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E8.1 Introduction to probability

Probability measures likelihood on a scale from 0 to 1 inclusive. A larger value means the event is more likely.

Probability	Meaning
$0$	impossible
between $0$ and $0.5$	unlikely
$0.5$	equally likely to occur or not occur
between $0.5$ and $1$	likely
$1$	certain

Convert values to the scale's form before locating or comparing them. For example, $\tfrac38=0.375$ lies between 0 and 0.5, while $65\%=0.65$ lies between 0.5 and 1. The larger probability is farther to the right.

The endpoints 0 and 1 are valid probabilities. No probability can be negative or exceed 1, and a probability describes likelihood rather than guaranteeing how often an event occurs in a short experiment.

An event is a set of outcomes and is usually named with a capital letter. $P(A)$ means the probability that event $A$ occurs.

Notation	Meaning
$A$	the event itself
$P(A)$	the probability that $A$ occurs
$A'$	the complementary event: $A$ does not occur
$P(A')$	the probability that $A$ does not occur

Let $A$ be the event 'a fair six-sided die shows an even number'. The favourable outcomes are 2, 4 and 6, so $P(A)=\tfrac36=\tfrac12$ . Event $A'$ is 'the die shows an odd number', and $P(A')=\tfrac12$ .

$P(A)$ is a number between 0 and 1; it is not $P$ multiplied by $A$ . Define the event before using its letter, and do not confuse $A'$ with a second unrelated event.

When individual outcomes are equally likely, divide the number of outcomes in the event by the total number of possible outcomes.

P(A)=\frac{\text{number of favourable outcomes}}{\text{total number of equally likely outcomes}}

Identify the complete outcome set, count each outcome once, select only those satisfying the event, then simplify the fraction or convert it to a decimal or percentage if required. Counts may come from a table, graph or Venn diagram.

A bag contains 4 black, 3 white and 5 yellow counters. If one counter is selected at random, there are 12 equally likely counters and 8 are not yellow. Therefore $P(\text{not yellow})=\tfrac8{12}=\tfrac23$ .

The favourable-over-total rule requires equally likely individual outcomes. Do not use the number of category labels as the denominator when categories contain different numbers of outcomes.

Event $A$ and its complement $A'$ cover every possible outcome exactly once, so their probabilities add to 1.

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

Given	Complement
$P(A)=\tfrac7{10}$	$P(A')=\tfrac3{10}$
$P(B)=0.8$	$P(B')=0.2$
$P(C)=18\%$	$P(C')=82\%$

With counts, subtract the event count from the total before dividing. In a table, graph or Venn diagram, include every region outside the event; those regions together form the complement.

$A'$ means every allowed outcome not in $A$ , not one chosen alternative. Subtract from 1 for fractions or decimals and from $100\%$ for percentages; the two probabilities must sum to one whole.

E8.2 Relative and expected frequencies

Relative frequency is the proportion of trials in which an event occurs. It gives an experimental estimate of the event's probability.

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

A spinner is used 80 times and lands on 5 on 13 occasions. The relative frequency of landing on 5 is $\tfrac{13}{80}=0.1625$ , so $0.1625$ is an estimate of the probability of landing on 5.

A different set of trials can give a different relative frequency. A larger, well-run sample usually gives a more stable estimate because one unusual result has less influence, but it does not guarantee the exact theoretical probability.

To combine experiments, add the event counts and add the trial counts, then divide the two totals. For example, results of 8 successes from 30 trials and 17 from 70 trials combine to $\tfrac{8+17}{30+70}=0.25$ . Do not take an unweighted average when the trial totals differ.

Relative frequency is evidence from observed results, not a certainty about the next result. 'Random' means an individual result is unpredictable; it does not mean that every outcome must have equal probability.

Expected frequency is the number of occurrences predicted from a probability over a stated number of trials or members of a population.

\text{expected frequency}=\text{probability of the event}\times\text{number of trials or population size}

A bag contains 7 red, 5 green and 2 pink counters. A counter is selected at random, replaced and the experiment is repeated 140 times. Since $P(\text{green})=\tfrac5{14}$ , the expected frequency of green is $\tfrac5{14}\times140=50$ . Replacement keeps the probability the same for each trial.

