Lesson 19 ~30 min Unit 4 · Probability +85 XP

Two-Stage Experiments

Unlock the critical difference between with-replacement and without-replacement experiments — the distinction exams love to test.

Today's hook: Drawing two cards — does it matter if you put the first card BACK before drawing the second? Absolutely. "With replacement" and "without replacement" create completely different probability landscapes — and exams love to test which one you're in.

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Think First

warm-up

Before you read on — quickly: A bag has 3 red and 2 blue marbles. You draw one marble, note its colour, then draw a second. Do you think it matters whether you put the first marble back before drawing again? Why might these be different situations?

Record your answer in your workbook.

The Big Idea

+5 XP

In two-stage experiments, with replacement means the first outcome is returned before the second draw — so both draws have the same probability. Without replacement means the first item is NOT returned, so the second draw's sample space is reduced.

Bag: 3 red (R), 2 blue (B) = 5 total. With replacement: After drawing first, return it. Second draw still from 5. P(R) = 3/5 each time. Without replacement: After drawing first, keep it. Second draw from only 4 remaining. Probabilities at stage 2 change depending on what was drawn at stage 1.

With: denominator stays same Without: denominator decreases

Check the key word

Questions will say "with replacement" or "without replacement". Read carefully every time.

Without: total decreases

If 5 items and no replacement: 2nd draw from 4. The denominator drops by 1 per draw.

With: stages independent

With replacement, each draw is independent — the first result doesn't change the second.

What You'll Master

objectives

Know

With replacement: denominator stays the same for each draw
Without replacement: denominator decreases by 1 per draw
These create different probability calculations

Understand

Why replacement makes stages independent
Why without replacement makes stage 2 depend on stage 1
When each situation applies in real life

Can Do

Draw two-stage tree diagrams with correct fractions on each branch
Calculate P(specific combined outcome) from a tree
Distinguish with vs without replacement and adjust calculations accordingly

Words You Need

vocabulary

Two-stage experimentAn experiment with two sequential selections or events.

With replacementThe first item is returned before the second draw — sample space stays the same.

Without replacementThe first item is NOT returned — second draw is from a smaller pool.

Independent eventsEvents where the outcome of one doesn't affect the other (with replacement).

Dependent eventsEvents where the outcome of the first affects the probabilities of the second (without replacement).

Multiplication ruleP(A then B) = P(A) × P(B given A). Multiply along tree branches.

Spot the Trap

heads-up

Wrong: Drawing 2 from 5 WITHOUT replacement, but keeping P(2nd draw) = 3/5. The denominator stays 5 — wrong!

Right: Without replacement: after the first draw, only 4 remain. Denominator for 2nd draw = 4. The numerator also changes based on what was drawn first.

Wrong: Only drawing the first stage of the tree, then trying to calculate probabilities for two-stage outcomes.

Right: Always draw the COMPLETE tree with both stages, then multiply probabilities along branches to find each outcome's probability.

Two-Stage: With Replacement

+5 XP

With replacement: the first item is returned, so every draw comes from the same pool. Both stages have identical probabilities. Outcomes are independent. Multiply probabilities along tree branches to find each combined outcome's probability.

Bag: 3 red (R), 2 blue (B) = 5 total. Draw 2 WITH replacement. P(R then B): Stage 1 P(R) = 3/5. Return the red marble. Stage 2 still from 5: P(B) = 2/5. P(R then B) = 3/5 × 2/5 = 6/25. Both stages have the SAME probabilities.

P(R then B) = 3/5 × 2/5 = 6/25

Same fractions each branch

With replacement: all stage-2 branches have identical fractions regardless of stage-1 outcome.

Multiply along branches

P(R then B) = P(R) × P(B) = 3/5 × 2/5. Multiply the fractions on each path.

All outcomes sum to 1

9/25 + 6/25 + 6/25 + 4/25 = 25/25 = 1. Always check this!

Two-Stage: Without Replacement

+5 XP

Without replacement: the first item is kept out, so stage 2 draws from a smaller pool. Denominator decreases. The numerator also changes depending on what was drawn first. These are dependent events.

Bag: 3 red (R), 2 blue (B) = 5 total. Draw 2 WITHOUT replacement. Stage 1: P(R) = 3/5. After drawing R: 2R, 2B left = 4 total. P(B at stage 2 | R first) = 2/4 = 1/2. P(R then B) = 3/5 × 2/4 = 6/20 = 3/10. Different from with-replacement answer of 6/25!

P(R then B) = 3/5 × 2/4 = 6/20 = 3/10

Denominator drops by 1

If you start with 5, the 2nd draw is from 4. The denominator decreases by 1 each draw.

Numerator depends on stage 1

After drawing R, there are 2 R and 2 B left. After drawing B, there are 3 R and 1 B left.

Different from "with"

P(RB) with replacement = 6/25 = 0.24. Without = 6/20 = 0.30. Different! Always check which applies.

Finding P(Specific Combined Outcome)

+5 XP

To find the probability of a specific path through a tree diagram, multiply the branch probabilities along the path. For events that can happen in multiple ways (like "one of each colour"), add the relevant paths.

