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Year 12 Methods (Unit 3 & 4) Refresher on probability and discrete random variables

Sample spaces and probability

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), the sample space is the set of all possible outcomes and an event is a subset of them. For equally likely outcomes, \(P(A)=\dfrac{\text{favourable}}{\text{total}}\). The complement, addition and mutually exclusive rules combine event probabilities, Venn diagrams and two-way tables organise the counts, and relative frequency estimates a probability from data.

A chance experiment (such as rolling a die) has a set of possible outcomes. The sample space is the set of all of them, and an event is any subset of the sample space — for example, 'rolling an even number' is the event \(\{2,4,6\}\).

When every outcome is equally likely, the probability of an event is

\(P(A)=\dfrac{\text{number of outcomes in }A}{\text{total number of outcomes}},\qquad 0\le P(A)\le 1.\)

The complement \(A'\) is 'not \(A\)'. Two events are mutually exclusive if they cannot both occur. Set notation writes \(A\cap B\) for 'A and B' (the intersection) and \(A\cup B\) for 'A or B' (the union), which a Venn diagram pictures as overlapping circles.

Key idea. \(P(A)=\dfrac{\text{favourable}}{\text{total}}\); \(P(A')=1-P(A)\); \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\); mutually exclusive \(\Rightarrow P(A\cap B)=0\).
Venn diagram of two overlapping eventsTwo overlapping circles A and B. The left-only region shows 7, the overlap shows 5, the right-only region shows 8, and 3 lies outside both circles. A B 7 5 8 3
Four regions: only \(A=7\), overlap \(A\cap B=5\), only \(B=8\), outside \(=3\)
Union of two events shadedTwo overlapping circles A and B with both circles shaded to represent the union A or B. A B A ∪ B (union)
\(A\cup B\): every outcome in \(A\), in \(B\), or in both

Probability of an event (equally likely outcomes):

\[P(A)=\dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}},\qquad 0\le P(A)\le 1\]
P(A)=favourabletotal

The complement and addition rules:

\[P(A')=1-P(A),\qquad P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
P(AB)=P(A)+P(B)P(AB)

Mutually exclusive events (no overlap) and relative frequency:

\[P(A\cap B)=0\ \Rightarrow\ P(A\cup B)=P(A)+P(B),\qquad P(A)\approx\dfrac{\text{times }A\text{ occurred}}{\text{number of trials}}\]
P(AB)=P(A)+P(B)
Two-way tables. A two-way (contingency) table breaks a group down by two attributes at once. Add across rows and down columns to a grand total, then read a probability as the relevant count divided by that total.

How to find a probability

  1. List the sample space. Identify all possible outcomes and how many there are; check whether they are equally likely.
  2. Count the favourable outcomes. For an equally likely space, \(P(A)=\dfrac{\text{favourable}}{\text{total}}\).
  3. Use a rule if events combine. For 'not \(A\)' use \(P(A')=1-P(A)\); for '\(A\) or \(B\)' use \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), dropping the overlap only when the events are mutually exclusive.
  4. Organise with a diagram or table. A Venn diagram or two-way table lays out the counts so each region or cell can be read directly.
From data. When outcomes are not equally likely, estimate a probability by relative frequency: divide the number of times the event occurred by the number of trials. The estimate improves as the number of trials grows.
Example 1 — Basic probability
A bag has \(3\) red and \(7\) yellow counters. Find \(P(\text{red})\).
Solution

Favourable outcomes over total outcomes.

\(P(\text{red})\)\(=\)\(\dfrac{3}{3+7}=\dfrac{3}{10}\)
P=310
Example 2 — Complement
A letter is chosen from \(\text{LEVEL}\). Find \(P(\text{not }E)\).
Solution

There are \(2\) \(E\)'s among \(5\) letters, so use the complement.

\(P(\text{not }E)\)\(=\)\(1-\dfrac{2}{5}\)
\(=\)\(\dfrac{3}{5}\)
P=35
Example 3 — Addition rule
\(P(A)=0.55\), \(P(B)=0.30\) and \(P(A\cap B)=0.15\). Find \(P(A\cup B)\).
Solution

Subtract the overlap once so it is not counted twice.

\(P(A\cup B)\)\(=\)\(0.55+0.30-0.15\)
\(=\)\(0.70\)
P=0.70
Example 4 — Relative frequency
A die is rolled \(250\) times and a \(6\) appears \(42\) times. Estimate \(P(6)\).
Solution

Relative frequency is the count divided by the number of trials.

\(P(6)\)\(\approx\)\(\dfrac{42}{250}=0.168\)
Bar chart of die-roll frequenciesA bar chart of the frequency of each score from 1 to 6 over 250 rolls: 38, 41, 44, 39, 46 and 42, with the score of 6 bar highlighted. 1 2 3 4 5 6
P(6)0.168

Common pitfalls

Do not add overlapping probabilities. When \(A\) and \(B\) can both happen, \(P(A)+P(B)\) double-counts the overlap; subtract \(P(A\cap B)\) once.
Only drop the overlap when mutually exclusive. \(P(A\cup B)=P(A)+P(B)\) is valid only when \(P(A\cap B)=0\).
Relative frequency is an estimate. It approaches the theoretical probability as trials increase, but is rarely exactly equal to it.

Frequently asked questions

How do you find the probability of an event?

For equally likely outcomes, divide the favourable outcomes by the total. A bag of \(3\) red and \(7\) yellow gives \(P(\text{red})=\dfrac{3}{10}=0.3\). Every probability lies between \(0\) and \(1\).

What is the complement rule?

\(A'\) is 'not \(A\)', and \(P(A')=1-P(A)\). If \(P(6)=\dfrac{1}{6}\), then \(P(\text{not }6)=\dfrac{5}{6}\). It is handy for 'at least one' problems.

What is the addition rule for probability?

\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). For \(P(A)=0.55\), \(P(B)=0.30\), \(P(A\cap B)=0.15\), you get \(P(A\cup B)=0.70\).

What does mutually exclusive mean?

Two events that cannot both happen, so \(P(A\cap B)=0\); then \(P(A\cup B)=P(A)+P(B)\). Rolling a \(5\) and a \(6\) on one die are mutually exclusive.

What is relative frequency?

The number of times an event happens divided by the number of trials. \(250\) rolls with \(42\) sixes give a relative frequency of \(\dfrac{42}{250}=0.168\).

How do Venn diagrams and two-way tables help?

They organise the counts — a Venn diagram into only \(A\), the overlap, only \(B\) and outside; a two-way table by two attributes — so a probability is the relevant count over the total.

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