Probability – Overview

Probability measures how likely something is to happen. It is always a number between $0$ and $1$.

The definition

$P(\text{event}) = \dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}$
(valid when all outcomes are equally likely)

A probability can never be negative or bigger than $1$. If you get $1.4$ or $-0.2$, you have made an arithmetic slip. Check immediately.

Three ways to write a probability: as a fraction $\tfrac{3}{8}$, a decimal $0.375$, or a percentage $37.5\%$. All are acceptable — but never write it as a ratio like "$3:8$", which means something different.

Word	Meaning
Outcome	One possible result of a trial, e.g. rolling a $4$.
Event	A set of outcomes you are interested in, e.g. "an even number".
Trial	One go of the experiment — one roll, one spin, one pick.
Sample space	The list of every possible outcome.
Mutually exclusive	Two events that cannot both happen at once.
Exhaustive	A set of events covering every possibility.
Independent	One event happening does not change the chance of the other.
Dependent	One event does change the chance of the other (e.g. no replacement).
Biased	Not all outcomes are equally likely.
Random	Every outcome has the same chance of being chosen.

Notation

$P(A)$ — the probability of $A$ · $P(A')$ — the probability of not $A$
$P(A \cap B)$ — $A$ and $B$ · $P(A \cup B)$ — $A$ or $B$ (or both)

Rule 1 — The total is 1

The probabilities of all possible outcomes add to $1$

Rule 2 — The complement

$P(\text{not } A) = 1 - P(A)$

Rule 3 — AND means multiply

For independent events: $P(A \text{ and } B) = P(A) \times P(B)$

Rule 4 — OR means add

For mutually exclusive events: $P(A \text{ or } B) = P(A) + P(B)$

The memory hook: AND — multiply. OR — add. Both rules have conditions attached (independence for AND, mutual exclusivity for OR), so check those before using them.

Worked Example — Using several rules together

A bag has $4$ red, $3$ blue and $5$ green counters. One counter is taken at random.

①Total $= 4 + 3 + 5 = 12$ counters.

②$P(\text{red}) = \dfrac{4}{12} = \dfrac{1}{3}$

③$P(\text{not red}) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$

④$P(\text{red or blue}) = \dfrac{4}{12} + \dfrac{3}{12} = \dfrac{7}{12}$ (mutually exclusive, so add)

Check: $\dfrac{4}{12} + \dfrac{3}{12} + \dfrac{5}{12} = 1$ ✓

The National Curriculum splits Probability into nine statements, usually labelled P1 to P9. Each has its own page with explanations, worked examples and ten practice questions.

P1Recording Outcomes: Tables and Frequency TreesTally charts, frequency tables, two-way tables and frequency trees.
P2Randomness, Fairness and Expected OutcomesPredicting how many times an event should happen in $n$ trials.
P3The Probability Scale and Relative FrequencyTheoretical vs experimental probability, and the language of likelihood.
P4Exhaustive and Mutually Exclusive EventsProbabilities summing to one, and the complement rule.
P5Experimental Probability and Sample SizeWhy larger samples converge on the theoretical value; detecting bias.
P6Listing Outcomes SystematicallyOrdered lists, product rule, Venn diagrams and set notation.
P7Sample Space DiagramsGrids for two dice, two spinners and other combined experiments.
P8Combined Events and Tree DiagramsIndependent and dependent events, with and without replacement.
P9Conditional ProbabilityUsing two-way tables, Venn diagrams and trees to find $P(A \text{ given } B)$.

Mistake 1 — Adding when you should multiply. "Two heads in a row" is $\tfrac{1}{2} \times \tfrac{1}{2} = \tfrac{1}{4}$, not $\tfrac{1}{2} + \tfrac{1}{2} = 1$.

Mistake 2 — Forgetting that the total changes. If a counter is taken and not replaced, the denominator drops by one for the second pick.

Mistake 3 — The gambler's fallacy. After five heads in a row, the next toss is still $\tfrac{1}{2}$. Coins have no memory.

Mistake 4 — Assuming equally likely outcomes. "It either rains or it doesn't, so $P = \tfrac{1}{2}$" is wrong. The two outcomes are not equally likely.

Mistake 5 — Cancelling too early in tree diagrams. Keep fractions with a common denominator until the very end, so the branches can be added.

