Compound events | Year 10 | Ascent Library

01 / LEARN

Understand the idea.

Keep complete outcomes

A compound outcome records every stage of an experiment. For a coin and a die, $(H, 4)$ identifies both the coin result and the die result. A complete sample space makes it possible to check which outcomes meet a combined condition.

“And” requires both events; the inclusive “or” allows either or both. When events overlap, adding their probabilities counts the overlap twice. Subtract it once: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ .

P (A \cup B) = P (A) + P (B) - P (A \cap B)

Equiprobable outcomes must be justified

Counting favourable outcomes over total outcomes works directly when the complete outcomes are equally likely. For independent fair dice, the ordered pairs are equally likely; $(1, 6)$ and $(6, 1)$ are different results even though they have the same sum.

The possible sums are not equally likely because they arise from different numbers of pairs. Use a systematic grid, list or tree, then count each complete outcome once. Fairness describes each device; independence justifies multiplying stage probabilities.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Coin and die

A fair coin and fair six-sided die are used. Find the probability of a head and an even number.

Worked solution

1
Count the complete outcomes.
$2 \times 6 = 12$
Under the independent fair-device model, each coin result can accompany any of the six equally likely die results.
2
List the outcomes satisfying both conditions.
${(H, 2), (H, 4), (H, 6)}$
Each favourable result has a head and an even die value.
3
Divide by the total count.
$P (head and even) = \frac{3}{12} = \frac{1}{4}$
The three favourable outcomes are among twelve equally likely complete outcomes.

Answer

$\frac{1}{4}$ .

EXAMPLE 02

An overlapping or event

A fair die is rolled. Find P(even or greater than 4).

Worked solution

1
Write the two event sets.
$E = {2, 4, 6}, G = {5, 6}$
These list the even results and results greater than four.
2
Combine them without repeating the overlap.
$E \cup G = {2, 4, 5, 6}$
The result six satisfies both conditions but is only one die outcome.
3
Calculate the union probability.
$P (E \cup G) = \frac{3 + 2 - 1}{6} = \frac{4}{6} = \frac{2}{3}$
Subtracting the single overlap avoids counting six twice.

Answer

$\frac{4}{6} = \frac{2}{3}$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

For one fair die, what is P(even and greater than 3)?

Show answer & reasoning

$\frac{1}{3}$ .

The outcomes satisfying both conditions are ${4, 6}$ . Thus $P = \frac{2}{6} = \frac{1}{3}$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

List the complete outcomes for two coin tosses.
2 marks
Two fair coins are tossed. Find $P (exactly one head)$ .
2 marks
A fair six-sided die is rolled. Find $P (odd or prime)$ .
2 marks
Two fair dice are rolled. Find $P (sum 7)$ .
3 marks
Find $P (at least one 6)$ on two fair dice.
3 marks
A fair coin is tossed independently of a fair six-sided die roll. Find $P (head or die shows 6)$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

List the complete outcomes for two coin tosses.
2 marks
Keep first and second toss positions distinct.
HH, HT, TH, TT.
Two fair coins are tossed. Find $P (exactly one head)$ .
2 marks
HT and TH are favourable: $\frac{2}{4}$ .
$\frac{1}{2}$ .
A fair six-sided die is rolled. Find $P (odd or prime)$ .
2 marks
$Odd = {1, 3, 5}$ , $prime = {2, 3, 5}$ .
$Union = {1, 2, 3, 5}$ : $\frac{4}{6}$ .
$\frac{2}{3}$ .
Two fair dice are rolled. Find $P (sum 7)$ .
3 marks
There are 36 ordered pairs.
Six pairs sum to 7, giving $\frac{6}{36}$ .
$\frac{1}{6}$ .
Find $P (at least one 6)$ on two fair dice.
3 marks
The complement has neither die equal to 6.
$1 - {(\frac{5}{6})}^{2} = \frac{11}{36}$ .
$\frac{11}{36}$ .
A fair coin is tossed independently of a fair six-sided die roll. Find $P (head or die shows 6)$ .
3 marks
Head contributes 6 outcomes, a 6 contributes 2, and (H,6) overlaps.
$\frac{6 + 2 - 1}{12} = \frac{7}{12}$ .
$\frac{7}{12}$ .

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Curriculum reference & lesson scope

Australian Curriculum v 9 / SA Curriculum

Compound events; systematic counting and the multiplication principle; factorials and simple arrangements; conditional language and restricted sample spaces; conditional probabilities from tables/trees; replacement and dependence; conditional versus unconditional probabilities; designing and interpreting simulations.

Official curriculum source ↗

In this lesson

Understand the idea.

Keep complete outcomes

Equiprobable outcomes must be justified

See the method in action.

Coin and die

Count the complete outcomes.

List the outcomes satisfying both conditions.

Divide by the total count.

An overlapping or event

Write the two event sets.

Combine them without repeating the overlap.

Calculate the union probability.

Try it yourself.

Check your understanding

Keep practising