This Year 9 probability lesson is written for pupils aged 13-14 and follows the National Curriculum. In short, it shows you how to use sample space diagrams, tables and Venn diagrams to find theoretical probabilities. Before you start, it helps to revise angles in polygons and similarity and enlargement, because this lesson builds on them.
Year 9 probability lesson: what you will learn
This lesson covers the probability scale, the complement, mutually exclusive events, sample spaces, sets and Venn diagrams, and expected outcomes.
Starter activity
Roll two dice 36 times and tally the totals. Which total do you think is the most common? Now make a table of all 36 ways the dice can land and count how many ways make each total. Does your table agree with your tally? Why might the results from only 36 rolls differ from the table?
Measuring chance
Definition: probability
Probability tells us how likely an event is, on a scale from 0 to 1. An impossible event has probability 0, a certain event has probability 1, and an event as likely to happen as not has probability \(\tfrac12\). Probabilities can be written as fractions, decimals or percentages.
Rule: equally likely outcomes
\(P(\text{event}) = \dfrac{\text{number of outcomes in the event}}{\text{total number of possible outcomes}}\)
Worked example 1: a spinner
A fair spinner has the numbers 1 to 8. Find \(P(\text{even})\) and \(P(\text{greater than } 5)\).
\(P(\text{even}) = \dfrac{4}{8} = \dfrac12\). \(P(\text{greater than } 5) = \dfrac{3}{8}\) (the numbers 6, 7 and 8). This is theoretical probability: it comes from counting the possible outcomes, not from doing an experiment.
Probabilities add up to 1
Key fact: the complement
The probabilities of all possible outcomes of an experiment add up to 1. So \(P(\text{event does not happen}) = 1 - P(\text{event happens})\).
Worked example 2: counters and rain
A bag has red, blue and green counters with \(P(\text{red}) = 0.3\) and \(P(\text{blue}) = 0.45\). Find \(P(\text{green})\). The chance of rain is 0.35; find the chance of no rain.
\(P(\text{green}) = 1 - 0.3 - 0.45 = 0.25\). The chance of no rain is \(1 - 0.35 = 0.65\).
Mutually exclusive events
Definition: mutually exclusive
Two events are mutually exclusive if they cannot happen at the same time (rolling a 2 and rolling a 5 on one throw of a die).
Rule: the addition rule
For mutually exclusive events A and B, \(P(A \text{ or } B) = P(A) + P(B)\). If \(P(A) = 0.3\) and \(P(B) = 0.2\), then \(P(A \text{ or } B) = 0.5\).
Sample space diagrams
Definition: sample space
When two events happen together, a sample space lists every possible outcome systematically in a table or grid. For two fair dice there are \(6 \times 6 = 36\) equally likely outcomes.
| + | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 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 |
Worked example 3: two dice
Two dice are added. Find \(P(\text{total }7)\), \(P(\text{total}\ge10)\) and \(P(\text{a double})\).
The total 7 appears 6 times, so \(P = \dfrac{6}{36} = \dfrac16\). A total of 10 or more appears 6 times (10 three times, 11 twice, 12 once), so the probability is also \(\dfrac16\). A double has probability \(\dfrac{6}{36} = \dfrac16\).
Worked example 4: coins and a spinner
A coin and a spinner numbered 1, 2, 3 are used together. Find \(P(\text{head and }3)\). For two coins find \(P(\text{at least one head})\).
The outcomes are H1, H2, H3, T1, T2, T3: six in total, so \(P(\text{head and } 3) = \dfrac16\). Two coins give HH, HT, TH, TT, so \(P(\text{at least one head}) = \dfrac34\).
Sets and Venn diagrams
Definition: sets and Venn notation
A set is a collection of items, written in curly brackets: \(\{2, 4, 6\}\). A Venn diagram shows sets as overlapping circles inside a rectangle that stands for the universal set \(\xi\).
- \(A \cap B\) is the intersection: items in both A and B (the overlap).
- \(A \cup B\) is the union: items in A or B or both.
- \(A'\) is the complement: items not in A.
- \(n(A)\) is the number of items in A.
Method: filling in a Venn diagram
- Fill in the overlap first.
- Subtract the overlap from each set total to get the "only" regions.
- Add all the regions and subtract from the total to find the number in neither set.
Worked example 5: football and hockey
In a class of 30, 18 play football (F), 12 play hockey (H) and 5 play both. Find the probabilities of neither, F or H, both, and football only.
Football only: \(18 - 5 = 13\). Hockey only: \(12 - 5 = 7\). In at least one: \(13 + 5 + 7 = 25\). Neither: \(30 - 25 = 5\). So \(P(\text{neither}) = \dfrac{5}{30} = \dfrac16\), \(P(F \cup H) = \dfrac{25}{30} = \dfrac56\), \(P(F \cap H) = \dfrac{5}{30} = \dfrac16\) and \(P(\text{football only}) = \dfrac{13}{30}\).
Rule: counting a union
\(n(A \cup B) = n(A) + n(B) - n(A \cap B)\). The overlap is counted twice when you add, so subtract it once.
Worked example 6: numbers 1 to 20
From the numbers 1 to 20, A is the even numbers and B the multiples of 3. Find \(n(A\cup B)\) and the probability that a random number is in A or B.
