Probability — chapter notes

A random experiment is one whose result cannot be predicted in advance, such as tossing a coin or throwing a die. Each possible result is an outcome, and the set of all outcomes is the sample space $S$.

• One coin: $S = \{H, T\}$.
• A coin tossed twice (order noted): $S = \{HH, HT, TH, TT\}$.
• One die: $S = \{1, 2, 3, 4, 5, 6\}$.
• One card from a pack: $52$ outcomes.

List the outcomes in a fixed order so none is missed or repeated.

An event is a set of outcomes of the sample space. An elementary event has exactly one outcome, a sure event contains every outcome, and an impossible event contains none.

Example: on one throw of a die, "an even prime" is $\{2\}$ (elementary), "a number less than $7$" is $S$ (sure) and "the number $7$" is $\{\}$ (impossible).

When all outcomes are equally likely,

$P(E) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$.

Example: a die and the event "a divisor of $8$": favourable $1, 2, 4$, so $P = \dfrac{3}{6} = \dfrac{1}{2}$.

Outcomes must be equally likely: in a bag of $3$ red and $1$ white ball the four balls are equally likely but the two colours are not, so $P(\text{white}) = \dfrac{1}{4}$.

For every event, $0 \le P(E) \le 1$. An impossible event has $P = 0$ and a sure event has $P = 1$.

The complement of $E$, written "not $E$", contains every outcome outside $E$. Together they fill the sample space, so

$P(E) + P(\text{not } E) = 1$, so $P(\text{not } E) = 1 - P(E)$.

Example: $5$ black and $13$ white tokens: $P(\text{not black}) = 1 - \dfrac{5}{18} = \dfrac{13}{18}$.

Coins. Tossing a coin once, twice or three times (order noted) gives $2$, $4$ or $8$ outcomes. For two tosses, "exactly one head" is $\{HT, TH\}$ with $P = \dfrac{1}{2}$.

Die. Six equally likely outcomes. A prime number: $2, 3, 5$, so $P = \dfrac{1}{2}$; a number at most $4$: $P = \dfrac{2}{3}$, so a number greater than $4$ has $P = \dfrac{1}{3}$.

A pack has $52$ cards: $4$ suits (spades, hearts, diamonds, clubs) of $13$ each; $26$ red and $26$ black; $12$ face cards (jack, queen, king) and $4$ aces.

Examples: $P(\text{queen}) = \dfrac{1}{13}$, $P(\text{club}) = \dfrac{1}{4}$, $P(\text{red face card}) = \dfrac{3}{26}$.

If cards are removed, recount: with all spades removed there are $39$ cards, $9$ face cards and $26$ red cards, so $P(\text{red}) = \dfrac{2}{3}$.

The sample space is the set of balls. The total is the sum of all the colours, and $P(\text{colour}) = \dfrac{\text{balls of that colour}}{\text{total balls}}$. The probabilities of all the colours add up to $1$.

Unknown number: if a bag has $8$ green and $y$ purple balls and $P(\text{purple}) = \dfrac{3}{5}$, then $\dfrac{y}{8 + y} = \dfrac{3}{5}$ gives $y = 12$. After balls are added or removed, recount favourable and total.

Write the favourable numbers out and count them. Example: cards $1$ to $45$: primes $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43$ ($14$ of them), so $P(\text{prime}) = \dfrac{14}{45}$. Remember that $1$ is not a prime.

If some cards are removed (for example all multiples of $7$ from $50$ cards), the sample space is only the cards that remain.

Not in the ICSE 2027 syllabus: compound events and the probability of combined events; every problem in this chapter is about a single event on one sample space.