With/without replacement | Year 9

01 / LEARN

Understand the idea.

Replacement restores the starting collection

With replacement, return the selected counter and mix before drawing again. A bag with 3 red and 2 blue counters then starts each draw with the same contents, so the red probability stays $\frac{3}{5}$ .

Without replacement, update the bag after the first draw. Removing a red leaves 2 red and 2 blue counters, making the next-red probability $\frac{2}{4}$ . Removing a blue instead would leave 3 red among 4 counters.

Multiply sequences; add different orders

For a sequence, multiply the first-stage probability by the second-stage probability for the bag that actually remains. Thus two reds without replacement have probability $\frac{3}{5} \times \frac{2}{4}$ , rather than using the unchanged red probability twice.

An event described without order may contain several ordered sequences. One red and one blue includes RB and BR. Calculate each path using its own remaining bag, then add because a single trial cannot follow both paths.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Two reds

A bag has 3 red and 2 blue counters. Draw twice at random without replacement. Find P(RR).

Worked solution

1
Find the chance of the first red.
$P (first R) = \frac{3}{5}$
Three of the five equally likely counters are red.
2
Update the bag after that red is removed.
$P (second R after first R) = \frac{2}{4}$
The removal leaves two red and two blue counters.
3
Multiply along the RR path.
$P (RR) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$
Both reds must occur in sequence, using the changed probability for the second draw.

Answer

$3 / 5 \times 2 / 4 = 3 / 10$ .

EXAMPLE 02

One of each

A bag contains 3 red and 2 blue counters. Draw twice at random without replacement. Find P(one red and one blue).

Worked solution

1
Identify both possible colour orders.
$E = {RB, BR}$
Exactly one counter of each colour can be drawn red first or blue first.
2
Calculate the red-then-blue path.
$P (RB) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$
After removing a red, both blue counters remain among four counters.
3
Calculate the blue-then-red path.
$P (BR) = \frac{2}{5} \times \frac{3}{4} = \frac{3}{10}$
After removing a blue, all three red counters remain among four counters.
4
Add the two path probabilities.
$P (one of each) = \frac{3}{10} + \frac{3}{10} = \frac{3}{5}$
RB and BR are mutually exclusive and cover every way to obtain one of each colour.

Answer

$3 / 5$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

A bag has 4 green and 1 yellow. After removing a green, what is the next-green probability?

Show answer & reasoning

$3 / 4$

Removing one green changes 4 green and 1 yellow into 3 green and 1 yellow. A random next draw therefore has green probability $\frac{3}{4}$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

A bag has 4 red and 2 blue counters. One counter is drawn at random and is red. After it is removed, find the next-blue probability.
2 marks
The bag has 4 red and 2 blue counters. Draw twice at random, replacing and mixing after the first draw. Find $P (RR)$ .
2 marks
A bag has 4 red and 2 blue counters. Draw twice at random without replacement. Find $P (RR)$ .
2 marks
A bag has 3 black and 1 white counter. Draw twice at random without replacement. Find $P (one of each)$ .
3 marks
Two cards are chosen without replacement from cards labelled 1, 2, 3. List all ordered outcomes.
2 marks
A bag has 5 red and 3 blue counters. Is $P (RR)$ greater with or without replacement? Explain.
3 marks

Read the worked solutions

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

A bag has 4 red and 2 blue counters. One counter is drawn at random and is red. After it is removed, find the next-blue probability.
2 marks
Two blue counters remain among five.
$\frac{2}{5}$
The bag has 4 red and 2 blue counters. Draw twice at random, replacing and mixing after the first draw. Find $P (RR)$ .
2 marks
$\frac{4}{6} \times \frac{4}{6} = \frac{16}{36} = \frac{4}{9}$ .
$\frac{4}{9}$
A bag has 4 red and 2 blue counters. Draw twice at random without replacement. Find $P (RR)$ .
2 marks
$\frac{4}{6} \times \frac{3}{5} = \frac{12}{30} = \frac{2}{5}$ .
$\frac{2}{5}$
A bag has 3 black and 1 white counter. Draw twice at random without replacement. Find $P (one of each)$ .
3 marks
BW gives $\frac{3}{4} \times \frac{1}{3} = \frac{1}{4}$ .
WB gives $\frac{1}{4} \times \frac{3}{3} = \frac{1}{4}$ .
Add to obtain $\frac{1}{2}$ .
$\frac{1}{2}$
Two cards are chosen without replacement from cards labelled 1, 2, 3. List all ordered outcomes.
2 marks
A label cannot be chosen twice.
Each first card has two second-card choices.
$12, 13, 21, 23, 31, 32$ .
A bag has 5 red and 3 blue counters. Is $P (RR)$ greater with or without replacement? Explain.
3 marks
With replacement: $\frac{25}{64}$ .
Without replacement: $\frac{5}{8} \times \frac{4}{7} = \frac{5}{14}$ .
After a red is kept out, the next-red chance decreases.
Greater with replacement.

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Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum Mathematics

Year 9 compound-event sample spaces, probabilities, experiments and simulations.

Official curriculum source ↗

In this lesson

Understand the idea.

Replacement restores the starting collection

Multiply sequences; add different orders

See the method in action.

Two reds

Find the chance of the first red.

Update the bag after that red is removed.

Multiply along the RR path.

One of each

Identify both possible colour orders.

Calculate the red-then-blue path.

Calculate the blue-then-red path.

Add the two path probabilities.

Try it yourself.

Check your understanding

Keep practising