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Cambridge IGCSE Mathematics — 0580 Extended

Topic 8.1: Probability — Introduction

Probability measures how likely an event is, on a scale from \(0\) (impossible) to \(1\) (certain). On Extended 0580 you write \(\mathrm{P}(A)\) for the probability of \(A\), and \(\mathrm{P}(A')\) for the probability that \(A\) does not happen. Give the answer as a fraction, a decimal or a percentage — the question will usually signal which.

The probability scale

An event that is as likely as not sits at \(0.5\). Nothing lives outside \([0, 1]\): a working of \(1.2\) or of \(-0.1\) has already gone wrong.

Probability scale from 0 to 1 with impossible, even chance and certain labelled, and example events pinned on the scale
Every probability on 0580 belongs on this scale. Place the event before you calculate.

Equally likely outcomes

When every outcome of an experiment is equally likely,

\[ \mathrm{P}(A) = \dfrac{n(A)}{n(\mathcal{E})} \]

Favourable outcomes on top, equally likely outcomes in the whole sample space \(\mathcal{E}\) on the bottom.

Method

  1. List the sample space, or count it (\(n(\mathcal{E})\)).
  2. Count the outcomes that match the event (\(n(A)\)).
  3. Write the fraction and cancel if you can.

A fair six-sided die is rolled. Find \(\mathrm{P}(\text{even})\).

Six dice faces with 2, 4 and 6 highlighted, giving P(even) = 3/6 = 1/2
Three even faces out of six. Cancel \(\mathrm{P}(\text{even}) = 3/6 = 1/2\).

A fair spinner is split into eight equal sectors. Three sectors are shaded. Find \(\mathrm{P}(\text{shaded})\).

Fair spinner with eight equal sectors, three shaded, so P(shaded) = 3/8
Equal sectors mean equally likely. \(\mathrm{P}(\text{shaded}) = 3/8\).

Notation and a bag of counters

Extended papers use \(\mathrm{P}(A)\) and \(\mathrm{P}(A')\). “At random” means each counter (or card, or student) is equally likely.

A bag contains 5 red, 3 blue and 2 green counters. One is taken at random. Find \(\mathrm{P}(\text{red})\).

Bag with 5 red, 3 blue and 2 green counters, so P(red) = 5/10 = 1/2
\(n(\mathcal{E}) = 10\), so \(\mathrm{P}(\text{red}) = 5/10 = 1/2\).

The complement

\(A'\) is “not \(A\)”. The two events split the whole sample space:

\[ \mathrm{P}(A') = 1 - \mathrm{P}(A) \]

\(\mathrm{P}(B) = 0.8\). Find \(\mathrm{P}(B')\).

Bar showing P(A) = 0.8 and P(A') = 0.2 adding to 1
\(\mathrm{P}(B') = 1 - 0.8 = 0.2\). The two parts add to 1.

Paper 2 (non-calculator)

Leave a complement as a fraction: if \(\mathrm{P}(A) = 3/8\), then \(\mathrm{P}(A') = 5/8\). Do not convert unless the question asks for a decimal or a percentage.

Try this

A fair spinner has 5 equal sectors; 2 are red. Find \(\mathrm{P}(\text{not red})\).

Show answer
Answer

  1. \(\mathrm{P}(\text{red}) = 2/5\)

    \[ \mathrm{P}(\text{not red}) = 1 - \dfrac{2}{5} = \dfrac{3}{5} \]

Exam Traps

  • \(\mathrm{P}(A')\) is \(1 - \mathrm{P}(A)\), not \(1/\mathrm{P}(A)\). The reciprocal is a different (and usually illegal) number.

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