Theoretical probability | Year 7

01 / LEARN

Understand the idea.

Build the probability fraction

Theoretical probability comes from a chance model. First check that the outcomes being counted are equally likely. Then count those satisfying the event and divide by the total count. Here, favourable means 'satisfies the condition', rather than 'a result you want'.

For an even result on a fair die, the favourable outcomes are ${2, 4, 6}$ . They make up $\frac{3}{6} = \frac{1}{2}$ of all faces. The denominator includes both favourable and unfavourable faces, because either kind can occur.

P (E) = \frac{favourable outcomes}{all equally likely outcomes}

Represent and check the answer

A fraction, decimal and percentage can describe the same chance: $\frac{1}{4} = 0.25 = 25 %$ . Simplify a probability fraction by dividing its numerator and denominator by the same nonzero number. Convert to a percentage by expressing the share out of one hundred.

Check that $0 \leq P (E) \leq 1$ . Zero means no possible outcome satisfies the event, while one means all do. A larger-than-one result signals a counting or denominator error; a probability cannot include more than all the possibilities.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A number-card condition

One card is selected at random from cards numbered 1 to 12. Find the probability of a multiple of 3.

Worked solution

1
List the multiples of three among the cards.
$E = {3, 6, 9, 12}$
These are the numbers from one to twelve divisible by three.
2
Divide the favourable count by the full count.
$P (E) = \frac{4}{12}$
Four cards satisfy the condition, and all twelve cards are equally likely to be selected.
3
Simplify the fraction.
$\frac{4}{12} = \frac{4 \div 4}{12 \div 4} = \frac{1}{3}$
Dividing both counts by four preserves their ratio.

Answer

$1 / 3$ .

EXAMPLE 02

An event described by “not”

A bag has 4 red, 3 green and 3 yellow equal-size counters. Find the chance that a random selection is not red.

Worked solution

1
Count all counters in the bag.
$4 + 3 + 3 = 10$
The denominator must include every colour that could be drawn.
2
Count the counters that are not red.
$3 + 3 = 6$
Not red includes both green and yellow, with no overlap between those colours.
3
Form and simplify the probability.
$P (not red) = \frac{6}{10} = \frac{3}{5}$
Six of the ten equally selectable counters satisfy the condition.
4
Express the chance as a percentage.
$\frac{3}{5} = \frac{60}{100} = 60 %$
A percentage records the equivalent share out of one hundred.

Answer

$3 / 5 = 60 %$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

One of six cards numbered 1 to 6 is selected at random. What is the chance of a number greater than 4?

Show answer & reasoning

$2 / 6 = 1 / 3$ .

Only $5$ and $6$ are greater than $4$ . These two favourable cards are among six equally likely cards, so $P (greater than 4) = \frac{2}{6} = \frac{1}{3}$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

A fair die is rolled. Find the probability of a 5.
2 marks
A fair die is rolled. Find the probability of an odd number.
2 marks
A card is selected at random from cards numbered 1 to 10. Find the probability of a factor of 10.
2 marks
A counter is chosen at random from a bag with 7 white and 3 black equal-size counters. Give the chance of black as a fraction, decimal and percentage.
2 marks
A fair die is rolled. Find the chance of a number less than 1 and the chance of a number at most 6.
2 marks
Which is greater: the chance of blue when a counter is chosen at random from 3 blue and 5 red equal-size counters, or an even result on a fair die?
3 marks

Read the worked solutions

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

A fair die is rolled. Find the probability of a 5.
2 marks
One face is labelled 5.
There are six equally likely faces.
$\frac{1}{6}$ .
A fair die is rolled. Find the probability of an odd number.
2 marks
The odd results are 1, 3 and 5.
Three of the six results are favourable.
$\frac{3}{6} = \frac{1}{2}$ .
A card is selected at random from cards numbered 1 to 10. Find the probability of a factor of 10.
2 marks
The factors in the set are 1, 2, 5 and 10.
Four of ten cards are favourable.
$\frac{4}{10} = \frac{2}{5}$ .
A counter is chosen at random from a bag with 7 white and 3 black equal-size counters. Give the chance of black as a fraction, decimal and percentage.
2 marks
There are ten counters altogether.
Three favourable counters give $\frac{3}{10}$ .
$\frac{3}{10} = 0.3 = 30 %$ .
A fair die is rolled. Find the chance of a number less than 1 and the chance of a number at most 6.
2 marks
No possible face is less than 1.
All six possible faces are at most 6.
Less than 1: 0. At most 6: 1.
Which is greater: the chance of blue when a counter is chosen at random from 3 blue and 5 red equal-size counters, or an even result on a fair die?
3 marks
Blue chance is $\frac{3}{8}$ .
The even chance is $\frac{1}{2} = \frac{4}{8}$ .
$\frac{4}{8} > \frac{3}{8}$ .
An even result on the die.

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Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Year 7 single-stage sample spaces, equally likely outcomes, probabilities, expected frequencies, chance experiments, simulations and relative frequencies.

Official curriculum source ↗

In this lesson

Understand the idea.

Build the probability fraction

Represent and check the answer

See the method in action.

A number-card condition

List the multiples of three among the cards.

Divide the favourable count by the full count.

Simplify the fraction.

An event described by “not”

Count all counters in the bag.

Count the counters that are not red.

Form and simplify the probability.

Express the chance as a percentage.

Try it yourself.

Check your understanding

Keep practising