Mathematics • Year 7 • Unit 4 • Lesson 19
Two-Stage Experiments
Build fluency with the critical distinction: with replacement (the pool stays the same, stages are independent) vs without replacement (the pool shrinks, stages are dependent). Multiply along the tree branches: P(A then B) = P(A) × P(B given A).
1. I do — fully worked example
Read every line. Same bag, two different procedures, very different answers.
Problem. A bag has 3 red (R) and 2 blue (B) marbles. Two marbles are drawn one at a time, WITH replacement. Find P(red then blue).
Step 1 — Stage 1 probability.
P(R on draw 1) = 3/5
Reason: 3 red marbles out of 5 total.
Step 2 — Put the red marble back. Stage 2 probability.
Pool is back to 5 (3R, 2B). P(B on draw 2) = 2/5
Reason: WITH replacement means the bag is reset — the second draw is independent of the first.
Step 3 — Multiply along the branch.
P(R then B) = 3/5 × 2/5 = 6/25 = 0.24
Answer: P(R then B) = 6/25 = 0.24.
2. We do — fill in the missing steps
Same bag (3 red, 2 blue), but now draw two marbles WITHOUT replacement. Find P(red then blue). Fill in each blank. 4 marks
Step 1 — Stage 1 probability.
P(R on draw 1) = ___ / ___
Step 2 — Adjust the pool after taking a red.
Remaining marbles: ___ red and ___ blue → ___ total.
Step 3 — Stage 2 probability with the new pool.
P(B on draw 2 | R first) = ___ / ___
Step 4 — Multiply along the branch.
P(R then B) = ___/___ × ___/___ = _______
Compare your answer to the with-replacement answer (6/25 = 0.24) from the I-do example.
3. You do — independent practice
For each question, state the procedure (with or without replacement), then show the multiplication line.
Foundation — with replacement
3.1 Bag: 4 red, 1 blue. Draw 2 WITH replacement. Find P(both red). 2 marks
3.2 A coin is flipped twice. Find P(heads then tails). (This is automatically with replacement.) 1 mark
3.3 A die is rolled twice. Find P(6 then 6). 2 marks
Standard — without replacement
3.4 Same bag as 3.1 (4 red, 1 blue). Draw 2 WITHOUT replacement. Find P(both red). 2 marks
3.5 Bag: 5 red, 3 green. Two marbles are taken out one at a time WITHOUT replacement. Find P(red then green). 2 marks
3.6 A standard 52-card deck. Two cards are dealt without replacement. Find P(both aces). 2 marks
Extension — push your thinking
3.7 Bag: 3 red, 2 blue. Two marbles drawn WITHOUT replacement. (i) Find P(both red). (ii) Find P(both blue). (iii) Show that P(both red) + P(both blue) + P(one of each) = 1. 3 marks
3.8 Same bag (3 red, 2 blue). Two draws WITH replacement. Compute P(both red) and P(both blue). Compare with 3.7 and explain in one sentence why "with replacement" gives slightly larger P(both same) values. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — Without replacement (We do)
Step 1: P(R) = 3/5.
Step 2: Remaining = 2 red, 2 blue, 4 total.
Step 3: P(B | R first) = 2/4.
Step 4: P(R then B) = 3/5 × 2/4 = 6/20 = 3/10 = 0.30. (Higher than 6/25 = 0.24 from with replacement.)
3.1 — Bag 4R 1B, with replacement, both red
P(R) = 4/5 on each draw. P(RR) = 4/5 × 4/5 = 16/25 = 0.64.
3.2 — Coin flipped twice, P(H then T)
P(H) × P(T) = 1/2 × 1/2 = 1/4 = 0.25.
3.3 — Die rolled twice, P(6 then 6)
1/6 × 1/6 = 1/36 ≈ 0.028.
3.4 — Bag 4R 1B, without replacement, both red
P(R on draw 1) = 4/5. Remaining 3R, 1B → 4 total. P(R on draw 2 | R first) = 3/4. P(RR) = 4/5 × 3/4 = 12/20 = 3/5 = 0.60.
3.5 — Bag 5R 3G, without replacement, P(red then green)
P(R) = 5/8. Remaining 4R, 3G → 7 total. P(G | R first) = 3/7. P(R then G) = 5/8 × 3/7 = 15/56 ≈ 0.268.
3.6 — Two aces dealt from 52
P(A on draw 1) = 4/52 = 1/13. After taking one ace: 3 aces in 51 cards. P(A on draw 2 | A first) = 3/51 = 1/17. P(AA) = 1/13 × 1/17 = 1/221 ≈ 0.0045.
3.7 — Bag 3R 2B, without replacement
(i) P(RR) = 3/5 × 2/4 = 6/20 = 3/10.
(ii) P(BB) = 2/5 × 1/4 = 2/20 = 1/10.
(iii) P(one of each) = P(RB) + P(BR) = (3/5 × 2/4) + (2/5 × 3/4) = 6/20 + 6/20 = 12/20 = 3/5. Sum = 3/10 + 1/10 + 6/10 = 10/10 = 1. ✓
3.8 — Same bag with replacement, compare
P(RR) = 3/5 × 3/5 = 9/25 = 0.36. P(BB) = 2/5 × 2/5 = 4/25 = 0.16. With replacement gives larger P(both same) because the pool is reset — taking a red does NOT reduce the chance of another red on the next draw, whereas without replacement that chance drops.