Introduction to Probability

What you will learn

• Use the $0$ to $1$ probability scale.
• Work out probabilities from equally likely outcomes.
• Use the complement, independent events and expected frequency.

Before you start: you should know fractions and decimals and the chance words. See Chance: Impossible to Certain.

The idea

Probability measures how likely something is to happen. Every probability is a number from $0$ to $1$:

• $0$ means impossible (rolling a $7$ on an ordinary die).
• $1$ means certain (the Sun rising tomorrow, for practical purposes).
• $0.5$ means an even chance (a fair coin landing heads).

Probabilities can be written as fractions, decimals or percentages: $\frac{1}{4}$, $0.25$ and $25\%$ all describe the same chance.

Outcomes and events

An outcome is a single possible result, such as rolling a $3$. An event is a set of outcomes you are interested in, such as “rolling an even number”, which includes $2$, $4$ and $6$. When all outcomes are equally likely:

$P(\text{event}) = \frac{\text{favourable outcomes}}{\text{total outcomes}}$

Two key rules

• Complements: something either happens or it doesn’t, so $P(\text{not } A) = 1 - P(A)$.
• Combining: for independent events (one doesn’t affect the other), multiply for “and”. For mutually exclusive events (they can’t both happen), add for “or”.

Worked examples

Example 1: A single event

A bag holds $5$ red, $3$ blue and $2$ green counters. One is picked at random. Find $P(\text{blue})$ and $P(\text{not green})$.

1. Total counters: $5 + 3 + 2 = 10$.
2. $P(\text{blue}) = \frac{3}{10} = 0.3$.
3. $P(\text{green}) = \frac{2}{10}$, so $P(\text{not green}) = 1 - \frac{2}{10} = \frac{8}{10} = 0.8$.

Answer: $0.3$ and $0.8$

Example 2: Independent events

A fair coin is flipped and a fair die is rolled. What is the probability of heads and a six?

1. $P(\text{heads}) = \frac{1}{2}$ and $P(\text{six}) = \frac{1}{6}$.
2. They are independent, so multiply: $\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$.

Answer: $\frac{1}{12}$

Example 3: Without replacement (tree diagram thinking)

From the same bag of $10$ counters, two are taken without replacement. Find the probability both are red.

1. First pick: $P(\text{red}) = \frac{5}{10}$.
2. Now $9$ counters remain, $4$ of them red: $P(\text{red second} \mid \text{red first}) = \frac{4}{9}$.
3. Multiply along the branch: $\frac{5}{10} \times \frac{4}{9} = \frac{20}{90} = \frac{2}{9}$.

Answer: $\frac{2}{9} \approx 0.22$

Expected frequency

If you roll a fair die $300$ times, you would expect about $300 \times \frac{1}{6} = 50$ sixes. Real results wobble around this figure, and the wobble shrinks, relatively, as the number of trials grows.

The “gambler’s fallacy”: after five heads in a row, a fair coin is still exactly $50\%$ likely to land heads. Coins have no memory.

Why it matters

Probability sits behind weather forecasts (“70% chance of rain”), insurance premiums, medical test results, game design, sports analytics and the spam filter in your inbox. Understanding it helps you weigh risks sensibly, such as why the lottery is a poor investment, or why a positive result from a rare-disease test may still be more likely a false alarm than not.