Replacement and dependence | Year 10

01 / LEARN

Understand the idea.

Replacement changes the second draw

With replacement, returning the selected counter restores the original composition. If the bag is remixed and each draw is independently random, its colour probabilities are unchanged. Without replacement, the next draw is from a smaller bag whose composition depends on the first result.

Write the remaining counts on each route. From three red and two blue, a first red leaves two red among four; a first blue leaves three red among four. Keeping the old numerator or denominator would misrepresent the second experiment.

Independence is a probability condition

Events are independent when learning one does not alter the chance of the other. When the condition has positive probability, $P (A ∣ B) = P (A)$ . Equivalently, the joint probability satisfies $P (A \cap B) = P (A) P (B)$ .

Dependence describes a probability change, rather than requiring a physical causal influence. Drawing without replacement makes the colour events dependent because information about the first removal changes the second colour chance. The multiplication rule still works when its second factor is conditional.

P (A \cap B) = P (A) P (B) for independent events

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Compare two-red probabilities

A bag contains 3 red and 2 blue. Compare P(RR) with and without replacement.

Worked solution

1
Use the restored bag for a replaced draw.
$P (R R with replacement) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$
Replacement and independent remixing restore the red probability before the second draw.
2
Update the bag when a red is not replaced.
$P (R_{2} ∣ R_{1}) = \frac{2}{4}$
One red has been removed from the five counters.
3
Multiply along the unreplaced path.
$P (R R without replacement) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$
The second conditional red probability is smaller, reducing the two-red probability.

Answer

9/25 with replacement;3/10 without.

EXAMPLE 02

Check from given probabilities

P(A)=0.4, P(B)=0.5 and P(A and B)=0.2. Are A and B independent?

Worked solution

1
Compute the product required for independence.
$P (A) P (B) = 0.4 \times 0.5 = 0.2$
Independent events must have joint probability equal to the product of their individual probabilities.
2
Compare with the supplied joint probability.
$P (A \cap B) = 0.2 = P (A) P (B)$
The supplied overlap satisfies the independence criterion exactly.
3
Check the equivalent conditional statement.
$P (A ∣ B) = \frac{0.2}{0.5} = 0.4 = P (A)$
Knowing B leaves A's probability unchanged, confirming independence.

Answer

Yes.

03 / YOUR TURN

Try it yourself.

Check your understanding

From 3 red and 2 blue without replacement, what is P(second red|first blue)?

Show answer & reasoning

3/4.

Given that the first counter was blue, four counters remain, three of them red. Thus $P (R_{2} ∣ B_{1}) = \frac{3}{4}$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

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 $P (A \cap B) = 0.18$ . Test independence.
2 marks
$P (A) = 0.5$ , $P (B) = 0.4$ and $P (A \cap B) = 0.3$ . 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.
$\frac{1}{2}$ .
A bag has 4 green and 3 yellow counters. Two counters are drawn at random with replacement. Find $P (two green)$ .
2 marks
${(\frac{4}{7})}^{2} = \frac{16}{49}$ .
$\frac{16}{49}$ .
A bag has 4 green and 3 yellow counters. Two counters are drawn at random without replacement. Find $P (two yellow)$ .
2 marks
$(\frac{3}{7}) (\frac{2}{6}) = \frac{1}{7}$ .
$\frac{1}{7}$ .
$P (A) = 0.3$ , $P (B) = 0.6$ and $P (A \cap B) = 0.18$ . Test independence.
2 marks
$0.3 \times 0.6 = 0.18$ .
Independent.
$P (A) = 0.5$ , $P (B) = 0.4$ and $P (A \cap B) = 0.3$ . Find $P (A ∣ B)$ and test independence.
3 marks
$\frac{0.3}{0.4} = 0.75$ .
0.75 differs from $P (A) = 0.5$ .
$\frac{3}{4}$ ; 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.

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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.

Replacement changes the second draw

Independence is a probability condition

See the method in action.

Compare two-red probabilities

Use the restored bag for a replaced draw.

Update the bag when a red is not replaced.

Multiply along the unreplaced path.

Check from given probabilities

Compute the product required for independence.

Compare with the supplied joint probability.

Check the equivalent conditional statement.

Try it yourself.

Check your understanding

Keep practising