P1, P2, P3, P4, P5 - Probability, relative frequency and expected outcomes

Probability describes uncertainty; frequency records what actually happened. P1–P3 connect experimental records, fair models and predictions, P4 explains how complete outcomes fit together, and P5 explains why more unbiased evidence usually improves an estimate. You will calculate chances and expected counts without mistaking a prediction for a promise.

Randomness and the probability scale

If a fair coin lands heads seven times in ten tosses, is it broken? Keep that question in mind as we separate a model's chance from an experiment's results.

You already know

A fraction describes a part of a whole. For example, $3/6=1/2=0.5=50\%$ ; these are different forms of the same number.

A random outcome is uncertain beforehand. Random does not mean that every category you name has equal chance: on a fair die, “6” and “not 6” have probabilities $1/6$ and $5/6$ .

Probability

A number from $0$ to $1$ measuring an event's chance. Impossible events have probability $0$ ; certain events have probability $1$ . Below $1/2$ is unlikely, $1/2$ is an even chance, and above $1/2$ is likely. An event groups one or more outcomes.

Fairness provides a reason to treat the elementary outcomes alike. A loaded die does not justify counting faces as though they had equal chances.

Equally likely outcomes

When all individual outcomes are equally likely,

P(\text{event})=\frac{\text{number of successful outcomes}}{\text{total number of outcomes}}

Not on the formula sheet: learn it.

First identify the individual outcomes, then decide which meet the words in the question.

Worked example 1: an even die score

Question. A fair six-sided die is rolled. Find the probability of an even number.

Plan. List the six equally likely faces and count the even ones.

\begin{aligned} \text{successful faces}&=\{2,4,6\} && \text{divisible by two}\\ P(\text{even})&=\frac36=\frac12 && \text{three of six faces} \end{aligned}

Why this step? Each face has the same chance, so counting faces measures the share of probability.

Answer. $1/2$ .

Check. The other three faces are odd, so even and odd share the chance equally.

Change only the event in example 2. “At least” includes the boundary number.

Worked example 2: including the boundary

Question. The same fair die is rolled. Find the probability of a score at least $4$ .

Plan. Keep the six-face sample space and select scores of $4$ or more.

\begin{aligned} \text{successful faces}&=\{4,5,6\} && \text{include four}\\ P(\text{at least }4)&=\frac36=\frac12 && \text{count successful faces} \end{aligned}

Why this step? “Greater than $4$ ” would omit $4$ and describe a different event.

Answer. $1/2$ .

Check. The three unsuccessful faces are $1,2,3$ .

A spinner needs the same care: ten equal sectors coloured four red, one green and five yellow give $P(\text{not yellow})=(4+1)/10=1/2$ .

Watch out

Wrong: “There are three colours, so not yellow has probability $2/3$ .” The colour groups are not equally likely. Count equal sectors instead: $5/10=1/2$ .

Now change the die-score boundary once more.

Exhaustive outcomes and complementary events

A missing probability can be found by subtracting the other probabilities from one, provided the categories cover everything without overlap.

Exhaustive and mutually exclusive

Exhaustive events cover all possibilities. Mutually exclusive events cannot occur together on one trial. Odd and even scores on a die have both properties; “even” and “greater than four” overlap at six.

The probabilities of all individual outcomes sum to one. Grouped events also sum to one when they form an exhaustive, mutually exclusive division. Exhaustive overlapping events need not have a sum of one because an outcome may be counted twice.

Complete outcomes and complements

For exhaustive, mutually exclusive events $A,B,C$ ,

P(A)+P(B)+P(C)=1

The complement $A'$ means “not $A$ ”, so

P(A')=1-P(A)

Not on the formula sheet: learn it.

Use the whole collection first, then find the particular complement requested.

Worked example 3: a missing game probability

Question. A game has exactly three outcomes $A,B,C$ , with probabilities $0.35,0.40,q$ . Find $q$ and $P(A')$ .

Plan. Subtract the known outcomes from one, then subtract $P(A)$ from one.

\begin{aligned} q&=1-0.35-0.40=0.25 && \text{complete the total}\\ P(A')&=1-0.35=0.65 && \text{all outcomes except A} \end{aligned}

Why this step? Exactly one of the three outcomes happens on each play.

Answer. $q=0.25$ , $P(A')=0.65$ .

Check. $0.35+0.40+0.25=1$ and $0.40+0.25=0.65$ .

Change only $P(B)$ , leaving $P(A)$ and the three-outcome structure unchanged.

Worked example 4: changing one known probability

Question. Now $P(A)=0.35$ , $P(B)=0.20$ and $P(C)=q$ . Find $q$ and $P(A')$ .

Plan. Repeat the subtraction with the new value of $P(B)$ .

\begin{aligned} q&=1-0.35-0.20=0.45 && \text{complete the total}\\ P(A')&=1-0.35=0.65 && \text{A has not changed} \end{aligned}

Why this step? Redistributing chance between $B$ and $C$ does not change their combined chance.

Answer. $q=0.45$ , $P(A')=0.65$ .

Check. $0.20+0.45=0.65$ .

For overlapping die events, list the successful faces once: even or greater than four gives $\{2,4,5,6\}$ , hence $4/6=2/3$ .

Watch out

Wrong: $P(\text{even or greater than }4)=3/6+2/6=5/6$ . This counts face $6$ twice. The four distinct successful faces give $4/6$ .

A win, draw and loss form a division without overlap. For win $0.28$ , draw $0.17$ , the loss chance is $0.55$ , and not losing has chance $0.45$ . Now change only $P(B)$ from example 4.

Recording outcomes and relative frequency

Frequency counts what happened; relative frequency compares that count with the total number of trials.

Relative frequency

\text{relative frequency}=\frac{\text{frequency of the event}}{\text{total number of trials}}

This estimates probability from experimental evidence. It does not establish the exact theoretical probability.

Not on the formula sheet: learn it.

Record every trial in a table, including categories with no results, and check the total.

Worked example 5: reading an experimental record

Question. This is a spinner's complete experimental record. Estimate the probability of red.

Outcome	Red	Blue	Green
Frequency	$22$	$31$	$27$

Plan. Total the complete record, then divide the red count by that total.

\begin{aligned} N&=22+31+27=80 && \text{all recorded spins}\\ \text{relative frequency of red}&=\frac{22}{80}=0.275 && \text{red share of trials} \end{aligned}

Why this step? The denominator includes unsuccessful spins as well as red ones.

Answer. The estimate is $0.275$ .

Check. $80\times0.275=22$ , recovering the red count.

Change only the requested event: combine two categories in the same record.

Worked example 6: combining recorded categories

Question. For the same red $22$ , blue $31$ , green $27$ record, estimate the probability of not red.

Plan. Add blue and green, then keep all $80$ trials as the denominator.

\begin{aligned} \text{not-red count}&=31+27=58 && \text{successful categories}\\ \text{relative frequency}&=\frac{58}{80}=0.725 && \text{divide by all trials} \end{aligned}

Why this step? “Not red” is one event containing two recorded outcomes.

Answer. $0.725$ .

Check. $0.275+0.725=1$ .

A frequency tree records two attributes in stages. Its labels are counts, and every pair of child counts adds to its parent. Draw the first split, split each subgroup again, and check every total.

Last Accessed

Papers 1, 2 and 3

P1, P2, P3, P4, P5 - Probability, relative frequency and expected outcomes

Randomness and the probability scale

Exhaustive outcomes and complementary events

Recording outcomes and relative frequency

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