Bag: 3R, 2B = 5. Without replacement. P(exactly one of each colour) = P(R then B) + P(B then R). = (3/5 × 2/4) + (2/5 × 3/4) = 6/20 + 6/20 = 12/20 = 3/5. Key: add paths that give the same final outcome.

Multiply along each path → Add paths with same outcome

Multiply within a path

For each single path (e.g. R then B), multiply the probabilities on the branches.

Add across paths

If your event can happen in multiple ways, add the probabilities of each path.

Rule: × then +

Multiply along branches. Add when combining paths. Never mix the operations.

Watch Me Solve It · 3 examples

Watch Me Solve It · P(red then blue) WITH replacement

+15 XP per step

PROBLEM

Bag: 3 red, 2 blue (5 total). Draw 2 marbles WITH replacement. Find P(red, then blue).

1

Set up Stage 1 probabilities

P(R) = 3/5 P(B) = 2/5 Total = 5

With replacement: the marble is returned before stage 2.
2

Set up Stage 2 probabilities (after R returned)

Pool is back to 5: P(B | R first) = 2/5 Same as before
3

Multiply along the R → B path

P(R then B) = 3/5 × 2/5 = 6/25 = 0.24

Independent stages: multiply the branch probabilities.

AnswerP(red then blue) WITH replacement = 6/25 = 0.24

Watch Me Solve It · P(red then blue) WITHOUT replacement

+15 XP per step

PROBLEM

Same bag: 3 red, 2 blue (5 total). Draw 2 WITHOUT replacement. Find P(red, then blue). Note the difference!

1

Stage 1 probabilities

P(R) = 3/5 P(B) = 2/5

Still drawing from 5 at stage 1.
2

Stage 2 after drawing R (NOT returned)

Remaining: 2R + 2B = 4 total. P(B | R first) = 2/4

One red was removed, so only 2 reds left. Pool size drops to 4.
3

Multiply along the R → B path

P(R then B) = 3/5 × 2/4 = 6/20 = 3/10 = 0.30

Compare: WITH replacement gave 6/25 = 0.24. Without gives 0.30 — higher because removing a red makes blue more likely next!

AnswerP(red then blue) WITHOUT replacement = 6/20 = 3/10 = 0.30

Watch Me Solve It · Grid/array method

+15 XP per step

PROBLEM

Bag has 2 red (R1, R2) and 2 blue (B1, B2) marbles. Draw 2 WITHOUT replacement. Use a grid to show all outcomes and find P(both same colour).

1

Set up the grid (rows = 1st draw, cols = 2nd draw)

Rows: R1, R2, B1, B2 Columns: R1, R2, B1, B2. But can't draw same marble twice.

Without replacement: can't draw R1 then R1 — it's already out.
2

Count valid outcomes

12 valid outcomes: {R1R2, R1B1, R1B2, R2R1, R2B1, R2B2, B1R1, B1R2, B1B2, B2R1, B2R2, B2B1}
3

Find P(both same colour)

Both red: {R1R2, R2R1} = 2. Both blue: {B1B2, B2B1} = 2. Total favourable = 4.

P(both same colour) = 4/12 = 1/3 ≈ 0.333.

AnswerP(both same colour) = 4/12 = 1/3 ≈ 0.333

Common Pitfalls

heads-up

Using the wrong denominator on the second draw

Without replacement from 5: the 2nd draw comes from 4, NOT 5. Students forget to reduce the denominator and get 3/5 × 2/5 instead of the correct 3/5 × 2/4.

Fix: Without replacement = denominator decreases by 1 per draw. Always adjust both numerator and denominator based on what was removed.

Not adjusting the numerator after stage 1

After drawing a RED marble (without replacement), there are now 2 reds and 2 blues left. Students sometimes still write P(R) = 3/4 for the second draw when it should be 2/4.

Fix: Track which marbles remain. After drawing 1 red from {3R, 2B}: pool becomes {2R, 2B}. Both numerator AND denominator must update.

Only drawing the first stage of the tree

Students draw stage 1 (two branches: R and B) and then try to calculate two-stage probabilities. Without the full two-stage tree, branch multiplication cannot be done correctly.

Fix: Always draw the COMPLETE tree. Every stage 1 outcome must have its own stage 2 branches with the correct (possibly different) probabilities.

Copy Into Your Books

With Replacement

Item returned before 2nd draw
Denominator stays the SAME
Stages are independent
P(A then B) = P(A) × P(B)

Without Replacement

Item NOT returned
Denominator decreases by 1
Stages are dependent
Numerator also changes based on stage 1

Tree Diagram Rules

Draw complete tree (both stages)
Label fractions on every branch
Multiply along a path
Add across paths for same outcome

Key Check

All outcome probabilities sum to 1
With ≠ Without (usually)
Real examples: dealing cards = without; rolling dice = with (each roll independent)

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Two-Stage Experiments

4 problems

Four drill problems. Work each, then reveal the answer.