\(n(A) = 10\), \(n(B) = 6\) (3, 6, 9, 12, 15, 18) and \(A \cap B = \{6, 12, 18\}\), so \(n(A \cap B) = 3\). Then \(n(A \cup B) = 10 + 6 - 3 = 13\), and the probability is \(\dfrac{13}{20}\).
Using probability to predict
Rule: expected number
The expected number of times an event happens is \(\text{probability} \times \text{number of trials}\). If \(P(\text{red}) = \tfrac14\) and a spinner is spun 60 times, you expect \(\tfrac14 \times 60 = 15\) reds. The real result will usually be close but not exactly equal.
Watch out: common mistakes
- Giving a probability bigger than 1 or a negative one.
- Counting outcomes as equally likely when they are not (the total of two dice is not equally likely: 7 is more likely than 2).
- Forgetting to subtract the overlap when counting a union, or putting the total of a set into the overlap region.
- Forgetting the items outside both circles.
- Not simplifying the fraction when asked.
Worked example 7: a two-way table
In a group of 50 pupils, 28 are girls. 15 of the girls and 9 of the boys have a pet. Find the probability that a pupil chosen at random is a boy without a pet.
Boys: \(50-28=22\). Boys with a pet: 9. Boys without a pet: \(22-9=13\). So the probability is \(\dfrac{13}{50}=0.26\).
Worked example 8: expected number
The probability that a train is late is 0.08. How many late trains are expected in 250 journeys?
Expected number \(=250\times0.08=20\) late trains. This is an estimate for the long run, not a promise: the actual number could be a little higher or lower.
Worked example 9: the union with the addition rule
\(P(A)=0.5\), \(P(B)=0.4\) and \(P(A\cap B)=0.2\). Find \(P(A\cup B)\) and \(P(\text{neither})\).
\(P(A\cup B)=0.5+0.4-0.2=0.7\). The 0.2 must be subtracted because it is counted in both \(P(A)\) and \(P(B)\). \(P(\text{neither})=1-0.7=0.3\).
Worked example 10: choosing between two outcomes
A bag has 3 red, 4 blue and 5 green counters. Find the probability of taking a blue or green counter.
There are 12 counters. Blue or green: \(4+5=9\). So \(P=\dfrac9{12}=\dfrac34\). Another way: \(1-P(\text{red})=1-\dfrac3{12}=\dfrac34\). Using both methods is a good check.
Note: write probabilities in the form you are asked for
A probability can be a fraction, a decimal or a percentage, but never greater than 1 and never negative. If a question gives decimals, answer in decimals. Simplify fractions. If you are asked for the expected number of times, multiply the probability by the number of trials.
Watch out: common mistakes
- "Even chance" does not mean two outcomes always have probability \(\frac12\): the outcomes must be equally likely.
- When two events can happen together, do not just add their probabilities; subtract the overlap.
- In a sample space, list outcomes in order so that you do not miss any: for two dice there must be \(6\times6=36\).
Did you know?
Probability began with a gambling puzzle. In 1654 the French mathematicians Blaise Pascal and Pierre de Fermat exchanged letters about how to share the prize in an unfinished game of chance. Their letters are often said to be the start of probability theory, which today guides insurance, weather forecasts and medical tests.
Theory and experiment
Definition: relative frequency
When outcomes are not equally likely, or when we do not know the probabilities, we can carry out an experiment. The relative frequency of an outcome is the number of times it happens divided by the number of trials. It is an estimate of the probability, and the estimate usually gets better as the number of trials increases.
Worked example 11: is the dice biased?
Leo rolls a dice 300 times and gets a six 78 times. Do you think the dice is fair?
The relative frequency of a six is \(78\div300=0.26\). For a fair dice the theoretical probability is \(\dfrac16\), which is about 0.17, so we would expect about \(300\times\dfrac16=50\) sixes. Getting 78 sixes is a lot more than 50, and 300 is a large number of trials, so the result suggests the dice is biased towards six. Leo cannot be completely sure, but he has good evidence. With only 12 rolls, a few extra sixes would not tell us very much.
Method: choosing how to list outcomes
- For one event, list the outcomes in a simple list or table of frequencies.
- For two events, such as two dice or a coin and a spinner, draw a sample space grid with one event across the top and the other down the side.
- For groups of people with two properties, such as sports or languages, use a Venn diagram and always fill in the overlap first.
- Count the outcomes you want and divide by the total, simplifying the fraction at the end.
Otto’s tip
In Venn diagram problems, always fill in the middle (the overlap) first, then work outwards. Check that every region adds to the total. Hoot hoot!
How to use this Year 9 probability lesson
First, read the key points so that you know where the Year 9 probability lesson is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the probability: sample spaces and Venn diagrams worksheet and check every question with the worksheet answers. When you feel ready, test yourself with the multiple choice questions or the quick quiz. Finally, sit the probability: sample spaces and Venn diagrams test and mark it with the test answers.
Also, the starters are a good warm-up, so you can keep the key facts fresh. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, this Year 9 probability lesson gives you everything you need for the chapter.
To sum up, regular practice works better than one long session. For instance, ten minutes a day is often enough. Besides, mistakes are useful, because they show what to revise next. Then, once a method feels easy, try a harder question. As a result, you build confidence step by step.
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