6.3 - Probability, relative frequency and expectation

Learn to describe chance, calculate probabilities from a model and estimate them from observations. Use probabilities to predict counts while allowing for uncertainty in what actually happens. These ideas apply at Foundation and Higher.

Language and the probability scale

Can a sensible prediction say that a batch will contain $1.4$ rejected items? You cannot count $1.4$ actual items. By the end, you will see why this can still be useful.

You already know

A fraction describes a part of a whole. You can write the same proportion as a fraction, decimal or percentage: $\frac14=0.25=25\%$ .

Before rolling a die, you do not know which face will appear. This is a random experiment. Random means uncertain; it does not mean that every result has the same chance. A damaged die could favour one face.

Outcome and event

An outcome is one possible recorded result, such as rolling $6$ . An event is one outcome or a collection of outcomes, such as rolling an even number: $2$ , $4$ or $6$ . Outcomes are equally likely when each has the same probability.

A probability measures how likely an event is. Write $P(A)$ for the probability of event $A$ . Every probability satisfies $0\leq P(A)\leq1$ .

Probability	Description
$0$	Impossible
Between $0$ and $\frac12$	Unlikely
$\frac12$	An even chance
Between $\frac12$ and $1$	Likely
$1$	Certain

On a fair ordinary die, each face from $1$ to $6$ is equally likely. A number greater than $4$ has probability $\frac26=\frac13$ , so it is unlikely. Getting $7$ is impossible, with probability $0$ . A number at most $6$ is certain, with probability $1$ .

Key idea

Likely does not mean certain. An event with probability $0.9$ can still fail to happen. A negative number or a number greater than $1$ cannot be a probability.

The endpoints describe impossibility or certainty under the stated model. Merely observing no successes in a few trials does not establish impossibility.

Probabilities from a theoretical model

A theoretical model describes possible outcomes and assumptions about their chances. For a fair spinner with eight equal sectors, each sector has probability $\frac18$ . The colours need not have equal probabilities.

Counting equally likely outcomes

When the individual outcomes are equally likely,

P(A)=\frac{\text{number of outcomes in }A}{\text{total number of possible outcomes}}

This rule is absent from both the Foundation and Higher sheets. Not in the formulae sheet: learn it.

For a bag of otherwise identical counters, mix them and choose without looking. The model gives each individual counter the same chance. Count counters, rather than colour names.

Worked example 1: equal sectors

Question. A fair spinner has eight equal sectors: three red, four blue and one green. Find the probability of red and of green.

Plan. Count the sectors for each event and divide by all eight sectors.

\begin{aligned} P(\text{red})&=\frac38 &&\text{three red sectors}\\ P(\text{green})&=\frac18 &&\text{one green sector} \end{aligned}

Why divide by $8$ ? The sectors are equally likely; the colour groups are not.

Answer. $\frac38$ for red and $\frac18$ for green.

Check. Including blue gives $\frac38+\frac48+\frac18=1$ , accounting for every sector once.

Change only the colouring: repaint one blue sector red. The sector sizes and total remain unchanged.

Worked example 2: one sector changes colour

Question. A fair spinner has eight equal sectors: four red, three blue and one green. Find the probability of red and of green.

Plan. Use the new colour counts with the same total.

\begin{aligned} P(\text{red})&=\frac48=\frac12 &&\text{new red count}\\ P(\text{green})&=\frac18 &&\text{green is unchanged} \end{aligned}

Why is green unchanged? Neither its sector count nor the total has changed.

Answer. $\frac12$ for red and $\frac18$ for green.

Check. $\frac48+\frac38+\frac18=1$ .

An unequal spinner needs information about sector sizes or probabilities; its labels alone do not give equal chances.

Watch out

$P(\text{red})=\frac13$ because there are three colours is wrong here. Count equally likely sectors: the second spinner gives $P(\text{red})=\frac48$ .

Now make the same counting decision with individual counters as the outcomes.

Estimating probability from data

You may have observations rather than a fair-device model. If $48$ of $160$ recorded bus journeys were late, their observed proportion late is $\frac{48}{160}$ . This can estimate the probability for another comparable journey.

Relative frequency

The relative frequency is the number of times an event occurred divided by the total number of trials or observations. It estimates a probability when the observations represent the situation being predicted.

Include unsuccessful trials in the denominator. A larger representative sample generally reduces random fluctuation in the proportion, but cannot correct a biased sample. Records collected only during severe weather might poorly represent ordinary journeys.

Worked example 3: estimating lateness

Question. Of $160$ recorded bus journeys, $48$ were late. Estimate the probability that another comparable journey is late. State one limitation.

Plan. Divide late journeys by all journeys, then consider whether conditions could change.

\begin{aligned} \text{Estimated }P(\text{late})&=\frac{48}{160} &&\text{occurrences over observations}\\ &=\frac3{10}=0.3 &&\text{simplify} \end{aligned}

Why use all $160$ ? Every journey was an opportunity to be late, including journeys that arrived on time.

Answer. $0.3$ . A changed timetable could make these records unsuitable for prediction.

Check. $48$ is $30\%$ of $160$ , below half the journeys.

Last Accessed

Number

Algebra

Shape, space and measure

Handling data

6.3 - Probability, relative frequency and expectation

Language and the probability scale

Probabilities from a theoretical model

Estimating probability from data

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