\{1, 2, 3, 4, 5, 6\}\).
An event is a set of outcomes, such as rolling an even number. When the outcomes are equally likely, the probability of an event is the number of favourable outcomes divided by the total number of outcomes.
Every probability lies between \(0\) and \(1\). It can be written as a fraction, a decimal or a percentage: \(\dfrac{2}{5} = 0.4 = 40\%\).
The complement of an event is the event not happening. An event and its complement have probabilities that add to \(1\).
Two events are independent if one has no effect on the other, such as a coin toss and a die roll. A coin has no memory: earlier results never change the next toss.
How to find a probability
- List the sample space, or work out how many outcomes there are. For two steps, pair every result of one with every result of the other.
- Check the outcomes are equally likely.
- Count the favourable outcomes. Write a list if a number property is involved, and count overlaps once.
- Divide favourable by total and simplify. Use \(1 - P\) for "not" or "at least one".
- Check the answer is between \(0\) and \(1\).
(i) \(P(\text{red})\)
(ii) \(P(\text{not red})\).
Count the total first. It goes on the bottom of every fraction.
| \(\text{total}\) | \(=\) | \(7 + 5 = 12\) |
| \(P(\text{red})\) | \(=\) | \(\mathbf{\dfrac{7}{12}}\) |
| \(P(\text{not red})\) | \(=\) | \(1 - \dfrac{7}{12} = \mathbf{\dfrac{5}{12}}\) |
Check: \(\dfrac{7}{12} + \dfrac{5}{12} = 1\). Not red here means blue.
List each set, then count the number in both only once.
| \(\text{multiples of } 3\) | \(:\) | \(3, 6, 9, 12, 15, 18\) |
| \(\text{multiples of } 5\) | \(:\) | \(5, 10, 15, 20\) |
| \(\text{in both}\) | \(:\) | \(15\) |
| \(\text{favourable}\) | \(=\) | \(6 + 4 - 1 = 9\) |
| \(P\) | \(=\) | \(\mathbf{\dfrac{9}{20}}\) |
Adding \(\dfrac{6}{20} + \dfrac{4}{20}\) would count \(15\) twice.
(i) List the sample space.
(ii) Find the probability of a tail and an even number.
(i) Pair every coin result with every spinner result.
| 1 | 2 | 3 | 4 | |
|---|---|---|---|---|
| H | H1 | H2 | H3 | H4 |
| T | T1 | T2 | T3 | T4 |
There are \(2 \times 4 = 8\) equally likely outcomes.
(ii) The favourable outcomes are T2 and T4.
| \(P(\text{T and even})\) | \(=\) | \(\dfrac{2}{8} = \mathbf{\dfrac{1}{4}}\) |
| \(\text{or}\) | \(:\) | \(\dfrac{1}{2} \times \dfrac{2}{4} = \dfrac{1}{4}\) |
The coin has no effect on the spinner, so the two probabilities can be multiplied.
| \(P(\text{blue})\) | \(=\) | \(\dfrac{1}{4}\) |
| \(\text{expected}\) | \(=\) | \(\dfrac{1}{4} \times 120 = \mathbf{30}\) |
This is what is expected, not a guarantee. A real set of \(120\) spins usually lands near \(30\), not exactly on it.
Common pitfalls
Frequently asked questions
How do you calculate probability?
When outcomes are equally likely, divide the number of favourable outcomes by the total number of outcomes. A bag of \(7\) red and \(5\) blue counters gives \(P(\text{red}) = \dfrac{7}{12}\).
What is a sample space?
The sample space is the list of all possible outcomes of a chance experiment. For tossing a coin and spinning a spinner numbered \(1\) to \(4\), it has \(8\) outcomes.
What is the complement of an event?
It is the event not happening. Its probability is \(1\) minus the probability of the event, because the two always add to \(1\).
How do you find the expected number of outcomes?
Multiply the probability by the number of trials. A spinner with a \(\dfrac{1}{4}\) chance of blue, spun \(120\) times, is expected to land on blue about \(30\) times.
Does a coin remember what it landed on before?
No. Each toss is independent, so the probability of a head is always \(\dfrac{1}{2}\), whatever happened on earlier tosses.