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Year 11 Methods (Unit 1 & 2) Probability

Sample Spaces And Probability

20 practice questions 1 video lesson Theory + worked examples

Understand sample spaces and probability for Queensland Year 11 Mathematical Methods (QCAA). The sample space lists every possible outcome, and when outcomes are equally likely, probability is simply favourable outcomes over the total.

You will learn to list a sample space, calculate theoretical probability, apply the complement rule, and count outcomes for two-stage experiments using dice arrays and coin trees — the groundwork for all later QCAA probability.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1 Topic 5), the sample space \(S\) is the set of all possible outcomes of an experiment. When outcomes are equally likely, the theoretical probability of an event \(E\) is \(P(E)=\dfrac{n(E)}{n(S)}\). This page also covers the complement rule \(P(A')=1-P(A)\) and counting two-stage experiments.

An experiment is a process with an uncertain result. Each result is an outcome, and the sample space \(S\) is the set of every possible outcome. An event is a subset of the sample space — the outcomes we are interested in.

When every outcome is equally likely, the theoretical probability of an event \(E\) is the fraction of outcomes that are favourable: \(P(E)=\dfrac{n(E)}{n(S)}\). Every probability lies between \(0\) (impossible) and \(1\) (certain).

The complement \(A'\) is ‘\(A\) does not happen’. Since \(A\) and \(A'\) together cover the whole sample space, their probabilities add to \(1\).

Count, then divide. Find \(n(S)\) (all outcomes) and \(n(E)\) (favourable outcomes), then \(P(E)=\dfrac{n(E)}{n(S)}\). Give the answer as an exact fraction or a decimal.
Sample space of a fair die with even outcomes shadedSix boxes numbered one to six; the even outcomes two, four and six are shaded. 1 2 3 4 5 6
The sample space of a fair die is \(S=\{1,2,3,4,5,6\}\). The even event \(\{2,4,6\}\) has \(3\) of the \(6\) outcomes, so \(P(\text{even})=\dfrac{1}{2}\).
Two-dice sum array with the outcomes giving seven shadedA six by six array of the sums of two dice; the six cells that total seven are shaded. + 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12
Two dice give \(6\times 6=36\) equally likely outcomes. The \(6\) shaded cells total \(7\), so \(P(\text{sum}=7)=\dfrac{6}{36}=\dfrac{1}{6}\).

For an equally likely sample space:

\[P(E)=\dfrac{n(E)}{n(S)}\]
P(E)=n(E)n(S)

The complement rule:

\[P(A')=1-P(A)\]
P(A′)=1-P(A)

For a two-stage equally likely experiment, count the outcomes with an array:

\[n(S)=n_1\times n_2\]
n(S)=n1×n2
Complement shortcut: for ‘at least one’ questions it is often easier to find \(P(\text{none})\) first and use \(P(\text{at least one})=1-P(\text{none})\).

How to find a theoretical probability

  1. List or count \(S\): write down every equally likely outcome, or use an array/tree for two stages, and find \(n(S)\).
  2. Count favourable \(n(E)\): pick out the outcomes that make the event happen.
  3. Divide: \(P(E)=\dfrac{n(E)}{n(S)}\), then simplify the fraction.
  4. Complement if easier: for ‘not’ or ‘at least one’, use \(P(A')=1-P(A)\).
Example 1 — Probability from a sample space
A fair six-sided die is rolled once. Find the probability of rolling an even number.
Solution

Sample space — list every outcome:

\(S\)\(=\)\(\{1,2,3,4,5,6\}\)
\(n(S)\)\(=\)\(6\)

Favourable event — the even numbers:

\(E\)\(=\)\(\{2,4,6\}\)
\(n(E)\)\(=\)\(3\)

Divide and simplify:

\(P(\text{even})\)\(=\)\(\dfrac{n(E)}{n(S)}\)
\(=\)\(\dfrac{3}{6}\)
\(=\)\(\dfrac{1}{2}\)

\(P(\text{even})=\dfrac{1}{2}\).

