Probability — Basics and Possibility Spaces

Probability measures how likely an event is on a scale from 0 (impossible) to 1 (certain).

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

Complementary probability: $P(A^{\prime })=1-P(A)$ (the probability that $A$ does not happen).

Exhaustive and mutually exclusive events (P4):

If events $A_1,A_2,\dots ,A_n$ cover all possible outcomes and cannot occur simultaneously:

$P(A_1)+P(A_2)+\cdots +P(A_n)=1$

Worked example — a bag contains red, blue, and green counters. $P(\text{red})=0.3$, $P(\text{blue})=0.5$. Find $P(\text{green})$.

$P(\text{green})=1-0.3-0.5=0.2$ ✓

Probability from a table: divide the frequency for the event by the total frequency.

Relative frequency estimates probability from experiments:

$\text{Relative frequency}=\frac{\text{frequency of outcome}}{\text{total number of trials}}$

Worked example — a biased coin is tossed 200 times; heads appears 130 times.

Relative frequency of heads $=130/200=0.65$

The law of large numbers (P5): as the number of trials increases, the relative frequency tends towards the true theoretical probability. A small sample may deviate significantly from theory; a large sample is more reliable.

Expected frequency (P2): if the probability of an event is $p$ and the experiment is performed $n$ times:

$\text{Expected frequency}=p\times n$

Worked example — $P(\text{six on a fair die})=1/6$. In 300 rolls, expected frequency $=300\times 1/6=50$ ✓

Two-way tables record outcomes across two categories simultaneously.

	Cats	Dogs	Total
Female	8	12	20
Male	6	14	20
Total	14	26	40

$P(\text{female cat})=8/40=0.2$; $P(\text{dog given male})=14/20=0.7$ (conditional — see combined probability lesson)

Frequency trees show how a sample is divided step by step. At each branch, the frequencies must add to the parent frequency.

Worked example — 60 students: 35 study French, of whom 20 also study German; 25 study neither (neither French nor German).

Tree: 60 total → 35 French, 25 not French. Of 35 French: 20 also German, 15 French only. Of 25 not French: 0 study German (given). Total studying German $=20$. ✓

A Venn diagram shows sets as overlapping circles within a rectangle (the universal set $\xi$).

Symbol	Meaning
$A\cup B$	$A$ or $B$ (or both) — union
$A\cap B$	$A$ and $B$ — intersection
$A^{\prime }$	not $A$ — complement
$n(A)$	number of elements in $A$

Worked example — 30 students: 18 play football ($F$), 12 play tennis ($T$), 5 play both.

$n(F\text{ only})=18-5=13$; $n(T\text{ only})=12-5=7$; $n(\text{neither})=30-13-5-7=5$

$P(F\cup T)=(13+5+7)/30=25/30=5/6$

$P(F\cap T)=5/30=1/6$ ✓

Sample space diagrams (grids) list all outcomes for two combined experiments.

Worked example — roll a fair die and flip a fair coin. Construct the sample space.

$\Omega =\{(1,H),(2,H),(3,H),(4,H),(5,H),(6,H),(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)\}$ — 12 equally likely outcomes.

$P(\text{even and heads})=3/12=1/4$ ✓

Tree diagrams show sequential experiments with branching.

Worked example — a bag has 3 red and 2 blue balls. Two balls are drawn with replacement.

Branch probabilities: $P(R)=3/5$, $P(B)=2/5$ at every draw (replacement restores the bag).

$P(RR)=3/5\times 3/5=9/25$

$P(\text{one of each colour})=P(RB)+P(BR)=\frac{3}{5}\times \frac{2}{5}+\frac{2}{5}\times \frac{3}{5}=\frac{12}{25}$ ✓

1. Forgetting to list all branches in a tree diagram

Every outcome at each stage must be shown — if an event has three possible outcomes, there must be three branches. Missing a branch makes the probabilities at that level sum to less than 1.

2. Adding instead of multiplying along a branch

Multiply probabilities along a branch (AND); add probabilities across branches for the same event (OR). The product of branch probabilities gives the probability of that specific path.

3. Venn diagrams — placing the intersection count in both circles

The overlap region shows students in both sets. The "circle-only" regions show students in one set but not the other. Placing the total for set $A$ in the $A$ circle (instead of subtracting the intersection) double-counts.

4. Expected frequency is not the same as guaranteed frequency

$300\times 1/6=50$ is the expected number of sixes, not the guaranteed number. In any real experiment, the actual count will vary around this value.

Mistake	Correction
"$P(A)+P(B)=1$ for any two events"	This is true only if $A$ and $B$ are complementary (exhaustive and mutually exclusive)
"Relative frequency = $\frac{\text{trials}}{\text{successes}}$"	Relative frequency = $\frac{\text{successes}}{\text{trials}}$
"P(two heads in a row) = 1/2 + 1/2 = 1"	$P(HH)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$ — multiply for AND