The idea, step by step
Replacement restores the starting collection
A bag contains 3 red and 2 blue counters. Draw at random, with every remaining counter equally likely. If you replace and mix the first counter before drawing again, the second draw still has red probability 3/5. The same model applies at each draw.
Without replacement, both the total and a category count can change. After taking out a red, there are 2 red and 2 blue left, so the next-red probability is 2/4.
Multiply sequences; add different orders
For two reds without replacement, multiply the chance of first red by the chance of second red after that removal: .
For one of each colour, include RB and BR. They are separate disjoint sequences. Work out both probabilities and add. Order may not matter to the event description, but it still matters when listing complete outcomes.
A common trap
Without replacement, change only the denominator.
Instead: After a red is removed, the total and the red count both decrease.
See the thinking
Two reds
A bag has 3 red and 2 blue counters. Draw twice at random without replacement. Find P(RR).
- First-red probability is 3/5.
- After a red, second-red probability is 2/4.
- Multiply along this path.
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).
- RB: .
- BR: .
- Add the separate orders.
Check your understanding
A bag has 4 green and 1 yellow. After removing a green, what is the next-green probability?
Show answer
Three green counters remain among four counters.
Your turn
6 questions · 14 marks · Work at your own pace.
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.
- 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
BW gives .
WB gives .
Add to obtain 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: .
Without replacement: .
After a red is kept out, the next-red chance decreases.
Greater with replacement.
Curriculum reference & lesson scope
Australian Curriculum v9 / SA Curriculum Mathematics
Year 9 compound-event sample spaces, probabilities, experiments and simulations.
Official curriculum source ↗