Mistake 6 — Missing outcomes when listing. Always list systematically, in a fixed order, so nothing is left out or counted twice.

Definition

$P = \dfrac{\text{favourable}}{\text{total}}$ when outcomes are equally likely.

The scale

Every probability lies between $0$ and $1$ inclusive.

Complement

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

AND

Multiply — for independent events.

OR

Add — for mutually exclusive events.

Expected number

$P \times$ number of trials.

Relative frequency

$\dfrac{\text{times it happened}}{\text{number of trials}}$.

Without replacement

Both the numerator and the denominator change.

Tree diagrams

Multiply along branches, add between them.

Always check

Do your probabilities add to $1$?

These ten questions sample the whole topic. If one type catches you out, follow the link in Section 4 to the page that covers it.

Question 1

A fair six-sided die is rolled. Find (a) $P(\text{a } 3)$, (b) $P(\text{an even number})$, (c) $P(\text{more than } 4)$.

▶ Show solution

(a) One favourable outcome out of six: $\dfrac{1}{6}$

(b) Even numbers are $2, 4, 6$: $\dfrac{3}{6} = \dfrac{1}{2}$

(c) More than $4$ means $5$ or $6$: $\dfrac{2}{6} = \dfrac{1}{3}$

Question 2

The probability that it rains tomorrow is $0.35$. Find the probability that it does not rain.

▶ Show solution

$P(\text{not rain}) = 1 - 0.35 = 0.65$

Question 3

A bag contains $7$ red and $5$ yellow beads. One is chosen at random. Find $P(\text{yellow})$.

▶ Show solution

Total $= 7 + 5 = 12$

$P(\text{yellow}) = \dfrac{5}{12}$

Question 4

A fair coin is tossed twice. Find the probability of getting two heads.

▶ Show solution

The tosses are independent, so multiply:

$P(\text{HH}) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$

Question 5

A spinner has probabilities $P(\text{red}) = 0.3$, $P(\text{blue}) = 0.45$ and $P(\text{green}) = x$. Find $x$.

▶ Show solution

All probabilities must add to $1$:

$0.3 + 0.45 + x = 1$

$x = 1 - 0.75 = 0.25$

Question 6

The probability of a biased coin landing heads is $0.6$. It is tossed $200$ times. How many heads would you expect?

▶ Show solution

Expected number $= P \times$ trials

$= 0.6 \times 200 = 120$ heads

Question 7

Two fair dice are rolled and the scores added. How many outcomes are there altogether, and what is $P(\text{total} = 7)$?

▶ Show solution

Outcomes: $6 \times 6 = 36$.

Totals of $7$: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — six ways.

$P(7) = \dfrac{6}{36} = \dfrac{1}{6}$

Question 8

A bag has $5$ black and $3$ white balls. Two are taken without replacement. Find $P(\text{both black})$.

▶ Show solution

First ball: $\dfrac{5}{8}$

Second ball: only $4$ black left out of $7$: $\dfrac{4}{7}$

$P(\text{both black}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$

Question 9

A spinner is spun $150$ times and lands on red $54$ times. Find the relative frequency of red, and comment on whether the spinner might be biased if red is one of five equal sections.

▶ Show solution

Relative frequency $= \dfrac{54}{150} = 0.36$

If the five sections were equal, we would expect $P(\text{red}) = \dfrac{1}{5} = 0.2$.

$0.36$ is well above $0.2$ over a reasonably large number of trials, so the spinner is probably biased towards red.

Question 10

In a class of $30$ students, $18$ study French, $14$ study German and $7$ study both.

(a) How many study neither? (b) A student is chosen at random; find $P(\text{French only})$. (c) Given that a student studies German, find the probability that they also study French.

▶ Show solution

(a) French only $= 18 - 7 = 11$; German only $= 14 - 7 = 7$.

Studying at least one $= 11 + 7 + 7 = 25$, so neither $= 30 - 25 = 5$.

(b) $P(\text{French only}) = \dfrac{11}{30}$

(c) This is conditional: restrict to the $14$ German students, of whom $7$ also do French.

$P(\text{French} \mid \text{German}) = \dfrac{7}{14} = \dfrac{1}{2}$