The idea, step by step
Replacement changes the second draw
Replacing a chosen counter restores the original composition before the next draw. With independent random draws from a bag of 3 red and 2 blue, P(red on each draw)=3/5. Without replacement, the bag has four counters on the second draw and its composition depends on the first colour.
After red first, the second red chance is 2/4. After blue first, it is 3/4. Label each branch from the bag contents on that route, not from a memorised unchanging fraction.
Independence is a probability condition
Events A and B are independent when learning B does not change the probability of A: P(A|B)=P(A), when the condition is possible. Equivalently P(A and B)=P(A)P(B).
Dependent does not mean that one event physically causes the other. It means the probability changes when the extra information is given. Drawing without replacement creates dependence even when the process is fair.
A common trap
Using the original total and colour counts for both draws without replacement.
Instead: Remove the first selected counter before calculating the second branch.
See the thinking
Compare two-red probabilities
A bag contains 3 red and 2 blue. Compare P(RR) with and without replacement.
- With replacement:(3/5)(3/5)=9/25.
- Without replacement:(3/5)(2/4)=3/10.
Check from given probabilities
P(A)=0.4, P(B)=0.5 and P(A and B)=0.2. Are A and B independent?
- P(A)P(B)=0.4×0.5=0.2.
- This equals the stated overlap probability.
Check your understanding
From 3 red and 2 blue without replacement, what is P(second red|first blue)?
Show answer
3/4.
Removing a blue leaves 3 red among 4 counters.
Your turn
6 questions · 14 marks · Work at your own pace.
Read the questions online
- A bag has 4 green and 3 yellow. After drawing a green without replacement, find the next green chance.2 marks
- A bag has 4 green and 3 yellow counters. Two counters are drawn at random with replacement. Find P(two green).2 marks
- A bag has 4 green and 3 yellow counters. Two counters are drawn at random without replacement. Find P(two yellow).2 marks
- P(A)=0.3, P(B)=0.6 and P(A and B)=0.18. Test independence.2 marks
- P(A)=0.5, P(B)=0.4 and P(A and B)=0.3. Find P(A|B) and test independence.3 marks
- Two fair coin tosses are independent. Does a first head make a second tail “due”? Explain.3 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- A bag has 4 green and 3 yellow. After drawing a green without replacement, find the next green chance.2 marks
3 green remain out of 6.
1/2. - A bag has 4 green and 3 yellow counters. Two counters are drawn at random with replacement. Find P(two green).2 marks
(4/7)^2=16/49.
16/49. - A bag has 4 green and 3 yellow counters. Two counters are drawn at random without replacement. Find P(two yellow).2 marks
(3/7)(2/6)=1/7.
1/7. - P(A)=0.3, P(B)=0.6 and P(A and B)=0.18. Test independence.2 marks
0.3×0.6=0.18.
Independent. - P(A)=0.5, P(B)=0.4 and P(A and B)=0.3. Find P(A|B) and test independence.3 marks
0.3/0.4=0.75.
0.75 differs from P(A)=0.5.
3/4; dependent. - Two fair coin tosses are independent. Does a first head make a second tail “due”? Explain.3 marks
The second toss remains equally likely to be H or T.
Previous outcomes do not rebalance the next independent trial.
No.
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 ↗