1 Bag: 4 red, 1 blue. Draw 2 WITH replacement. Find P(both red).

P(R) = 4/5 each draw. P(RR) = 4/5 × 4/5 = 16/25.P(both red) = 16/25 = 0.64
2 Same bag. Draw 2 WITHOUT replacement. Find P(both red).

P(R first) = 4/5. After removing 1 red: 3R + 1B left (4 total). P(R second) = 3/4. P(RR) = 4/5 × 3/4 = 12/20 = 3/5.P(both red) = 3/5 = 0.60
3 Are the two answers from Q1 and Q2 the same? Why not?

No. 16/25 ≠ 3/5. Without replacement, drawing a red marble first makes it slightly harder to draw another red (fewer reds remain). With replacement, the pool always resets, giving a higher probability.16/25 = 0.64 vs 3/5 = 0.60 — different!
4 Give a real-life example where you must use WITHOUT replacement.

Any situation where items are NOT put back: dealing playing cards from a deck, selecting students from a class for a committee, picking raffle tickets from a barrel.Cards, raffle tickets, team selection — anything without return

Complete in your workbook.

Quick Check · 5 questions

Drawing WITH replacement from a bag of 5 marbles means the 2nd draw comes from:

+10 XP

Bag: 3 red, 2 blue. Draw 2 WITHOUT replacement. P(both red) =

+10 XP

Two cards are dealt from a standard 52-card deck. This is an example of:

+10 XP

To find the probability of a specific two-stage outcome from a tree diagram, you:

+10 XP

Bag: 2 red, 2 blue (4 total). Draw 2 WITHOUT replacement. P(both red) =

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

Apply Medium 3 MARKS

Q6. A bag has 4 green and 2 yellow marbles. Two marbles are drawn WITHOUT replacement. Draw a tree diagram and find P(one of each colour).

Draw the tree in your workbook.

Understand Easy 2 MARKS

Q7. A coin is flipped twice. Explain why this is WITH replacement even though there's no bag. What makes coin flipping "with replacement"?

Answer in your workbook.

Reason Hard 4 MARKS

Q8. Bag: 3 red, 2 blue. Draw 2 WITH replacement. (a) Complete the tree diagram with probabilities on every branch. (b) Find P(at least one blue). (c) Is P(blue then red) the same as P(red then blue)? Explain.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. A — With replacement: marble returned, pool stays at 5.

2. C — P(RR) = 3/5 × 2/4 = 6/20 = 3/10.

3. B — Cards are not returned to the deck. Without replacement.

4. D — Multiply branch probabilities along the path.

5. C — P(R first) = 2/4. After drawing: 1R+2B remain. P(R second) = 1/3. P(RR) = 2/4 × 1/3 = 2/12 = 1/6.

Show Your Working Model Answers

Q6 (3 marks): Total = 6. P(G first) = 4/6 = 2/3. After G: 3G+2Y left (5). P(Y second | G first) = 2/5 [1]. P(GY) = 2/3 × 2/5 = 4/15 [1]. P(YG) = 2/6 × 4/5 = 1/3 × 4/5 = 4/15. P(one of each) = 4/15 + 4/15 = 8/15 [1].

Q7 (2 marks): Each coin flip is a fresh, independent event — the coin doesn't "remember" the previous result [1]. The outcomes (H or T) have the same probabilities on every flip, just as with replacement gives the same pool size [1].

Q8 (4 marks): (a) 4 branches: RR (9/25), RB (6/25), BR (6/25), BB (4/25) [1]. (b) P(at least one blue) = 1 − P(no blue) = 1 − 9/25 = 16/25 [1]. (c) P(BR) = 2/5 × 3/5 = 6/25. P(RB) = 3/5 × 2/5 = 6/25. They ARE the same [1]. This is because with replacement, stages are independent and multiplication is commutative [1].

Stretch Challenge · +25 XP, +10 coins

The Raffle Problem

A raffle barrel contains 5 winning tickets and 15 losing tickets (20 total). Two tickets are drawn WITHOUT replacement. (a) Find P(both winning). (b) Find P(first wins, second loses). (c) Find P(at least one win). (d) Would your answers change if tickets were drawn WITH replacement? Show the comparison.

Reveal solution

Without replacement: (a) P(WW) = 5/20 × 4/19 = 20/380 = 1/19. (b) P(WL) = 5/20 × 15/19 = 75/380 = 15/76. (c) P(at least one win) = 1 − P(LL) = 1 − (15/20 × 14/19) = 1 − 210/380 = 170/380 = 17/38. With replacement: (a) 5/20 × 5/20 = 25/400 = 1/16. (b) 5/20 × 15/20 = 75/400 = 3/16. These are different because replacement resets the pool.

Quick Review

With replacement

Denominator stays the same. Independent stages.

Without replacement

Denominator decreases. Dependent stages.

Multiply along paths

P(A then B) = P(A) × P(B given A)

Add across paths

Add when event can happen multiple ways

Check sum = 1

All outcome probabilities must total 1

Real life

Cards, raffles = without. Dice, coins = with.

Interactive: Replacement Comparison

See the difference between with and without replacement side by side. Adjust the bag contents and watch how the tree diagrams and probabilities change.

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