Conditional probabilities from tables/trees | Year 10

01 / LEARN

Understand the idea.

Tables preserve the group counts

A two-way table makes conditional probabilities visible through subgroup totals. An internal cell counts an overlap; its row and column margins count the two possible reference groups. Select the denominator from the condition, rather than from the event mentioned first.

In the club table the overlap is eight students, but drama and music have different totals. Thus $P (music ∣ drama) = \frac{8}{18}$ , while reversing the condition gives $\frac{8}{20}$ . The shared numerator does not make the questions equivalent.

Invented membership data for 60 students.
Music membership	Drama	Not drama	Total
Music	8	12	20
Not music	10	30	40
Total	18	42	60

Branches are conditional on the earlier route

On a probability tree, each branch is conditional on the route already taken. The outgoing labels from any node must sum to one. Multiply along a complete path to obtain the probability of all its successive events occurring.

If several distinct paths satisfy the question, add their probabilities because those complete outcomes are mutually exclusive. “Exactly one red” includes red-then-blue and blue-then-red, while excluding two reds. Without replacement, update both the total and relevant colour counts.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A conditional club count

Using the table, find P(music|drama).

Invented membership data for 60 students.
Music membership	Drama	Not drama	Total
Music	8	12	20
Not music	10	30	40
Total	18	42	60

Worked solution

1
Restrict to drama members.
$18 students$
The condition drama selects the drama-column total.
2
Locate the joint membership count.
$8 students$
The music-and-drama cell counts those satisfying both characteristics.
3
Divide within the conditioned group.
$P (music ∣ drama) = \frac{8}{18} = \frac{4}{9}$
Eight of the eighteen drama students also belong to music.

Answer

8/18=4/9.

EXAMPLE 02

Exactly one red

Two counters are drawn without replacement from 3 red and 2 blue. Find P(exactly one red).

Worked solution

1
Calculate the red-then-blue path.
$P (R B) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$
After a red counter is removed, two of the four remaining counters are blue.
2
Calculate the blue-then-red path.
$P (B R) = \frac{2}{5} \times \frac{3}{4} = \frac{3}{10}$
After a blue counter is removed, three of the four remaining counters are red.
3
Add the qualifying paths.
$P (exactly one red) = \frac{3}{10} + \frac{3}{10} = \frac{3}{5}$
RB and BR cannot both happen in one two-draw trial, and together cover every exactly-one-red result.

Answer

3/5.

03 / YOUR TURN

Try it yourself.

Check your understanding

In the tree, find P(two blue).

Show answer & reasoning

1/10.

For two blues, use the blue branch then its updated blue branch: $\frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Using the club table, find $P (drama ∣ music)$ .
2 marks
Invented membership data for 60 students.
Music membership Drama Not drama Total
Music 8 12 20
Not music 10 30 40
Total 18 42 60
Find $P (not drama ∣ not music)$ .
2 marks
Invented membership data for 60 students.
Music membership Drama Not drama Total
Music 8 12 20
Not music 10 30 40
Total 18 42 60
Find $P (music and drama)$ .
2 marks
Invented membership data for 60 students.
Music membership Drama Not drama Total
Music 8 12 20
Not music 10 30 40
Total 18 42 60
Use the tree to find $P (two red)$ .
2 marks
Use the tree to find $P (at least one red)$ .
3 marks
Explain how to check that a completed four-leaf tree is consistent.
3 marks

Read the worked solutions

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

Using the club table, find $P (drama ∣ music)$ .
2 marks
Invented membership data for 60 students.
Music membership Drama Not drama Total
Music 8 12 20
Not music 10 30 40
Total 18 42 60
$\frac{8}{20} = \frac{2}{5}$ .
$\frac{2}{5}$ .
Find $P (not drama ∣ not music)$ .
2 marks
Invented membership data for 60 students.
Music membership Drama Not drama Total
Music 8 12 20
Not music 10 30 40
Total 18 42 60
$\frac{30}{40} = \frac{3}{4}$ .
$\frac{3}{4}$ .
Find $P (music and drama)$ .
2 marks
Invented membership data for 60 students.
Music membership Drama Not drama Total
Music 8 12 20
Not music 10 30 40
Total 18 42 60
8 of all 60 do both; $\frac{8}{60} = \frac{2}{15}$ .
$\frac{2}{15}$ .
Use the tree to find $P (two red)$ .
2 marks
$(\frac{3}{5}) (\frac{2}{4}) = \frac{3}{10}$ .
$\frac{3}{10}$ .
Use the tree to find $P (at least one red)$ .
3 marks
The only excluded outcome is BB with probability $\frac{1}{10}$ .
$1 - \frac{1}{10} = \frac{9}{10}$ .
$\frac{9}{10}$ .
Explain how to check that a completed four-leaf tree is consistent.
3 marks
For this tree the leaf probabilities are $\frac{3}{10}, \frac{3}{10}, \frac{3}{10}, \frac{1}{10}$ .
Their sum is 1.
Every split totals 1, and all leaf probabilities total 1.

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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.

Tables preserve the group counts

Branches are conditional on the earlier route

See the method in action.

A conditional club count

Restrict to drama members.

Locate the joint membership count.

Divide within the conditioned group.

Exactly one red

Calculate the red-then-blue path.

Calculate the blue-then-red path.

Add the qualifying paths.

Try it yourself.

Check your understanding

Keep practising