Probability

Every probability is a number between $0$ and $1$ inclusive:

• $0$ means the event is impossible (rolling a $7$ on a normal $1$–$6$ die).
• $1$ means the event is certain (rolling a number less than $7$).
• $0.5$ (or $\tfrac{1}{2}$) means an even chance (a fair coin landing on heads).

You can write probabilities as fractions, decimals or percentages — they all describe the same thing. So $P(\text{heads}) = \tfrac{1}{2} = 0.5 = 50\%$.

A few quick sanity checks save marks. A probability bigger than $1$ is impossible — recheck your arithmetic. A negative probability is impossible too. And if you're using percentages, $P(A) + P(\text{not } A)$ must add to $100\%$ exactly.

When every outcome is equally likely, you can work probabilities out using counting:

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

Rolling a fair $1$–$6$ die, the outcomes are $\{1, 2, 3, 4, 5, 6\}$ — six equally likely results. So $P(\text{rolling a 4}) = \tfrac{1}{6}$, and $P(\text{rolling an even number}) = \tfrac{3}{6} = \tfrac{1}{2}$ because there are three even outcomes.

Some quick examples:

• A fair coin: $P(\text{heads}) = \tfrac{1}{2}$.
• A standard pack of $52$ cards: $P(\text{red card}) = \tfrac{26}{52} = \tfrac{1}{2}$.
• A bag with $3$ red, $5$ blue and $2$ green balls: $P(\text{blue}) = \tfrac{5}{10} = \tfrac{1}{2}$.

The phrase "equally likely" is doing a lot of work. If a spinner is not fair — say it has three big sectors and two tiny ones — the simple counting formula doesn't apply, and you have to use the experimental approach in the next section instead.

When the theoretical probability isn't obvious — a bent coin, an uneven spinner, the chance of rain — you can estimate it by carrying out an experiment:

$\text{experimental probability} = \frac{\text{number of times the event happened}}{\text{number of trials}}.$

This is also called the relative frequency. The more trials you carry out, the closer the experimental probability gets to the true probability — that pattern is the law of large numbers.

Suppose you flip a possibly-biased coin $200$ times and get $130$ heads:

$P(\text{heads}) \approx \frac{130}{200} = 0.65.$

So the coin looks biased — a fair coin would land on heads close to $100$ times out of $200$, not $130$. With more trials you'd become more confident in that estimate.

You can also use relative frequency to predict: if a $200$-trial experiment gave $0.65$ heads, then in $1000$ flips you'd expect about $1000 \times 0.65 = 650$ heads. That's a useful real-life skill — quality-control, weather forecasting and sport stats all use it.

A sample space diagram is a table that lists every possible outcome when something happens twice (or two events happen together). They're brilliant for two-dice problems, two-spinner problems and similar.

Rolling two fair dice and adding the scores gives a sample space of $36$ outcomes (six results on each die, $6 \times 6 = 36$). The table below shows the sum in each cell:

To find $P(\text{sum} = 7)$, count the cells with the value $7$ — there are $6$ of them. So $P(\text{sum} = 7) = \tfrac{6}{36} = \tfrac{1}{6}$. Similarly, $P(\text{sum} = 12) = \tfrac{1}{36}$, and $P(\text{sum} \ge 10) = \tfrac{6}{36} = \tfrac{1}{6}$ (the cells $10, 11, 11, 12, 10, 10 \ldots$ — count them off the diagram).

Sample space diagrams work for any pair of equally-likely experiments: two spinners, a coin and a die, a card and a coin. List, count, divide.

Two events are mutually exclusive if they can't happen at the same time. Rolling a $3$ and rolling a $5$ on a single die are mutually exclusive — one die can't show both at once. Rolling a $3$ and rolling an even number are not mutually exclusive when you think across many rolls, but on a single roll a $3$ excludes evens because $3$ isn't even.

For mutually exclusive events $A$ and $B$:

$P(A \text{ or } B) = P(A) + P(B).$

So on one roll of a fair die, $P(3 \text{ or } 5) = \tfrac{1}{6} + \tfrac{1}{6} = \tfrac{2}{6} = \tfrac{1}{3}$.

The complement of an event $A$ is "$A$ does not happen" — written $A'$ or $\text{not } A$. The event and its complement are always mutually exclusive and together cover every outcome, so:

$P(A) + P(\text{not } A) = 1 \quad \Longrightarrow \quad P(\text{not } A) = 1 - P(A).$

This trick saves time. If $P(\text{at least one head in three flips}) = \tfrac{7}{8}$, then $P(\text{no heads}) = 1 - \tfrac{7}{8} = \tfrac{1}{8}$. Sometimes the complement is the easier event to count first.

A tree diagram is a way of laying out a multi-stage experiment so you can follow each possible sequence. Each branch shows an outcome and its probability, and the probabilities on every set of branches sum to $1$.

Imagine flipping a coin twice. The first set of branches: $\text{H}$ (prob $0.5$) or $\text{T}$ (prob $0.5$). After each, the second set of branches: $\text{H}$ ($0.5$) or $\text{T}$ ($0.5$).

The two rules are:

• Multiply along the branches to get the probability of a path: $P(\text{HH}) = 0.5 \times 0.5 = 0.25$.
• Add the probabilities of separate paths that give the event you want: $P(\text{exactly one head}) = P(\text{HT}) + P(\text{TH}) = 0.25 + 0.25 = 0.5$.

A quick check at the end: all the path probabilities should sum to $1$. $0.25 + 0.25 + 0.25 + 0.25 = 1$ ✓.

Probability is one of the most useful corners of mathematics:

• Genetics in science uses tree diagrams to predict inherited traits.
• Games and sport rely on probability for fair design and predicting outcomes.
• Insurance and finance quote probabilities every time they price risk.
• At IGCSE you'll extend tree diagrams to dependent events, add Venn diagrams for sets, and meet conditional probability.

For Checkpoint, the formula stays simple — favourable over total, path multiplication, and complement subtraction. Get those three moves into your fingertips and most probability questions tidy themselves up.