Elementary Probability

Lessons

Senior 1 · Mathematics

Elementary Probability

5 lessons

Lesson 4 of 5

Theoretical probability

25 min

Probability without an experiment

In Lesson 3 you had to toss a bottle top hundreds of times to estimate a probability. Yet nobody tosses a coin a thousand times to find $P(\text{heads})$ : a coin has two faces of the same shape and weight, so there is no reason for one to come up more than the other. When all the outcomes of an experiment are equally likely like this, a probability can be calculated by counting, before the experiment is ever done.

Definition

Equally likely outcomes: outcomes that all have the same chance of happening.

Fair (unbiased): describes a coin, die or spinner whose outcomes are equally likely. A fair die has the same chance of showing each of its six faces.

At random: chosen so that every item has the same chance of being chosen, for example by picking a bead from a bag without looking.

Definition

Theoretical probability: when all the outcomes in the sample space $S$ are equally likely, the probability of an event $E$ is

P(E) = \frac{n(E)}{n(S)} = \frac{\text{number of outcomes in } E}{\text{total number of possible outcomes}}.

The outcomes in $E$ are called the favourable outcomes.

Example: for a fair die, $P(\text{even}) = \frac{n(\{2, 4, 6\})}{n(\{1, 2, 3, 4, 5, 6\})} = \frac{3}{6} = \frac{1}{2}$ .

Important

Why the formula works. If $n (S)$ outcomes are equally likely, they share the certainty of 1 equally, so each has probability $\frac{1}{n(S)}$ . An event with $n (E)$ of these outcomes has $n (E)$ such shares:

P(E) = n(E) \times \frac{1}{n(S)} = \frac{n(E)}{n(S)}.

Because $0 \le n(E) \le n(S)$ , this also shows again that $0 \le P(E) \le 1$ .

Worked example 1: a fair die

A fair die is rolled. Find (a) $P(\text{a prime number})$ , (b) $P(\text{a multiple of 3})$ , (c) $P(\text{a number greater than 6})$ , (d) $P(\text{a number less than 7})$ .

Solution $S = \{1, 2, 3, 4, 5, 6\}$ , so $n (S) = 6$ .

Primes: $\{2, 3, 5\}$ , so $P = \frac{3}{6} = \frac{1}{2}$ .
Multiples of 3: ${3, 6}$ , so $P = \frac{2}{6} = \frac{1}{3}$ .
No outcome is greater than 6, so $P = \frac{0}{6} = 0$ : impossible.
Every outcome is less than 7, so $P = \frac{6}{6} = 1$ : certain.

Worked example 2: beads in a bag

Ten beads of the same size in a bag: 5 red, 3 blue, 2 green.

One bead is taken at random. Find the probability that it is (a) blue, (b) red, (c) not red, (d) red or green, (e) yellow.

Solution Each of the 10 beads is equally likely to be picked, so $n (S) = 10$ .

$P(\text{blue}) = \frac{3}{10} = 0.3$ .
$P(\text{red}) = \frac{5}{10} = \frac{1}{2}$ .
Not red means blue or green: $3 + 2 = 5$ beads, so $P = \frac{5}{10} = \frac{1}{2}$ .
$5 + 2 = 7$ beads, so $P = \frac{7}{10} = 0.7$ .
There are no yellow beads: $P = 0$ .

Worked example 3: letter cards

The ten letter cards of NYARUGENGE, one letter on each card.

The letters of NYARUGENGE are written on ten cards, one letter per card. One card is picked at random. Find (a) $P(\text{N})$ , (b) $P(\text{a vowel})$ , (c) $P(\text{Y})$ .

Solution Each card is equally likely, so $n (S) = 10$ , the number of cards.

Two cards show N: $P(\text{N}) = \frac{2}{10} = \frac{1}{5}$ .
The vowel cards are A, U, E, E: $P = \frac{4}{10} = \frac{2}{5}$ .
One card shows Y: $P(\text{Y}) = \frac{1}{10}$ .

In Unit 1 the set of letters of NYARUGENGE had only 7 elements, because a set lists each letter once. For probability, count every card, because each card can be picked.

Worked example 4: two dice

The 5 outcomes with a total of 8 are shaded.

Two fair dice are rolled. Find the probability of (a) a total of 8, (b) a double, (c) a total of 10 or more.

Solution The 36 outcomes in the grid are equally likely, so $n (S) = 36$ .

Five shaded cells: $P(\text{total } 8) = \frac{5}{36}$ .
Six doubles, down the diagonal: $P(\text{double}) = \frac{6}{36} = \frac{1}{6}$ .
Totals 10, 11 and 12 fill $3 + 2 + 1 = 6$ cells: $P = \frac{6}{36} = \frac{1}{6}$ .

