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Probability

Grade 8
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\( P(\text{event}) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}} \)

Probability measures how likely something is to happen, on a scale from impossible to certain.

In this lesson you will learn to:

1. The Probability Scale

\( 0 \;\leq\; P(\text{event}) \;\leq\; 1 \)

Every probability is a number between 0 and 1.

• \(P=0\) — impossible (rolling a 7 on a normal die)
• \(P=\tfrac{1}{2}\) — even chance (a coin landing heads)
• \(P=1\) — certain (the sun rising tomorrow)

\( \text{impossible} \rightarrow \text{unlikely} \rightarrow \text{even} \rightarrow \text{likely} \rightarrow \text{certain} \)

Everyday words map onto the scale:

\( 0 \quad\rightarrow\quad \text{between } 0 \text{ and } \tfrac{1}{2} \quad\rightarrow\quad \tfrac{1}{2} \quad\rightarrow\quad \text{between } \tfrac{1}{2} \text{ and } 1 \quad\rightarrow\quad 1 \)
An event with probability \(0.9\) is very likely but not certain.

\( 0 \leq P \leq 1 \)

Key idea: a probability can never be negative and never bigger than 1. If your answer is \(1.2\) or \(-0.3\), something went wrong.

2. Listing Outcomes

\( \{1;\ 2;\ 3;\ 4;\ 5;\ 6\} \)

The sample space is the list of ALL possible outcomes.

• A die: \(1; 2; 3; 4; 5; 6\) — 6 outcomes
• A coin: heads; tails — 2 outcomes
• This spinner with 4 equal sectors: A; B; C; D — 4 outcomes

\( \text{fair} = \text{every outcome has the same chance} \)

Outcomes are equally likely when each has the same chance — a fair die, a fair coin, a spinner with equal sectors.

A spinner with one big red sector and three small blue ones is NOT equally likely — red has a bigger chance.

\( \text{even numbers on a die: } \{2;\ 4;\ 6\} \)

A favourable outcome is one that matches the event you care about.

Event: “roll an even number” → favourable outcomes: 2, 4, 6 → that is 3 of the 6 outcomes.

\( \text{list ALL outcomes first} \)

Key idea: before calculating anything, list the whole sample space and count the favourable outcomes. Most probability errors come from missing an outcome.

3. Calculating Probability

\( P(\text{event}) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}} \)

For equally likely outcomes:

\( P(\text{even on a die}) = \dfrac{3}{6} = \dfrac{1}{2} \)
Always simplify the fraction.

\( P(6) = \dfrac{1}{6} \qquad P(\text{more than } 4) = \dfrac{2}{6} = \dfrac{1}{3} \)

\( P(6) \): only one favourable outcome out of 6.
\( P(\text{more than } 4) \): favourable = \(\{5; 6\}\), so \( \tfrac{2}{6} = \tfrac{1}{3} \).

\( P(\text{red}) = \dfrac{4}{10} = \dfrac{2}{5} \)

A bag holds 4 red, 5 blue and 1 green bead (10 in total).

\( P(\text{red}) = \tfrac{4}{10} = \tfrac{2}{5} \)
\( P(\text{green}) = \tfrac{1}{10} \)
\( P(\text{red or blue}) = \tfrac{9}{10} \)

\( P = \dfrac{\text{favourable}}{\text{total}} \)

Key idea: count favourable, count total, divide, simplify. The formula only works when outcomes are equally likely.

4. Fractions, Decimals & %

\( \dfrac{1}{2} = 0.5 = 50\% \)

The same probability can be written three ways:

\( \tfrac{1}{2} = 0.5 = 50\% \)
\( \tfrac{1}{4} = 0.25 = 25\% \)
\( \tfrac{3}{10} = 0.3 = 30\% \)

\( P(\text{even}) = \dfrac{1}{2} = 0.5 = 50\% \)

For a fair die:

\( P(\text{even}) = \tfrac{3}{6} = \tfrac{1}{2} = 0.5 = 50\% \)
\( P(6) = \tfrac{1}{6} \approx 0.17 \approx 16.7\% \)
Use whichever form the question asks for.

\( \text{fraction} \leftrightarrow \text{decimal} \leftrightarrow \text{percentage} \)

Key idea: divide the fraction to get the decimal; multiply the decimal by 100 to get the percentage. All three say the same thing.

5. Relative Frequency

\( \text{relative frequency} = \dfrac{\text{times it happened}}{\text{number of trials}} \)

Relative frequency comes from actually DOING the experiment.

Toss a coin 50 times; heads comes up 28 times.
Relative frequency of heads \( = \tfrac{28}{50} = 0.56 \)

\( 0.56 \ \text{(experiment)} \quad \text{vs} \quad 0.5 \ \text{(theory)} \)

Theoretical probability says heads should be \(0.5\). The experiment gave \(0.56\). That is normal!

The MORE trials you do, the CLOSER relative frequency gets to the theoretical probability.
10 tosses can easily give 70% heads; 1000 tosses will be very close to 50%.

\( \dfrac{412}{600} \approx 0.69 \)

Relative frequency can expose an UNFAIR object. A die rolled 600 times shows a six 412 times:

Relative frequency \( \approx 0.69 \), but theory says \( \tfrac{1}{6} \approx 0.17 \).
With this many trials, such a big gap means the die is almost certainly loaded.

\( \text{more trials} \Rightarrow \text{closer to theory} \)

Key idea: theoretical probability predicts; relative frequency measures. They meet when the number of trials gets large.

6. Complementary Events

\( P(\text{not A}) = 1 - P(\text{A}) \)

The complement of an event is everything else that could happen.

\( P(6) = \tfrac{1}{6} \quad\Rightarrow\quad P(\text{not } 6) = 1 - \tfrac{1}{6} = \tfrac{5}{6} \)

\( P(\text{not red}) = 1 - \dfrac{2}{5} = \dfrac{3}{5} \)

From the bag with \( P(\text{red}) = \tfrac{2}{5} \):

\( P(\text{not red}) = 1 - \tfrac{2}{5} = \tfrac{3}{5} \)
Check: the 5 blue + 1 green = 6 of the 10 beads are not red, and \( \tfrac{6}{10} = \tfrac{3}{5} \). It matches.

\( P(\text{red}) + P(\text{blue}) + P(\text{green}) = 1 \)

The probabilities of ALL the outcomes in a sample space always add up to exactly 1:

\( \tfrac{4}{10} + \tfrac{5}{10} + \tfrac{1}{10} = \tfrac{10}{10} = 1 \)
This is a powerful check on your working.

\( P(\text{A}) + P(\text{not A}) = 1 \)

Key idea: when “not” appears in a question, subtract from 1 — it is usually far quicker than counting the outcomes directly.

\[ P = \dfrac{\text{favourable}}{\text{total}} \qquad P(\text{not A}) = 1 - P(\text{A}) \]

You now know how to:


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