Conditional language and restricted sample spaces | Year 10

01 / LEARN

Understand the idea.

The condition changes the population

A conditional probability, $P (A ∣ B)$ , asks for the chance of $A$ after restricting attention to outcomes in $B$ . In an equally likely finite sample space, count outcomes satisfying both conditions and divide by the number satisfying the given condition.

The condition must be possible: $P (B) > 0$ . The general formula $P (A ∣ B) = \frac{P (A \cap B)}{P (B)}$ expresses the same restriction using probabilities. Do not divide by the original total once the reference group has changed.

P (A ∣ B) = \frac{P (A \cap B)}{P (B)}, P (B) > 0

Read the order of the events

Read the condition after the vertical bar first. For $P (walk ∣ girl)$ , keep only girls; for $P (girl ∣ walk)$ , keep only walkers. Both use the same overlap cell, but their denominators can differ.

Before calculating, name the restricted group in words and locate its total. If that group is empty, the conditional probability is undefined by this formula. Otherwise, check that the favourable count cannot exceed the restricted total.

Invented survey of 60 students; every student is in exactly one travel category.
Travel	Girl	Boy	Total
Cycle	12	8	20
Walk	18	22	40
Total	30	30	60

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Restrict a die result

A fair die result is known to be odd. Find the probability it is prime.

Worked solution

1
Restrict the die outcomes.
$B = {1, 3, 5}$
Knowing the result is odd removes the even outcomes; the remaining outcomes stay equally likely.
2
Identify the prime results within that set.
$A \cap B = {3, 5}$
One is not prime because a prime has exactly two positive divisors.
3
Use the restricted total.
$P (prime ∣ odd) = \frac{2}{3}$
Two of the three possible odd results are prime.

Answer

2/3.

EXAMPLE 02

A survey condition

Using the supplied table, find P(walk|girl).

Invented survey of 60 students; every student is in exactly one travel category.
Travel	Girl	Boy	Total
Cycle	12	8	20
Walk	18	22	40
Total	30	30	60

Worked solution

1
Identify the given group.
$30 girls$
The condition after the bar is girl, so use the girl-column total.
2
Count the walkers in that group.
$18 girls walk$
The walk-and-girl cell gives the required overlap.
3
Form and simplify the conditional probability.
$P (walk ∣ girl) = \frac{18}{30} = \frac{3}{5}$
The denominator counts girls rather than walkers or all students.

Answer

3/5.

03 / YOUR TURN

Try it yourself.

Check your understanding

Using the table, find P(girl|walk).

Show answer & reasoning

9/20.

Given walk, restrict to the forty walkers. Eighteen are girls, so $P (girl ∣ walk) = \frac{18}{40} = \frac{9}{20}$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

A fair six-sided die result is known to exceed 3. Find $P (even)$ .
2 marks
A fair six-sided die result is known to be even. Find $P (greater than 3)$ .
2 marks
Find $P (cycle ∣ girl)$ from the table.
2 marks
Invented survey of 60 students; every student is in exactly one travel category.
Travel Girl Boy Total
Cycle 12 8 20
Walk 18 22 40
Total 30 30 60
Find $P (girl ∣ cycle)$ from the table.
2 marks
Invented survey of 60 students; every student is in exactly one travel category.
Travel Girl Boy Total
Cycle 12 8 20
Walk 18 22 40
Total 30 30 60
Using the travel table, explain why $P (cycle ∣ girl)$ and $P (girl ∣ cycle)$ have different denominators.
3 marks
Invented survey of 60 students; every student is in exactly one travel category.
Travel Girl Boy Total
Cycle 12 8 20
Walk 18 22 40
Total 30 30 60
$P (A \cap B) = 0.12$ and $P (B) = 0.3$ . Find $P (A ∣ B)$ .
2 marks

Read the worked solutions

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

A fair six-sided die result is known to exceed 3. Find $P (even)$ .
2 marks
Restrict to 4,5,6;4 and 6 are even.
$\frac{2}{3}$ .
A fair six-sided die result is known to be even. Find $P (greater than 3)$ .
2 marks
Restrict to 2,4,6;4 and 6 exceed 3.
$\frac{2}{3}$ .
Find $P (cycle ∣ girl)$ from the table.
2 marks
Invented survey of 60 students; every student is in exactly one travel category.
Travel Girl Boy Total
Cycle 12 8 20
Walk 18 22 40
Total 30 30 60
$\frac{12}{30} = \frac{2}{5}$ .
$\frac{2}{5}$ .
Find $P (girl ∣ cycle)$ from the table.
2 marks
Invented survey of 60 students; every student is in exactly one travel category.
Travel Girl Boy Total
Cycle 12 8 20
Walk 18 22 40
Total 30 30 60
$\frac{12}{20} = \frac{3}{5}$ .
$\frac{3}{5}$ .
Using the travel table, explain why $P (cycle ∣ girl)$ and $P (girl ∣ cycle)$ have different denominators.
3 marks
Invented survey of 60 students; every student is in exactly one travel category.
Travel Girl Boy Total
Cycle 12 8 20
Walk 18 22 40
Total 30 30 60
Girls total 30, cyclists total 20.
The same overlap 12 is divided by different totals.
They condition on different groups.
$P (A \cap B) = 0.12$ and $P (B) = 0.3$ . Find $P (A ∣ B)$ .
2 marks
$\frac{0.12}{0.3} = 0.4$ .
0.4.

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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.

The condition changes the population

Read the order of the events

See the method in action.

Restrict a die result

Restrict the die outcomes.

Identify the prime results within that set.

Use the restricted total.

A survey condition

Identify the given group.

Count the walkers in that group.

Form and simplify the conditional probability.

Try it yourself.

Check your understanding

Keep practising