Theory and experiment side by side

Theory says that each face of a fair die has probability $\frac{1}{6}$ , so in 60 rolls each number is expected $\frac{1}{6} \times 60 = 10$ times. A class rolled a die 60 times:

60 rolls of a fair die. The theory expects each number 10 times; the experiment comes close but not exact.

The six came up 13 times, a relative frequency of $\frac{13}{60} \approx 0.217$ , against the theoretical $\frac{1}{6} \approx 0.167$ . This does not show that the die is unfair: 60 rolls are not many. As in Lesson 3, the more times the die is rolled, the closer each relative frequency is expected to come to $\frac{1}{6}$ . When outcomes are equally likely, theory gives the exact value that experiments estimate; when they are not (the bottle top), experiment is the only way.

Warning

Only for equally likely outcomes

"A football match can be won, drawn or lost, so $P(\text{win}) = \frac{1}{3}$ " is wrong. The three results are not equally likely; it depends on the two teams.
With two dice, the totals 2 to 12 are 11 results, but $P(\text{total } 12)$ is $\frac{1}{36}$ , not $\frac{1}{11}$ . Count the 36 equally likely ordered pairs instead.
Simplify fractions where you can, but a correct unsimplified answer such as $\frac{6}{36}$ is still correct.

Tip

Write the denominator first: " $n(S) = \ldots$ ", the total number of equally likely outcomes. Then count the favourable outcomes on top. Finish with a sense check: is the answer between 0 and 1, and does its size match your feeling of how likely the event is?

Math Antics: Basic Probability (11 min). Watch how probability is written as the number of favourable outcomes over the total number of outcomes, and how it can also be written as a decimal or a percentage.

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Practice

Give each answer as a fraction in its simplest form. Questions 6 and 7 are harder.

A fair die is rolled. Find (a) $P (4)$ , (b) $P(\text{odd})$ , (c) $P(\text{a factor of 6})$ .
A bag holds 4 red, 6 white and 2 black balls. One is taken at random. Find (a) $P(\text{white})$ , (b) $P(\text{not black})$ .
A spinner has 10 equal sectors numbered 1 to 10. Find $P(\text{a multiple of 4})$ and $P(\text{a number greater than 7})$ .
A letter is picked at random from the cards spelling MUSANZE. Find $P(\text{a vowel})$ .
Two fair dice are rolled. Find (a) $P(\text{total } 5)$ , (b) $P(\text{total } 7)$ .
Two coins are tossed. Find $P(\text{exactly one head})$ and $P(\text{at least one head})$ .
A class of 40 learners has 22 girls. Six girls and five boys wear glasses. A learner is chosen at random. Find the probability that the learner is (a) a boy, (b) wears glasses, (c) a boy who does not wear glasses.

Show answersHide answers

(a) $\frac{1}{6}$ (b) $\frac{3}{6} = \frac{1}{2}$ (c) factors of 6 are 1, 2, 3, 6: $\frac{4}{6} = \frac{2}{3}$
(a) $\frac{6}{12} = \frac{1}{2}$ (b) $\frac{10}{12} = \frac{5}{6}$
Multiples of 4: 4 and 8, so $\frac{2}{10} = \frac{1}{5}$ ; greater than 7: 8, 9, 10, so $\frac{3}{10}$ .
MUSANZE has 7 cards with vowels U, A, E: $\frac{3}{7}$ .
(a) $(1, 4), (2, 3), (3, 2), (4, 1)$ : $\frac{4}{36} = \frac{1}{9}$ (b) $\frac{6}{36} = \frac{1}{6}$
$S = \{HH, HT, TH, TT\}$ : exactly one head $\frac{2}{4} = \frac{1}{2}$ ; at least one head $\frac{3}{4}$ .
(a) 18 boys: $\frac{18}{40} = \frac{9}{20}$ (b) $6 + 5 = 11$ : $\frac{11}{40}$ (c) $18 - 5 = 13$ : $\frac{13}{40}$

Sources and credits

7.5: Basic Concepts of Probability (theoretical probability), from Contemporary Mathematics by Donna Kirk and contributors, OpenStax, provided by LibreTexts, licensed under CC BY 4.0 (licence read on the page, 26 September 2026). Adapted by Ganzaa.
Video: Math Antics - Basic Probability, mathantics, embedded from YouTube.

The diagrams, the Rwandan examples, the practice items and the quiz questions were written by Ganzaa.

1Probability without an experiment

2Theory and experiment side by side

Probability without an experiment

Theory and experiment side by side