Rolling a fair die: even outcomesDie faces one to six with two, four and six shaded as the favourable even outcomes. 1 2 3 4 5 6
Example 2 — The complement rule
A bag holds \(5\) red, \(3\) blue and \(2\) green balls. One ball is drawn at random. Find the probability the ball is not blue.
Solution

Total outcomes:

\(n(S)\)\(=\)\(5+3+2\)
\(=\)\(10\)

Probability of blue:

\(P(\text{blue})\)\(=\)\(\dfrac{3}{10}\)

Apply the complement rule:

\(P(\text{not blue})\)\(=\)\(1-P(\text{blue})\)
\(=\)\(1-\dfrac{3}{10}\)
\(=\)\(\dfrac{7}{10}\)

\(P(\text{not blue})=\dfrac{7}{10}\).

Ten balls in a bag with the blue balls shadedFive red, three blue and two green balls; the three blue balls are shaded. R R R R R B B B G G
Example 3 — Two-dice array
Two fair dice are rolled and their scores added. Find the probability the sum is \(7\).
Solution

Count the sample space with a \(6\times 6\) array:

\(n(S)\)\(=\)\(6\times 6\)
\(=\)\(36\)

List the outcomes that total \(7\):

\(E\)\(=\)\(\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}\)
\(n(E)\)\(=\)\(6\)

Divide and simplify:

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

\(P(\text{sum}=7)=\dfrac{1}{6}\).

Two-dice sum array for a total of sevenSix by six array of two-dice sums; the six shaded cells total seven out of thirty six outcomes. + 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12
Example 4 — Three coins and a complement
Three fair coins are tossed. Find the probability of getting exactly two heads, and the probability of getting at least one head.
Solution

Sample space of three coins:

\(n(S)\)\(=\)\(2\times 2\times 2\)
\(=\)\(8\)

Exactly two heads — list the outcomes:

\(E\)\(=\)\(\{HHT,\,HTH,\,THH\}\)
\(P(\text{two heads})\)\(=\)\(\dfrac{3}{8}\)

At least one head — use the complement of ‘no heads’:

\(P(TTT)\)\(=\)\(\dfrac{1}{8}\)
\(P(\text{at least one head})\)\(=\)\(1-\dfrac{1}{8}\)
\(=\)\(\dfrac{7}{8}\)

\(P(\text{two heads})=\dfrac{3}{8}\) and \(P(\text{at least one head})=\dfrac{7}{8}\).

Tossing three fair coinsA tree of three coin tosses giving eight equally likely outcomes; three of them have exactly two heads. 1/2 H 1/2 H 1/2 H HHH 1/2 T HHT 1/2 T 1/2 H HTH 1/2 T HTT 1/2 T 1/2 H 1/2 H THH 1/2 T THT 1/2 T 1/2 H TTH 1/2 T TTT

Common pitfalls

Using unequal outcomes. \(P(E)=\dfrac{n(E)}{n(S)}\) only works when every outcome is equally likely. ‘Sum \(=2\)’ and ‘sum \(=7\)’ are not equally likely, so count the \(36\) dice pairs, not the \(11\) possible totals.
Probability outside \([0,1]\). A probability can never be negative or bigger than \(1\). If you get \(\dfrac{7}{6}\), recheck your counts.
Forgetting to simplify. \(\dfrac{6}{36}\) is correct but should be written \(\dfrac{1}{6}\).
Not using the complement. ‘At least one’ is much faster as \(1-P(\text{none})\) than adding many cases.

Frequently asked questions

What is a sample space?

The set of every possible outcome of an experiment, written \(S\). For one die, \(S=\{1,2,3,4,5,6\}\).

How do you calculate theoretical probability?

When outcomes are equally likely, \(P(E)=\dfrac{n(E)}{n(S)}\): favourable outcomes divided by total outcomes.

What is the complement rule?

\(P(A')=1-P(A)\). The probability an event does not happen is one minus the probability it does.

How many outcomes are there when you roll two dice?

\(6\times 6=36\) equally likely ordered pairs. An array is the easiest way to count them.

Can a probability be greater than 1?

No. Every probability lies in \([0,1]\); a value above \(1\) or below \(0\) means a counting error.

When should I use the complement?

For ‘not’ or ‘at least one’ events: find \(P(\text{none})\) and subtract from \(1\).

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