If the probability is unknown, estimate it with relative frequency first. In a representative sample, 36 of 240 people have a feature, so the estimated probability is $\tfrac{36}{240}=0.15$ . For a population of 1600, the expected number is $0.15\times1600=240$ .

Term	Meaning
fair	the relevant outcomes have equal probabilities
biased	the mechanism favours some relevant outcomes, so their probabilities are not equal
random	an individual result cannot be predicted with certainty; probabilities need not be equal

An expected frequency is a long-run prediction, not a guarantee of the actual count. Keep the probability unrounded until the final multiplication. A calculation may give a non-integer expectation even though an observed count must be a whole number; round only when the context or question requires it.

E8.3 Probability of combined events

A combined event joins conditions or stages. Choose a representation that lists every possible outcome once, then combine only the outcomes that satisfy the event.

Representation	Best use	Probability method
sample space diagram	two stages fit rows and columns	mark favourable cells; count cells only when they are equally likely
two-set Venn diagram	outcomes belong to $A$ , $B$ , both or neither	$P(A\cap B)$ uses the overlap; $P(A\cup B)$ uses every region in $A$ or $B$ , counting the overlap once
tree diagram	outcomes occur in stages	multiply along each required path; add probabilities of alternative paths

P(\text{one path})=\text{product of its branch probabilities},\qquad P(\text{alternative paths})=\text{sum of their path probabilities}

Selection	What happens before the next branch?
with replacement	the selected item returns, so totals and probabilities reset
without replacement	the selected item stays out, so the total falls by 1 and the relevant category may also fall by 1

A bag has 5 green and 3 blue counters. With replacement, $P(\text{two green})=\tfrac58\times\tfrac58=\tfrac{25}{64}$ . Without replacement, after one green there are 4 green among 7 counters, so $P(\text{two green})=\tfrac58\times\tfrac47=\tfrac5{14}$ .

On a tree diagram, write each outcome at the end of its branch and its probability beside the branch. Check that probabilities leaving the same point add to 1. In a Venn diagram, 'and' means the intersection $\cap$ and inclusive 'or' means the union $\cup$ ; the overlap must not be counted twice.

Do not use a fixed second-stage denominator in a without-replacement problem. Conditional probability notation $P(A\mid B)$ and its formulas belong to E8.4, not this card.

E8.4 Conditional probability

A condition tells you that only part of the original sample space is still possible. Ignore every outcome outside that condition, then compare the required outcomes with the new, smaller total.

First identify the outcomes that satisfy the given condition. Use all of them as the new denominator. Within that restricted group, count or weight the outcomes that also satisfy the target event; these form the numerator.

Representation	Restricted denominator	Required numerator
Venn diagram	every region inside the set named by the condition	the part of those regions also satisfying the target
table or sample space	the relevant row, column or selected cells	target cells within that restricted total
tree diagram	total probability of every complete path consistent with the condition	total probability of the consistent path or paths that also meet the target

One number is selected from $\{1,2,4\}$ and one from $\{3,4,6\}$ , with all nine ordered pairs equally likely. Given that their sum is odd, only $(1,4),(1,6),(2,3),(4,3)$ remain. Exactly one has first number 4, so the required probability is $\tfrac14$ .

A Venn diagram shows 18 students in set $F$ and 10 in the overlap of $F$ and $T$ . If the chosen student is known to be in $F$ , the sample space contains only those 18 students. Ten also belong to $T$ , so the probability is $\tfrac{10}{18}=\tfrac59$ .

A tree shows 60% of items follow route A and 40% route B. The faulty rates are 5% on A and 10% on B. Faulty path weights are $0.6\times0.05=0.03$ and $0.4\times0.10=0.04$ . Given that an item is faulty, only these paths remain, so the probability it followed B is $\tfrac{0.04}{0.03+0.04}=\tfrac47$ .

The condition changes the denominator, not just the numerator. This syllabus requires calculation from Venn diagrams, trees and tables, but does not require special conditional-probability notation or formulas.

8. Probability

Exam analysis

Most tested topics

E8.1 Introduction to probability

Read and use the probability scale

Use probability notation for events

Calculate the probability of one event

Find the probability of a complementary event

E8.2 Relative and expected frequencies

Use relative frequency to estimate probability

Calculate expected frequency and interpret fair, biased and random

E8.3 Probability of combined events

Calculate combined-event probabilities

E8.4 Conditional probability

Calculate probability after a condition is known